MATH 122 - TRIGONOMETRY -
Trigonometric Functions
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the following trigonometric equation for 0≤x≤2π:sin2(x) + 2 cos(x)−
1 = 0.
Solution
Step 1: We can rewrite the equation as sin2(x)+2 cos(x)−1 = 0 by substituting
sin2(x) = 1−cos2(x). Step 2: Substituting the above relation into the equation,
we get 1−cos2(x) + 2 cos(x)−1 = 0. Step 3: Simplifying further, we get
−cos2(x) + 2 cos(x)=0. Step 4: Factoring out a cos(x), we have cos(x)(2 −
cos(x)) = 0. Step 5: Setting each factor to zero, we get cos(x) = 0 or cos(x) = 2.
Step 6: Since the cosine function has a range of −1≤cos(x)≤1, there are no
solutions for cos(x) = 2. Step 7: Therefore, the only solution is cos(x) = 0. Step
8: The solutions for cos(x)=0in the interval 0≤x≤2πoccur at x=π
2and
x=3π
2. Step 9: Thus, the solutions to the trigonometric equation are x=π
2
and x=3π
2.
Question 2
Question
Let f(x) = cos(2x)and g(x) = sin(x). Find the values of xin the interval [0,2π]
that satisfy the equation f(x) = g(x).
Solution
Step 1: We are given f(x) = cos(2x)and g(x) = sin(x). The equation we
are trying to solve is f(x) = g(x). Step 2: Substitute f(x)and g(x)into the
equation:
cos(2x) = sin(x)
Step 3: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 4: Substitute the double angle identity for cos(2x):
2 cos2(x)−1 = sin(x)
Step 5: Since cos2(x) = 1 −sin2(x), substitute this into the equation:
2(1 −sin2(x)) −1 = sin(x)
Step 6: Simplify the equation:
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 7: Let u= sin(x), then the equation becomes a quadratic in u:
2u2+u−1 = 0
Step 8: Solve the quadratic equation for u:
u=−1±√12−4(2)(−1)
2(2)
u=−1±√9
4
u=−1±3
4
Step 9: Solve for u:
u1=−1+3
4=2
4=1
2
u2=−1−3
4=−4
4=−1
Step 10: Recall that u= sin(x), so x= arcsin(1
2)=π
6and x= arcsin(−1) =
−π
2. Step 11: Since we are looking for solutions in the interval [0,2π], we only
consider x=π
6. Step 12: Thus, the solution to the equation f(x) = g(x)in the
interval [0,2π]is x=π
6.
Question 3
Question
Solve the equation 2 sin2(x)−√3 sin(x)−1 = 0 for 0≤x≤2π.
2
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
2u2−√3u−1 = 0.
Step 2: Solve the quadratic equation by factoring or using the quadratic
formula. In this case, we will use the quadratic formula:
u=−(−√3) ±√(−√3)2−4(2)(−1)
2(2) .
u=√3±√7
4.
Step 3: Therefore, we have two possible values for u:
u1=√3 + √7
4and u2=√3−√7
4.
Step 4: Now, we recall that u= sin(x), so we need to find the angles xthat
satisfy the given values of u.
Step 5: For u1=√3+√7
4, we have x= sin−1(√3+√7
4).
Step 6: For u2=√3−√7
4, we have x= sin−1(√3−√7
4).
Step 7: Therefore, the solutions to the equation 2 sin2(x)−√3 sin(x)−1 = 0
for 0≤x≤2πare:
x= sin−1(√3 + √7
4),sin−1(√3−√7
4).
Question 4
Question
Find all solutions to the equation sin(2x) = cos(x)for 0≤x < 2π.
Solution
Step 1: Recall the double angle formula for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Rewrite the given equation using the double angle formula: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x < 2πare x=π
6
and x=5π
6.
Step 6: Check these solutions in the original equation: - For x=π
6,
sin(2·π
6)= sin(π
3)=√3
2and cos(π
6)=√3
2. Since they are not equal, π
6is
3
not a solution. - For x=5π
6,sin(2·5π
6)= sin(5π
3)=−√3
2and cos(5π
6)=−√3
2.
They are equal, so 5π
6is a valid solution.
Step 7: Therefore, the only solution to the equation sin(2x) = cos(x)for
0≤x < 2πis x=5π
6.
Question 5
Question
Prove the trigonometric identity:
cot(θ)−tan(θ) = −2 csc(2θ)
Solution
To prove the identity cot(θ)−tan(θ) = −2 csc(2θ), we will start by expressing
cot(θ)and tan(θ)in terms of sine and cosine functions, and csc(2θ)in terms of
sine and cosine functions.
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Expressing csc(2θ)
csc(2θ) = 1
sin(2θ)
=1
2 sin(θ) cos(θ)
=1
2 sin(θ) cos(θ)
=1
2·1
sin(θ) cos(θ)
=1
2·2
sin(2θ)(Using double angle formula)
= csc(2θ)
Step 3: Proving the identity We will now substitute the expressions we
4
derived into the given identity:
cot(θ)−tan(θ) = −2 csc(2θ)
cos(θ)
sin(θ)−sin(θ)
cos(θ)=−2·1
2 sin(θ) cos(θ)
cos2(θ)−sin2(θ)
sin(θ) cos(θ)=−1
sin(θ) cos(θ)
cos(2θ)
sin(2θ)=−1
sin(2θ)
cos(2θ) = −1
The last step is true since cos(π) = −1. Therefore, we have proved the trigono-
metric identity cot(θ)−tan(θ) = −2 csc(2θ).
Question 6
Question
Solve the equation sin(2x)−cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We can rewrite the equation as sin(2x) = cos(x).
Step 2: Recall the double-angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 3: Substituting this identity into our equation, we get 2 sin(x) cos(x) =
cos(x).
Step 4: Dividing both sides by cos(x)(which is not zero in the given interval),
we obtain 2 sin(x) = 1.
Step 5: Solving for sin(x), we have sin(x) = 1
2.
Step 6: From the unit circle, we know that sin(30◦) = 1
2.
Step 7: So, the solutions for xare x= 30◦and x= 150◦within the given
interval.
Step 8: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0◦≤x≤360◦are x= 30◦and x= 150◦.
Question 7
Question
Let f(x) = 2 sin(x) + 3 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
5
Solution
Step 1: Amplitude The amplitude of a function of the form f(x) = asin(bx) +
ccos(dx)is given by √a2+c2. In this case, a= 2 and c= 3, so the amplitude
of f(x)is √22+ 32=√4 + 9 = √13.
Step 2: Period For a function of the form f(x) = asin(bx) + ccos(dx), the
period is given by 2π
|b|if bis the coefficient of xin the sine term, or 2π
|d|if dis
the coefficient of xin the cosine term. Since there are no coefficient before xon
either trigonometric function term in f(x), we have 2π
1= 2πas the period of
f(x).
Step 3: Phase Shift To determine the phase shift of a function f(x) =
asin(bx) + ccos(dx), we first need to express it in a different form. In this
case, f(x) = 2 sin(x) + 3 cos(x) = √13 (2
√13 sin(x) + 3
√13 cos(x)). By rewriting
2
√13 sin(x) + 3
√13 cos(x)as sin(x−α)where α= arctan(3
2), we can determine
that the phase shift of f(x)is −α=−arctan(3
2).
Step 4: Vertical Shift The vertical shift of a function is simply the constant
term in the function. In this case, the constant term is 0, so the vertical shift
of f(x)is 0.
Question 8
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: Rewrite the angle as a sum of two special angles.
5π
12 =π
3+π
4
Step 2: Use the sum-to-product identity for sine.
sin (π
3+π
4)= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Calculate the sine and cosine values of π
3and π
4.
sin (π
3)=√3
2,cos (π
3)=1
2,sin (π
4)=√2
2,cos (π
4)=√2
2
Step 4: Substitute the values into the formula.
sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)=√3
2·√2
2+1
2·√2
2
6
Step 5: Simplify the expression.
√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
Question 9
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for (f◦g)(x)and
simplify the result.
Solution
Step 1: Find (f◦g)(x)by substituting g(x) = cos(x)into f(x) = sin(2x).
(f◦g)(x) = f(g(x)) = f(cos(x)) = sin(2 ·cos(x))
Step 2: Use the double angle identity for sine, which states that sin(2θ) =
2 sin(θ) cos(θ), to simplify sin(2 ·cos(x)).
sin(2 ·cos(x)) = 2 sin(cos(x)) cos(cos(x))
Step 3: Further simplify 2 sin(cos(x)) cos(cos(x)) since sin(cos(x)) and cos(cos(x))
are not standard trigonometric functions.
No further simplification possible
Therefore, the expression for (f◦g)(x)is 2 sin(cos(x)) cos(cos(x)).
Question 10
Question
Find the exact value of sin (7π
12 ).
Solution
Step 1: Convert 7π
12 to an angle within the unit circle.
Step 2: Since 7π
12 =π
12 +3π
4, we can rewrite this as 15◦+ 135◦in degrees.
Step 3: Identify the reference angle in the first quadrant by subtracting π
from π
12 , which gives us π
12 −π=−11π
12 .
Step 4: Use the reference angle and quadrant information to determine the
sign of sin in the second quadrant. Since sin is positive in the second quadrant,
the value of sin(7π
12 )will also be positive.
7
Step 5: Now, calculate the sine of the reference angle 11π
12 in the second
quadrant: sin (11π
12 )= sin (π−π
12 )= sin (π
12 ).
Step 6: The exact value of sin (π
12 )can be found using the angle sum identity
for sine: sin (π
12 )= sin (π
6−π
3)= sin (π
6)cos (π
3)−cos (π
6)sin (π
3).
Step 7: Substitute the values of sine and cosine for π
6and π
3:sin (π
12 )=
1
2·√3
2−√3
2·1
2=√3−√3
4=0
4= 0.
Step 8: Therefore, the exact value of sin (7π
12 )is 0.
Question 11
Question
Let f(x) = sin2(3x) + cos(3x). Find the maximum and minimum values of f(x)
for x∈[0,π
6].
Solution
Step 1: First, let’s find the critical points of f(x)in the interval [0,π
6]by taking
the derivative and solving for xwhen f′(x) = 0.
Step 2: Find f′(x):
f′(x) = 2 sin(3x) cos(3x)−3 sin(3x)
Step 3: Set f′(x) = 0 and solve for x:
2 sin(3x) cos(3x)−3 sin(3x) = 0
sin(3x)(2 cos(3x)−3) = 0
Now, we find the critical points by solving:
sin(3x) = 0 or 2 cos(3x)−3 = 0
Step 4: Solve sin(3x) = 0 for xin [0,π
6]:
sin(3x) = 0 =⇒3x= 0, π, 2π, . . .
Since x∈[0, π/6], the solution is x= 0.
Step 5: Solve 2 cos(3x)−3 = 0 for xin [0,π
6]:
2 cos(3x)−3 = 0 =⇒cos(3x) = 3
2
There are no solutions in the interval [0,π
6].
Step 6: Evaluate f(x)at the critical point and endpoints to find the maxi-
mum and minimum values:
f(0) = sin2(0) + cos(0) = 0 + 1 = 1
f(π
6)= sin2(π
2) + cos(π
2)= 1 + 0 = 1
Step 7: Therefore, the maximum value of f(x)in the interval [0,π
6]is 1, and
the minimum value is also 1.
8
Question 12
Question
Find the exact value of sin (11π
12 ).
Solution
Step 1: Recognize that 11π
12 is not one of the common angles on the unit circle,
so we need to express it in terms of more common angles.
Step 2: We can rewrite 11π
12 as 3π
4+π
6. This is because 11π
12 =9π
12 +2π
12 =3π
4+π
6.
Step 3: Now, since we know that 3π
4and π
6are common angles, we can use
the sum-to-product identities to find the exact value of sin (11π
12 ).
Step 4: The sum-to-product identity for sine is: sin(A+B) = sin Acos B+
cos Asin B.
Step 5: Applying the sum-to-product identity, we have:
sin (11π
12 )= sin (3π
4+π
6)
= sin (3π
4)cos (π
6)+ cos (3π
4)sin (π
6)
Step 6: Using the unit circle, we find that sin (3π
4)=−√2
2,cos (π
6)=√3
2,
cos (3π
4)=−√2
2, and sin (π
6)=1
2.
Step 7: Substituting these values into our expression, we get:
sin (11π
12 )=−√2
2·√3
2−√2
2·1
2
=−√6
4−√2
4
=−√2 + √6
4
Therefore, the exact value of sin (11π
12 )is −√2+√6
4.
Question 13
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
9
Solution
Step 1: Recall the trigonometric identity sin(3x) = sin(π−3x).
Step 2: Rewrite the equation as sin(π−3x) = cos(2x).
Step 3: Since sin(π−θ) = sin(θ)and cos(π−θ) = −cos(θ), we have
sin(3x) = −cos(3x) = cos(2x).
Step 4: Set up the equation cos(2x) = −cos(3x).
Step 5: Use the double angle formula for cosine: cos(2x) = 2 cos2(x)−1.
Step 6: Substitute this into the equation: 2 cos2(x)−1 = −cos(3x).
Step 7: Use the triple angle formula for cosine: cos(3x) = 4 cos3(x)−3 cos(x).
Step 8: Substitute this into the equation: 2 cos2(x)−1 = −4 cos3(x) +
3 cos(x).
Step 9: Rearrange the equation to get 4 cos3(x)+2 cos2(x)+3 cos(x)−1 = 0.
Step 10: The equation is now a cubic equation in terms of cos(x). Solving
it may require the use of numerical methods like Newton’s method.
Question 14
Question
Determine the exact value of cos (11π
6).
Solution
Step 1: Recall the unit circle where cos (11π
6)is located. The angle 11π
6is in the
fourth quadrant, corresponding to the point (−√3
2,1
2)on the unit circle.
Step 2: Since cos (11π
6)corresponds to the x-coordinate of the point on the
unit circle, cos (11π
6)=−√3
2.
Therefore, the exact value of cos (11π
6)is −√3
2.
Question 15
Question
Solve the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= cos(x). The equation becomes 2u2+u−1 = 0.
Step 2: Now, we need to solve this quadratic equation for u. We can factor
it as (2u−1)(u+ 1) = 0.
Step 3: Set each factor equal to zero:
{2u−1 = 0
u+ 1 = 0
10
Step 4: Solve for uin each case:
{2u= 1
u=1
2
or {u=−1
Step 5: Remember, u= cos(x), so:
{cos(x) = 1
2
cos(x) = −1
Step 6: Solve each equation for x:
{x=π
3or x=5π
3
x=π
Therefore, the solutions to the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤
x≤2πare x=π
3, π, 5π
3.
Question 16
Question
Let f(x) = 2 sin(x)−√3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Finding the amplitude The general form of a trigonometric function
is f(x) = Asin(Bx −C) + Dcos(Ex −F). The amplitude of f(x)is given by
A=√A2+D2. In this case, A= 2 and D=−√3. Therefore, the amplitude is
amplitude =√22+ (−√3)2=√4 + 3 = √7.
Step 2: Finding the period The period of a general trigonometric function
is given by 2π/B or 2π/E. In this case, the period of f(x)is
period =2π
1= 2π.
Step 3: Finding the phase shift To find the phase shift, we need to solve the
equations Bx −C= 0 and Ex −F= 0. In this case, the phase shift for f(x)is
phase shift =C
B=−π/6
1=−π
6.
Step 4: Finding the vertical shift The vertical shift of a trigonometric func-
tion is the value of D in the general form f(x) = Asin(Bx −C)+Dcos(Ex −F).
In this case, the vertical shift is
vertical shift =D=−√3.
11
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
−π
6, and the vertical shift is −√3.
Question 17
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double-angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: Substitute cos(2x) = 1 −2 sin2(x)into the equation sin(x) = cos(2x)
to obtain sin(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to get 2 sin2(x) + sin(x)−1 = 0.
Step 4: Factor the quadratic equation to get (2 sin(x)−1)(sin(x) + 1) = 0.
Step 5: Set each factor to zero to find the possible solutions.
• For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which gives x=π
6,5π
6in the
interval [0,2π].
• For sin(x)+1 = 0, we have sin(x) = −1, which gives x=3π
2in the interval
[0,2π].
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation cos(2x) = √2
2for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute cos(2x) = √2
2into the double angle identity:
2 cos2(x)−1 = √2
2
Step 3: Add 1 to both sides:
2 cos2(x) = √2
2+ 1
12
Step 4: Simplify the right side:
2 cos2(x) = √2+2
2
Step 5: Divide by 2 to isolate cos2(x):
cos2(x) = √2+2
4
Step 6: Take the square root of both sides:
cos(x) = ±√√2+2
4
Step 7: We know that cos(x) = √2
2corresponds to the angle x=π
4in the
unit circle.
Step 8: To find the other solution, we can consider the symmetry of the
cosine function. The cosine function is positive in the first and fourth quadrants.
Hence, the other solution is x=7π
4.
Therefore, the solutions to the equation cos(2x) = √2
2for 0≤x≤2πare
x=π
4and x=7π
4.
Question 19
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: We will use the angle addition identity tan(A+B) = tan A+tan B
1−tan Atan B.
Step 2: Let A=π
3and B=π
4. Then, 5π
12 =π
3+π
4.
Step 3: We know that tan(π/3) = √3and tan(π/4) = 1.
Step 4: Substitute these values into the angle addition formula:
tan (π
3+π
4)=√3+1
1−√3·1=√3+1
1−√3
Step 5: To rationalize the denominator, multiply by the conjugate of the
denominator:
=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3)
Step 6: Simplify the expression:
=√3 + √3+1+3
1−3=−2(√3 + 1) = −2√3−2
Step 7: Therefore, tan (5π
12 =−2√3−2.
13
Question 20
Question
Let f(x) = 2 sin(x) + √3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Amplitude
The amplitude of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by √A2+D2. In this case, A= 2 and D=√3, so the amplitude is
√22+ (√3)2=√4 + 3 = √7.
Step 2: Period
For a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C), the period
is given by 2π
|B|. In this case, since B= 1 (implicitly, since sin(x)and cos(x)
both have a coefficient of 1), the period is
2π
1= 2π.
Step 3: Phase Shift
The phase shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by C
B. Here, C= 0 (since there is no xterm inside the sine or cosine
functions), so the phase shift is 0
1= 0.
Step 4: Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is simply D. In this case, the vertical shift is
D=√3.
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
0, and the vertical shift is √3.
Question 21
Question
Find all solutions to the equation cos(2x) = sin(x)for 0≤x≤2π.
14
Solution
Step 1: We will use the double-angle formula for cosine to rewrite cos(2x)in
terms of cos(x)and sin(x).
cos(2x) = 2 cos2(x)−1
Step 2: Substitute the expression for cos(2x)into the original equation.
2 cos2(x)−1 = sin(x)
Step 3: Recall that cos2(x) = 1 −sin2(x). Substitute this into the equation.
2(1 −sin2(x)) −1 = sin(x)
Step 4: Simplify the equation.
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 5: We can treat this as a quadratic equation in terms of sin(x). Let
sin(x) = yto simplify the equation.
2y2+y−1 = 0
Step 6: Solve the quadratic equation for y.
y=−1±√12−4(2)(−1)
2(2)
y=−1±√9
4
Step 7: Solve for y:
y1=−1+3
4=2
4=1
2
y2=−1−3
4=−4
4=−1
Step 8: Recall that sin(x) = y. So, we have sin(x) = 1
2and sin(x) = −1.
Solve for xin the interval 0≤x≤2π.
For sin(x) = 1
2,x=π
6,5π
6
For sin(x) = −1,x=3π
2
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
6,5π
6,3π
2.
Question 22
Question
Solve the equation sin2(x) + cos2(x) = 5
4for 0≤x≤2π.
15
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x)=1. Therefore, we
have
sin2(x) + cos2(x) = 1 = 4
4.
Step 2: Subtracting 4
4from both sides of the given equation, we get
5
4−4
4= 1.
Step 3: Simplifying the left side of the equation gives us
1
4= 1.
Step 4: Since 1
4= 1, there are no solutions to the given equation sin2(x) +
cos2(x) = 5
4for 0≤x≤2π.
Question 23
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double angle formula for cosine: cos(2x)=1−2 sin2(x).
Therefore, we can rewrite the equation as sin(x) = 1 −2 sin2(x).
Step 2: Rearranging the equation, we get 2 sin2(x) + sin(x)−1 = 0. Let
u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 3: Solve the quadratic equation 2u2+u−1=0by factoring or us-
ing the quadratic formula. The solutions are u=−1±√12−4(2)(−1)
2(2) =−1±√9
4.
Therefore, u=−1+3
4= 1 or u=−1−3
4=−1/2.
Step 4: Since u= sin(x), the solutions are sin(x) = 1 and sin(x) = −1/2.
Step 5: For sin(x) = 1,x=π
2. For sin(x) = −1/2,x=7π
6.
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
2and x=7π
6.
Question 24
Question
Find all solutions to the equation cos(3x) = −1
2for 0◦≤x < 360◦.
16
Solution
Step 1: Recall the values of θfor which cos(θ) = −1
2. We have θ= 120◦and
θ= 240◦.
Step 2: Now, we need to solve the equation cos(3x) = −1
2. Since cos(3x) =
cos(2x+x), we can rewrite the equation as cos(2x+x) = −1
2.
Step 3: Using the sum formula for cosine, we have cos(2x) cos(x)−sin(2x) sin(x) =
−1
2.
Step 4: Recall the double angle formulas for cosine and sine: cos(2θ) =
2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ).
Step 5: Substitute these formulas into the equation from Step 3 to get
[2 cos2(x)−1] cos(x)−[2 sin(x) cos(x)] sin(x) = −1
2.
Step 6: Simplify the equation to obtain 2 cos3(x)−cos(x)−2 sin2(x) cos(x) =
−1
2.
Step 7: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1.
Step 8: Substitute sin2(x) = 1 −cos2(x)into the equation from Step 6 to
get 2 cos3(x)−cos(x)−2(1 −cos2(x)) cos(x) = −1
2.
Step 9: Simplify and solve the resulting cubic equation 2 cos3(x)−cos(x)−
2 cos(x) + 2 cos3(x) = −1
2.
Step 10: Combining like terms gives 4 cos3(x)−3 cos(x) + 1
2= 0.
Step 11: Use the solutions for cos(θ) = −1
2to solve the cubic equation in
Step 10: x= 120◦and x= 240◦.
Step 12: Therefore, the solutions to the equation cos(3x) = −1
2for 0◦≤x <
360◦are x= 40◦,80◦,120◦,160◦,200◦,240◦,280◦,320◦.
Question 25
Question
Find all solutions to the equation sin(2x) = cos(x)in the interval [0,2π].
Solution
Step 1: Recall the double angle identity for sine and the Pythagorean identity
for cosine.
sin(2x) = 2 sin(x) cos(x)and cos2(x) + sin2(x) = 1
Step 2: Substitute the double angle identity into the equation sin(2x) =
cos(x).
2 sin(x) cos(x) = cos(x)
Step 3: Rearrange the equation to get cos(x)terms on one side.
2 sin(x) cos(x)−cos(x) = 0
17
Step 4: Factor out a cos(x).
cos(x)(2 sin(x)−1) = 0
Step 5: Set each factor equal to zero and solve for x. For cos(x) = 0:
cos(x) = 0 ⇒x=π
2,3π
2
Step 6: For 2 sin(x)−1 = 0:
sin(x) = 1
2⇒x=π
6,5π
6
Step 7: Combine all solutions in the interval [0,2π].
x=π
6,π
2,5π
6,3π
2
18
Solution
Step 1: We are given f(x) = cos(2x)and g(x) = sin(x). The equation we
are trying to solve is f(x) = g(x). Step 2: Substitute f(x)and g(x)into the
equation:
cos(2x) = sin(x)
Step 3: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 4: Substitute the double angle identity for cos(2x):
2 cos2(x)−1 = sin(x)
Step 5: Since cos2(x) = 1 −sin2(x), substitute this into the equation:
2(1 −sin2(x)) −1 = sin(x)
Step 6: Simplify the equation:
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 7: Let u= sin(x), then the equation becomes a quadratic in u:
2u2+u−1 = 0
Step 8: Solve the quadratic equation for u:
u=−1±√12−4(2)(−1)
2(2)
u=−1±√9
4
u=−1±3
4
Step 9: Solve for u:
u1=−1+3
4=2
4=1
2
u2=−1−3
4=−4
4=−1
Step 10: Recall that u= sin(x), so x= arcsin(1
2)=π
6and x= arcsin(−1) =
−π
2. Step 11: Since we are looking for solutions in the interval [0,2π], we only
consider x=π
6. Step 12: Thus, the solution to the equation f(x) = g(x)in the
interval [0,2π]is x=π
6.
Question 3
Question
Solve the equation 2 sin2(x)−√3 sin(x)−1 = 0 for 0≤x≤2π.
2
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
2u2−√3u−1 = 0.
Step 2: Solve the quadratic equation by factoring or using the quadratic
formula. In this case, we will use the quadratic formula:
u=−(−√3) ±√(−√3)2−4(2)(−1)
2(2) .
u=√3±√7
4.
Step 3: Therefore, we have two possible values for u:
u1=√3 + √7
4and u2=√3−√7
4.
Step 4: Now, we recall that u= sin(x), so we need to find the angles xthat
satisfy the given values of u.
Step 5: For u1=√3+√7
4, we have x= sin−1(√3+√7
4).
Step 6: For u2=√3−√7
4, we have x= sin−1(√3−√7
4).
Step 7: Therefore, the solutions to the equation 2 sin2(x)−√3 sin(x)−1 = 0
for 0≤x≤2πare:
x= sin−1(√3 + √7
4),sin−1(√3−√7
4).
Question 4
Question
Find all solutions to the equation sin(2x) = cos(x)for 0≤x < 2π.
Solution
Step 1: Recall the double angle formula for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Rewrite the given equation using the double angle formula: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x < 2πare x=π
6
and x=5π
6.
Step 6: Check these solutions in the original equation: - For x=π
6,
sin(2·π
6)= sin(π
3)=√3
2and cos(π
6)=√3
2. Since they are not equal, π
6is
3
not a solution. - For x=5π
6,sin(2·5π
6)= sin(5π
3)=−√3
2and cos(5π
6)=−√3
2.
They are equal, so 5π
6is a valid solution.
Step 7: Therefore, the only solution to the equation sin(2x) = cos(x)for
0≤x < 2πis x=5π
6.
Question 5
Question
Prove the trigonometric identity:
cot(θ)−tan(θ) = −2 csc(2θ)
Solution
To prove the identity cot(θ)−tan(θ) = −2 csc(2θ), we will start by expressing
cot(θ)and tan(θ)in terms of sine and cosine functions, and csc(2θ)in terms of
sine and cosine functions.
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Expressing csc(2θ)
csc(2θ) = 1
sin(2θ)
=1
2 sin(θ) cos(θ)
=1
2 sin(θ) cos(θ)
=1
2·1
sin(θ) cos(θ)
=1
2·2
sin(2θ)(Using double angle formula)
= csc(2θ)
Step 3: Proving the identity We will now substitute the expressions we
4
derived into the given identity:
cot(θ)−tan(θ) = −2 csc(2θ)
cos(θ)
sin(θ)−sin(θ)
cos(θ)=−2·1
2 sin(θ) cos(θ)
cos2(θ)−sin2(θ)
sin(θ) cos(θ)=−1
sin(θ) cos(θ)
cos(2θ)
sin(2θ)=−1
sin(2θ)
cos(2θ) = −1
The last step is true since cos(π) = −1. Therefore, we have proved the trigono-
metric identity cot(θ)−tan(θ) = −2 csc(2θ).
Question 6
Question
Solve the equation sin(2x)−cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We can rewrite the equation as sin(2x) = cos(x).
Step 2: Recall the double-angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 3: Substituting this identity into our equation, we get 2 sin(x) cos(x) =
cos(x).
Step 4: Dividing both sides by cos(x)(which is not zero in the given interval),
we obtain 2 sin(x) = 1.
Step 5: Solving for sin(x), we have sin(x) = 1
2.
Step 6: From the unit circle, we know that sin(30◦) = 1
2.
Step 7: So, the solutions for xare x= 30◦and x= 150◦within the given
interval.
Step 8: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0◦≤x≤360◦are x= 30◦and x= 150◦.
Question 7
Question
Let f(x) = 2 sin(x) + 3 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
5
Solution
Step 1: Amplitude The amplitude of a function of the form f(x) = asin(bx) +
ccos(dx)is given by √a2+c2. In this case, a= 2 and c= 3, so the amplitude
of f(x)is √22+ 32=√4 + 9 = √13.
Step 2: Period For a function of the form f(x) = asin(bx) + ccos(dx), the
period is given by 2π
|b|if bis the coefficient of xin the sine term, or 2π
|d|if dis
the coefficient of xin the cosine term. Since there are no coefficient before xon
either trigonometric function term in f(x), we have 2π
1= 2πas the period of
f(x).
Step 3: Phase Shift To determine the phase shift of a function f(x) =
asin(bx) + ccos(dx), we first need to express it in a different form. In this
case, f(x) = 2 sin(x) + 3 cos(x) = √13 (2
√13 sin(x) + 3
√13 cos(x)). By rewriting
2
√13 sin(x) + 3
√13 cos(x)as sin(x−α)where α= arctan(3
2), we can determine
that the phase shift of f(x)is −α=−arctan(3
2).
Step 4: Vertical Shift The vertical shift of a function is simply the constant
term in the function. In this case, the constant term is 0, so the vertical shift
of f(x)is 0.
Question 8
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: Rewrite the angle as a sum of two special angles.
5π
12 =π
3+π
4
Step 2: Use the sum-to-product identity for sine.
sin (π
3+π
4)= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Calculate the sine and cosine values of π
3and π
4.
sin (π
3)=√3
2,cos (π
3)=1
2,sin (π
4)=√2
2,cos (π
4)=√2
2
Step 4: Substitute the values into the formula.
sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)=√3
2·√2
2+1
2·√2
2
6
Step 5: Simplify the expression.
√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
Question 9
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for (f◦g)(x)and
simplify the result.
Solution
Step 1: Find (f◦g)(x)by substituting g(x) = cos(x)into f(x) = sin(2x).
(f◦g)(x) = f(g(x)) = f(cos(x)) = sin(2 ·cos(x))
Step 2: Use the double angle identity for sine, which states that sin(2θ) =
2 sin(θ) cos(θ), to simplify sin(2 ·cos(x)).
sin(2 ·cos(x)) = 2 sin(cos(x)) cos(cos(x))
Step 3: Further simplify 2 sin(cos(x)) cos(cos(x)) since sin(cos(x)) and cos(cos(x))
are not standard trigonometric functions.
No further simplification possible
Therefore, the expression for (f◦g)(x)is 2 sin(cos(x)) cos(cos(x)).
Question 10
Question
Find the exact value of sin (7π
12 ).
Solution
Step 1: Convert 7π
12 to an angle within the unit circle.
Step 2: Since 7π
12 =π
12 +3π
4, we can rewrite this as 15◦+ 135◦in degrees.
Step 3: Identify the reference angle in the first quadrant by subtracting π
from π
12 , which gives us π
12 −π=−11π
12 .
Step 4: Use the reference angle and quadrant information to determine the
sign of sin in the second quadrant. Since sin is positive in the second quadrant,
the value of sin(7π
12 )will also be positive.
7
Step 5: Now, calculate the sine of the reference angle 11π
12 in the second
quadrant: sin (11π
12 )= sin (π−π
12 )= sin (π
12 ).
Step 6: The exact value of sin (π
12 )can be found using the angle sum identity
for sine: sin (π
12 )= sin (π
6−π
3)= sin (π
6)cos (π
3)−cos (π
6)sin (π
3).
Step 7: Substitute the values of sine and cosine for π
6and π
3:sin (π
12 )=
1
2·√3
2−√3
2·1
2=√3−√3
4=0
4= 0.
Step 8: Therefore, the exact value of sin (7π
12 )is 0.
Question 11
Question
Let f(x) = sin2(3x) + cos(3x). Find the maximum and minimum values of f(x)
for x∈[0,π
6].
Solution
Step 1: First, let’s find the critical points of f(x)in the interval [0,π
6]by taking
the derivative and solving for xwhen f′(x) = 0.
Step 2: Find f′(x):
f′(x) = 2 sin(3x) cos(3x)−3 sin(3x)
Step 3: Set f′(x) = 0 and solve for x:
2 sin(3x) cos(3x)−3 sin(3x) = 0
sin(3x)(2 cos(3x)−3) = 0
Now, we find the critical points by solving:
sin(3x) = 0 or 2 cos(3x)−3 = 0
Step 4: Solve sin(3x) = 0 for xin [0,π
6]:
sin(3x) = 0 =⇒3x= 0, π, 2π, . . .
Since x∈[0, π/6], the solution is x= 0.
Step 5: Solve 2 cos(3x)−3 = 0 for xin [0,π
6]:
2 cos(3x)−3 = 0 =⇒cos(3x) = 3
2
There are no solutions in the interval [0,π
6].
Step 6: Evaluate f(x)at the critical point and endpoints to find the maxi-
mum and minimum values:
f(0) = sin2(0) + cos(0) = 0 + 1 = 1
f(π
6)= sin2(π
2) + cos(π
2)= 1 + 0 = 1
Step 7: Therefore, the maximum value of f(x)in the interval [0,π
6]is 1, and
the minimum value is also 1.
8
Question 12
Question
Find the exact value of sin (11π
12 ).
Solution
Step 1: Recognize that 11π
12 is not one of the common angles on the unit circle,
so we need to express it in terms of more common angles.
Step 2: We can rewrite 11π
12 as 3π
4+π
6. This is because 11π
12 =9π
12 +2π
12 =3π
4+π
6.
Step 3: Now, since we know that 3π
4and π
6are common angles, we can use
the sum-to-product identities to find the exact value of sin (11π
12 ).
Step 4: The sum-to-product identity for sine is: sin(A+B) = sin Acos B+
cos Asin B.
Step 5: Applying the sum-to-product identity, we have:
sin (11π
12 )= sin (3π
4+π
6)
= sin (3π
4)cos (π
6)+ cos (3π
4)sin (π
6)
Step 6: Using the unit circle, we find that sin (3π
4)=−√2
2,cos (π
6)=√3
2,
cos (3π
4)=−√2
2, and sin (π
6)=1
2.
Step 7: Substituting these values into our expression, we get:
sin (11π
12 )=−√2
2·√3
2−√2
2·1
2
=−√6
4−√2
4
=−√2 + √6
4
Therefore, the exact value of sin (11π
12 )is −√2+√6
4.
Question 13
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
9
Solution
Step 1: Recall the trigonometric identity sin(3x) = sin(π−3x).
Step 2: Rewrite the equation as sin(π−3x) = cos(2x).
Step 3: Since sin(π−θ) = sin(θ)and cos(π−θ) = −cos(θ), we have
sin(3x) = −cos(3x) = cos(2x).
Step 4: Set up the equation cos(2x) = −cos(3x).
Step 5: Use the double angle formula for cosine: cos(2x) = 2 cos2(x)−1.
Step 6: Substitute this into the equation: 2 cos2(x)−1 = −cos(3x).
Step 7: Use the triple angle formula for cosine: cos(3x) = 4 cos3(x)−3 cos(x).
Step 8: Substitute this into the equation: 2 cos2(x)−1 = −4 cos3(x) +
3 cos(x).
Step 9: Rearrange the equation to get 4 cos3(x)+2 cos2(x)+3 cos(x)−1 = 0.
Step 10: The equation is now a cubic equation in terms of cos(x). Solving
it may require the use of numerical methods like Newton’s method.
Question 14
Question
Determine the exact value of cos (11π
6).
Solution
Step 1: Recall the unit circle where cos (11π
6)is located. The angle 11π
6is in the
fourth quadrant, corresponding to the point (−√3
2,1
2)on the unit circle.
Step 2: Since cos (11π
6)corresponds to the x-coordinate of the point on the
unit circle, cos (11π
6)=−√3
2.
Therefore, the exact value of cos (11π
6)is −√3
2.
Question 15
Question
Solve the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= cos(x). The equation becomes 2u2+u−1 = 0.
Step 2: Now, we need to solve this quadratic equation for u. We can factor
it as (2u−1)(u+ 1) = 0.
Step 3: Set each factor equal to zero:
{2u−1 = 0
u+ 1 = 0
10
Step 4: Solve for uin each case:
{2u= 1
u=1
2
or {u=−1
Step 5: Remember, u= cos(x), so:
{cos(x) = 1
2
cos(x) = −1
Step 6: Solve each equation for x:
{x=π
3or x=5π
3
x=π
Therefore, the solutions to the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤
x≤2πare x=π
3, π, 5π
3.
Question 16
Question
Let f(x) = 2 sin(x)−√3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Finding the amplitude The general form of a trigonometric function
is f(x) = Asin(Bx −C) + Dcos(Ex −F). The amplitude of f(x)is given by
A=√A2+D2. In this case, A= 2 and D=−√3. Therefore, the amplitude is
amplitude =√22+ (−√3)2=√4 + 3 = √7.
Step 2: Finding the period The period of a general trigonometric function
is given by 2π/B or 2π/E. In this case, the period of f(x)is
period =2π
1= 2π.
Step 3: Finding the phase shift To find the phase shift, we need to solve the
equations Bx −C= 0 and Ex −F= 0. In this case, the phase shift for f(x)is
phase shift =C
B=−π/6
1=−π
6.
Step 4: Finding the vertical shift The vertical shift of a trigonometric func-
tion is the value of D in the general form f(x) = Asin(Bx −C)+Dcos(Ex −F).
In this case, the vertical shift is
vertical shift =D=−√3.
11
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
−π
6, and the vertical shift is −√3.
Question 17
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double-angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: Substitute cos(2x) = 1 −2 sin2(x)into the equation sin(x) = cos(2x)
to obtain sin(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to get 2 sin2(x) + sin(x)−1 = 0.
Step 4: Factor the quadratic equation to get (2 sin(x)−1)(sin(x) + 1) = 0.
Step 5: Set each factor to zero to find the possible solutions.
• For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which gives x=π
6,5π
6in the
interval [0,2π].
• For sin(x)+1 = 0, we have sin(x) = −1, which gives x=3π
2in the interval
[0,2π].
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation cos(2x) = √2
2for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute cos(2x) = √2
2into the double angle identity:
2 cos2(x)−1 = √2
2
Step 3: Add 1 to both sides:
2 cos2(x) = √2
2+ 1
12
Step 4: Simplify the right side:
2 cos2(x) = √2+2
2
Step 5: Divide by 2 to isolate cos2(x):
cos2(x) = √2+2
4
Step 6: Take the square root of both sides:
cos(x) = ±√√2+2
4
Step 7: We know that cos(x) = √2
2corresponds to the angle x=π
4in the
unit circle.
Step 8: To find the other solution, we can consider the symmetry of the
cosine function. The cosine function is positive in the first and fourth quadrants.
Hence, the other solution is x=7π
4.
Therefore, the solutions to the equation cos(2x) = √2
2for 0≤x≤2πare
x=π
4and x=7π
4.
Question 19
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: We will use the angle addition identity tan(A+B) = tan A+tan B
1−tan Atan B.
Step 2: Let A=π
3and B=π
4. Then, 5π
12 =π
3+π
4.
Step 3: We know that tan(π/3) = √3and tan(π/4) = 1.
Step 4: Substitute these values into the angle addition formula:
tan (π
3+π
4)=√3+1
1−√3·1=√3+1
1−√3
Step 5: To rationalize the denominator, multiply by the conjugate of the
denominator:
=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3)
Step 6: Simplify the expression:
=√3 + √3+1+3
1−3=−2(√3 + 1) = −2√3−2
Step 7: Therefore, tan (5π
12 =−2√3−2.
13
Question 20
Question
Let f(x) = 2 sin(x) + √3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Amplitude
The amplitude of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by √A2+D2. In this case, A= 2 and D=√3, so the amplitude is
√22+ (√3)2=√4 + 3 = √7.
Step 2: Period
For a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C), the period
is given by 2π
|B|. In this case, since B= 1 (implicitly, since sin(x)and cos(x)
both have a coefficient of 1), the period is
2π
1= 2π.
Step 3: Phase Shift
The phase shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by C
B. Here, C= 0 (since there is no xterm inside the sine or cosine
functions), so the phase shift is 0
1= 0.
Step 4: Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is simply D. In this case, the vertical shift is
D=√3.
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
0, and the vertical shift is √3.
Question 21
Question
Find all solutions to the equation cos(2x) = sin(x)for 0≤x≤2π.
14
Solution
Step 1: We will use the double-angle formula for cosine to rewrite cos(2x)in
terms of cos(x)and sin(x).
cos(2x) = 2 cos2(x)−1
Step 2: Substitute the expression for cos(2x)into the original equation.
2 cos2(x)−1 = sin(x)
Step 3: Recall that cos2(x) = 1 −sin2(x). Substitute this into the equation.
2(1 −sin2(x)) −1 = sin(x)
Step 4: Simplify the equation.
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 5: We can treat this as a quadratic equation in terms of sin(x). Let
sin(x) = yto simplify the equation.
2y2+y−1 = 0
Step 6: Solve the quadratic equation for y.
y=−1±√12−4(2)(−1)
2(2)
y=−1±√9
4
Step 7: Solve for y:
y1=−1+3
4=2
4=1
2
y2=−1−3
4=−4
4=−1
Step 8: Recall that sin(x) = y. So, we have sin(x) = 1
2and sin(x) = −1.
Solve for xin the interval 0≤x≤2π.
For sin(x) = 1
2,x=π
6,5π
6
For sin(x) = −1,x=3π
2
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
6,5π
6,3π
2.
Question 22
Question
Solve the equation sin2(x) + cos2(x) = 5
4for 0≤x≤2π.
15
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x)=1. Therefore, we
have
sin2(x) + cos2(x) = 1 = 4
4.
Step 2: Subtracting 4
4from both sides of the given equation, we get
5
4−4
4= 1.
Step 3: Simplifying the left side of the equation gives us
1
4= 1.
Step 4: Since 1
4= 1, there are no solutions to the given equation sin2(x) +
cos2(x) = 5
4for 0≤x≤2π.
Question 23
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double angle formula for cosine: cos(2x)=1−2 sin2(x).
Therefore, we can rewrite the equation as sin(x) = 1 −2 sin2(x).
Step 2: Rearranging the equation, we get 2 sin2(x) + sin(x)−1 = 0. Let
u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 3: Solve the quadratic equation 2u2+u−1=0by factoring or us-
ing the quadratic formula. The solutions are u=−1±√12−4(2)(−1)
2(2) =−1±√9
4.
Therefore, u=−1+3
4= 1 or u=−1−3
4=−1/2.
Step 4: Since u= sin(x), the solutions are sin(x) = 1 and sin(x) = −1/2.
Step 5: For sin(x) = 1,x=π
2. For sin(x) = −1/2,x=7π
6.
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
2and x=7π
6.
Question 24
Question
Find all solutions to the equation cos(3x) = −1
2for 0◦≤x < 360◦.
16
Solution
Step 1: Recall the values of θfor which cos(θ) = −1
2. We have θ= 120◦and
θ= 240◦.
Step 2: Now, we need to solve the equation cos(3x) = −1
2. Since cos(3x) =
cos(2x+x), we can rewrite the equation as cos(2x+x) = −1
2.
Step 3: Using the sum formula for cosine, we have cos(2x) cos(x)−sin(2x) sin(x) =
−1
2.
Step 4: Recall the double angle formulas for cosine and sine: cos(2θ) =
2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ).
Step 5: Substitute these formulas into the equation from Step 3 to get
[2 cos2(x)−1] cos(x)−[2 sin(x) cos(x)] sin(x) = −1
2.
Step 6: Simplify the equation to obtain 2 cos3(x)−cos(x)−2 sin2(x) cos(x) =
−1
2.
Step 7: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1.
Step 8: Substitute sin2(x) = 1 −cos2(x)into the equation from Step 6 to
get 2 cos3(x)−cos(x)−2(1 −cos2(x)) cos(x) = −1
2.
Step 9: Simplify and solve the resulting cubic equation 2 cos3(x)−cos(x)−
2 cos(x) + 2 cos3(x) = −1
2.
Step 10: Combining like terms gives 4 cos3(x)−3 cos(x) + 1
2= 0.
Step 11: Use the solutions for cos(θ) = −1
2to solve the cubic equation in
Step 10: x= 120◦and x= 240◦.
Step 12: Therefore, the solutions to the equation cos(3x) = −1
2for 0◦≤x <
360◦are x= 40◦,80◦,120◦,160◦,200◦,240◦,280◦,320◦.
Question 25
Question
Find all solutions to the equation sin(2x) = cos(x)in the interval [0,2π].
Solution
Step 1: Recall the double angle identity for sine and the Pythagorean identity
for cosine.
sin(2x) = 2 sin(x) cos(x)and cos2(x) + sin2(x) = 1
Step 2: Substitute the double angle identity into the equation sin(2x) =
cos(x).
2 sin(x) cos(x) = cos(x)
Step 3: Rearrange the equation to get cos(x)terms on one side.
2 sin(x) cos(x)−cos(x) = 0
17
Step 4: Factor out a cos(x).
cos(x)(2 sin(x)−1) = 0
Step 5: Set each factor equal to zero and solve for x. For cos(x) = 0:
cos(x) = 0 ⇒x=π
2,3π
2
Step 6: For 2 sin(x)−1 = 0:
sin(x) = 1
2⇒x=π
6,5π
6
Step 7: Combine all solutions in the interval [0,2π].
x=π
6,π
2,5π
6,3π
2
18
Solution
Step 1: We are given f(x) = cos(2x)and g(x) = sin(x). The equation we
are trying to solve is f(x) = g(x). Step 2: Substitute f(x)and g(x)into the
equation:
cos(2x) = sin(x)
Step 3: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 4: Substitute the double angle identity for cos(2x):
2 cos2(x)−1 = sin(x)
Step 5: Since cos2(x) = 1 −sin2(x), substitute this into the equation:
2(1 −sin2(x)) −1 = sin(x)
Step 6: Simplify the equation:
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 7: Let u= sin(x), then the equation becomes a quadratic in u:
2u2+u−1 = 0
Step 8: Solve the quadratic equation for u:
u=−1±√12−4(2)(−1)
2(2)
u=−1±√9
4
u=−1±3
4
Step 9: Solve for u:
u1=−1+3
4=2
4=1
2
u2=−1−3
4=−4
4=−1
Step 10: Recall that u= sin(x), so x= arcsin(1
2)=π
6and x= arcsin(−1) =
−π
2. Step 11: Since we are looking for solutions in the interval [0,2π], we only
consider x=π
6. Step 12: Thus, the solution to the equation f(x) = g(x)in the
interval [0,2π]is x=π
6.
Question 3
Question
Solve the equation 2 sin2(x)−√3 sin(x)−1 = 0 for 0≤x≤2π.
2
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
2u2−√3u−1 = 0.
Step 2: Solve the quadratic equation by factoring or using the quadratic
formula. In this case, we will use the quadratic formula:
u=−(−√3) ±√(−√3)2−4(2)(−1)
2(2) .
u=√3±√7
4.
Step 3: Therefore, we have two possible values for u:
u1=√3 + √7
4and u2=√3−√7
4.
Step 4: Now, we recall that u= sin(x), so we need to find the angles xthat
satisfy the given values of u.
Step 5: For u1=√3+√7
4, we have x= sin−1(√3+√7
4).
Step 6: For u2=√3−√7
4, we have x= sin−1(√3−√7
4).
Step 7: Therefore, the solutions to the equation 2 sin2(x)−√3 sin(x)−1 = 0
for 0≤x≤2πare:
x= sin−1(√3 + √7
4),sin−1(√3−√7
4).
Question 4
Question
Find all solutions to the equation sin(2x) = cos(x)for 0≤x < 2π.
Solution
Step 1: Recall the double angle formula for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Rewrite the given equation using the double angle formula: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x < 2πare x=π
6
and x=5π
6.
Step 6: Check these solutions in the original equation: - For x=π
6,
sin(2·π
6)= sin(π
3)=√3
2and cos(π
6)=√3
2. Since they are not equal, π
6is
3
not a solution. - For x=5π
6,sin(2·5π
6)= sin(5π
3)=−√3
2and cos(5π
6)=−√3
2.
They are equal, so 5π
6is a valid solution.
Step 7: Therefore, the only solution to the equation sin(2x) = cos(x)for
0≤x < 2πis x=5π
6.
Question 5
Question
Prove the trigonometric identity:
cot(θ)−tan(θ) = −2 csc(2θ)
Solution
To prove the identity cot(θ)−tan(θ) = −2 csc(2θ), we will start by expressing
cot(θ)and tan(θ)in terms of sine and cosine functions, and csc(2θ)in terms of
sine and cosine functions.
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Expressing csc(2θ)
csc(2θ) = 1
sin(2θ)
=1
2 sin(θ) cos(θ)
=1
2 sin(θ) cos(θ)
=1
2·1
sin(θ) cos(θ)
=1
2·2
sin(2θ)(Using double angle formula)
= csc(2θ)
Step 3: Proving the identity We will now substitute the expressions we
4
derived into the given identity:
cot(θ)−tan(θ) = −2 csc(2θ)
cos(θ)
sin(θ)−sin(θ)
cos(θ)=−2·1
2 sin(θ) cos(θ)
cos2(θ)−sin2(θ)
sin(θ) cos(θ)=−1
sin(θ) cos(θ)
cos(2θ)
sin(2θ)=−1
sin(2θ)
cos(2θ) = −1
The last step is true since cos(π) = −1. Therefore, we have proved the trigono-
metric identity cot(θ)−tan(θ) = −2 csc(2θ).
Question 6
Question
Solve the equation sin(2x)−cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We can rewrite the equation as sin(2x) = cos(x).
Step 2: Recall the double-angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 3: Substituting this identity into our equation, we get 2 sin(x) cos(x) =
cos(x).
Step 4: Dividing both sides by cos(x)(which is not zero in the given interval),
we obtain 2 sin(x) = 1.
Step 5: Solving for sin(x), we have sin(x) = 1
2.
Step 6: From the unit circle, we know that sin(30◦) = 1
2.
Step 7: So, the solutions for xare x= 30◦and x= 150◦within the given
interval.
Step 8: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0◦≤x≤360◦are x= 30◦and x= 150◦.
Question 7
Question
Let f(x) = 2 sin(x) + 3 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
5
Solution
Step 1: Amplitude The amplitude of a function of the form f(x) = asin(bx) +
ccos(dx)is given by √a2+c2. In this case, a= 2 and c= 3, so the amplitude
of f(x)is √22+ 32=√4 + 9 = √13.
Step 2: Period For a function of the form f(x) = asin(bx) + ccos(dx), the
period is given by 2π
|b|if bis the coefficient of xin the sine term, or 2π
|d|if dis
the coefficient of xin the cosine term. Since there are no coefficient before xon
either trigonometric function term in f(x), we have 2π
1= 2πas the period of
f(x).
Step 3: Phase Shift To determine the phase shift of a function f(x) =
asin(bx) + ccos(dx), we first need to express it in a different form. In this
case, f(x) = 2 sin(x) + 3 cos(x) = √13 (2
√13 sin(x) + 3
√13 cos(x)). By rewriting
2
√13 sin(x) + 3
√13 cos(x)as sin(x−α)where α= arctan(3
2), we can determine
that the phase shift of f(x)is −α=−arctan(3
2).
Step 4: Vertical Shift The vertical shift of a function is simply the constant
term in the function. In this case, the constant term is 0, so the vertical shift
of f(x)is 0.
Question 8
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: Rewrite the angle as a sum of two special angles.
5π
12 =π
3+π
4
Step 2: Use the sum-to-product identity for sine.
sin (π
3+π
4)= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Calculate the sine and cosine values of π
3and π
4.
sin (π
3)=√3
2,cos (π
3)=1
2,sin (π
4)=√2
2,cos (π
4)=√2
2
Step 4: Substitute the values into the formula.
sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)=√3
2·√2
2+1
2·√2
2
6
Step 5: Simplify the expression.
√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
Question 9
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for (f◦g)(x)and
simplify the result.
Solution
Step 1: Find (f◦g)(x)by substituting g(x) = cos(x)into f(x) = sin(2x).
(f◦g)(x) = f(g(x)) = f(cos(x)) = sin(2 ·cos(x))
Step 2: Use the double angle identity for sine, which states that sin(2θ) =
2 sin(θ) cos(θ), to simplify sin(2 ·cos(x)).
sin(2 ·cos(x)) = 2 sin(cos(x)) cos(cos(x))
Step 3: Further simplify 2 sin(cos(x)) cos(cos(x)) since sin(cos(x)) and cos(cos(x))
are not standard trigonometric functions.
No further simplification possible
Therefore, the expression for (f◦g)(x)is 2 sin(cos(x)) cos(cos(x)).
Question 10
Question
Find the exact value of sin (7π
12 ).
Solution
Step 1: Convert 7π
12 to an angle within the unit circle.
Step 2: Since 7π
12 =π
12 +3π
4, we can rewrite this as 15◦+ 135◦in degrees.
Step 3: Identify the reference angle in the first quadrant by subtracting π
from π
12 , which gives us π
12 −π=−11π
12 .
Step 4: Use the reference angle and quadrant information to determine the
sign of sin in the second quadrant. Since sin is positive in the second quadrant,
the value of sin(7π
12 )will also be positive.
7
Step 5: Now, calculate the sine of the reference angle 11π
12 in the second
quadrant: sin (11π
12 )= sin (π−π
12 )= sin (π
12 ).
Step 6: The exact value of sin (π
12 )can be found using the angle sum identity
for sine: sin (π
12 )= sin (π
6−π
3)= sin (π
6)cos (π
3)−cos (π
6)sin (π
3).
Step 7: Substitute the values of sine and cosine for π
6and π
3:sin (π
12 )=
1
2·√3
2−√3
2·1
2=√3−√3
4=0
4= 0.
Step 8: Therefore, the exact value of sin (7π
12 )is 0.
Question 11
Question
Let f(x) = sin2(3x) + cos(3x). Find the maximum and minimum values of f(x)
for x∈[0,π
6].
Solution
Step 1: First, let’s find the critical points of f(x)in the interval [0,π
6]by taking
the derivative and solving for xwhen f′(x) = 0.
Step 2: Find f′(x):
f′(x) = 2 sin(3x) cos(3x)−3 sin(3x)
Step 3: Set f′(x) = 0 and solve for x:
2 sin(3x) cos(3x)−3 sin(3x) = 0
sin(3x)(2 cos(3x)−3) = 0
Now, we find the critical points by solving:
sin(3x) = 0 or 2 cos(3x)−3 = 0
Step 4: Solve sin(3x) = 0 for xin [0,π
6]:
sin(3x) = 0 =⇒3x= 0, π, 2π, . . .
Since x∈[0, π/6], the solution is x= 0.
Step 5: Solve 2 cos(3x)−3 = 0 for xin [0,π
6]:
2 cos(3x)−3 = 0 =⇒cos(3x) = 3
2
There are no solutions in the interval [0,π
6].
Step 6: Evaluate f(x)at the critical point and endpoints to find the maxi-
mum and minimum values:
f(0) = sin2(0) + cos(0) = 0 + 1 = 1
f(π
6)= sin2(π
2) + cos(π
2)= 1 + 0 = 1
Step 7: Therefore, the maximum value of f(x)in the interval [0,π
6]is 1, and
the minimum value is also 1.
8
Question 12
Question
Find the exact value of sin (11π
12 ).
Solution
Step 1: Recognize that 11π
12 is not one of the common angles on the unit circle,
so we need to express it in terms of more common angles.
Step 2: We can rewrite 11π
12 as 3π
4+π
6. This is because 11π
12 =9π
12 +2π
12 =3π
4+π
6.
Step 3: Now, since we know that 3π
4and π
6are common angles, we can use
the sum-to-product identities to find the exact value of sin (11π
12 ).
Step 4: The sum-to-product identity for sine is: sin(A+B) = sin Acos B+
cos Asin B.
Step 5: Applying the sum-to-product identity, we have:
sin (11π
12 )= sin (3π
4+π
6)
= sin (3π
4)cos (π
6)+ cos (3π
4)sin (π
6)
Step 6: Using the unit circle, we find that sin (3π
4)=−√2
2,cos (π
6)=√3
2,
cos (3π
4)=−√2
2, and sin (π
6)=1
2.
Step 7: Substituting these values into our expression, we get:
sin (11π
12 )=−√2
2·√3
2−√2
2·1
2
=−√6
4−√2
4
=−√2 + √6
4
Therefore, the exact value of sin (11π
12 )is −√2+√6
4.
Question 13
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
9
Solution
Step 1: Recall the trigonometric identity sin(3x) = sin(π−3x).
Step 2: Rewrite the equation as sin(π−3x) = cos(2x).
Step 3: Since sin(π−θ) = sin(θ)and cos(π−θ) = −cos(θ), we have
sin(3x) = −cos(3x) = cos(2x).
Step 4: Set up the equation cos(2x) = −cos(3x).
Step 5: Use the double angle formula for cosine: cos(2x) = 2 cos2(x)−1.
Step 6: Substitute this into the equation: 2 cos2(x)−1 = −cos(3x).
Step 7: Use the triple angle formula for cosine: cos(3x) = 4 cos3(x)−3 cos(x).
Step 8: Substitute this into the equation: 2 cos2(x)−1 = −4 cos3(x) +
3 cos(x).
Step 9: Rearrange the equation to get 4 cos3(x)+2 cos2(x)+3 cos(x)−1 = 0.
Step 10: The equation is now a cubic equation in terms of cos(x). Solving
it may require the use of numerical methods like Newton’s method.
Question 14
Question
Determine the exact value of cos (11π
6).
Solution
Step 1: Recall the unit circle where cos (11π
6)is located. The angle 11π
6is in the
fourth quadrant, corresponding to the point (−√3
2,1
2)on the unit circle.
Step 2: Since cos (11π
6)corresponds to the x-coordinate of the point on the
unit circle, cos (11π
6)=−√3
2.
Therefore, the exact value of cos (11π
6)is −√3
2.
Question 15
Question
Solve the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= cos(x). The equation becomes 2u2+u−1 = 0.
Step 2: Now, we need to solve this quadratic equation for u. We can factor
it as (2u−1)(u+ 1) = 0.
Step 3: Set each factor equal to zero:
{2u−1 = 0
u+ 1 = 0
10
Step 4: Solve for uin each case:
{2u= 1
u=1
2
or {u=−1
Step 5: Remember, u= cos(x), so:
{cos(x) = 1
2
cos(x) = −1
Step 6: Solve each equation for x:
{x=π
3or x=5π
3
x=π
Therefore, the solutions to the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤
x≤2πare x=π
3, π, 5π
3.
Question 16
Question
Let f(x) = 2 sin(x)−√3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Finding the amplitude The general form of a trigonometric function
is f(x) = Asin(Bx −C) + Dcos(Ex −F). The amplitude of f(x)is given by
A=√A2+D2. In this case, A= 2 and D=−√3. Therefore, the amplitude is
amplitude =√22+ (−√3)2=√4 + 3 = √7.
Step 2: Finding the period The period of a general trigonometric function
is given by 2π/B or 2π/E. In this case, the period of f(x)is
period =2π
1= 2π.
Step 3: Finding the phase shift To find the phase shift, we need to solve the
equations Bx −C= 0 and Ex −F= 0. In this case, the phase shift for f(x)is
phase shift =C
B=−π/6
1=−π
6.
Step 4: Finding the vertical shift The vertical shift of a trigonometric func-
tion is the value of D in the general form f(x) = Asin(Bx −C)+Dcos(Ex −F).
In this case, the vertical shift is
vertical shift =D=−√3.
11
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
−π
6, and the vertical shift is −√3.
Question 17
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double-angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: Substitute cos(2x) = 1 −2 sin2(x)into the equation sin(x) = cos(2x)
to obtain sin(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to get 2 sin2(x) + sin(x)−1 = 0.
Step 4: Factor the quadratic equation to get (2 sin(x)−1)(sin(x) + 1) = 0.
Step 5: Set each factor to zero to find the possible solutions.
• For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which gives x=π
6,5π
6in the
interval [0,2π].
• For sin(x)+1 = 0, we have sin(x) = −1, which gives x=3π
2in the interval
[0,2π].
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation cos(2x) = √2
2for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute cos(2x) = √2
2into the double angle identity:
2 cos2(x)−1 = √2
2
Step 3: Add 1 to both sides:
2 cos2(x) = √2
2+ 1
12
Step 4: Simplify the right side:
2 cos2(x) = √2+2
2
Step 5: Divide by 2 to isolate cos2(x):
cos2(x) = √2+2
4
Step 6: Take the square root of both sides:
cos(x) = ±√√2+2
4
Step 7: We know that cos(x) = √2
2corresponds to the angle x=π
4in the
unit circle.
Step 8: To find the other solution, we can consider the symmetry of the
cosine function. The cosine function is positive in the first and fourth quadrants.
Hence, the other solution is x=7π
4.
Therefore, the solutions to the equation cos(2x) = √2
2for 0≤x≤2πare
x=π
4and x=7π
4.
Question 19
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: We will use the angle addition identity tan(A+B) = tan A+tan B
1−tan Atan B.
Step 2: Let A=π
3and B=π
4. Then, 5π
12 =π
3+π
4.
Step 3: We know that tan(π/3) = √3and tan(π/4) = 1.
Step 4: Substitute these values into the angle addition formula:
tan (π
3+π
4)=√3+1
1−√3·1=√3+1
1−√3
Step 5: To rationalize the denominator, multiply by the conjugate of the
denominator:
=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3)
Step 6: Simplify the expression:
=√3 + √3+1+3
1−3=−2(√3 + 1) = −2√3−2
Step 7: Therefore, tan (5π
12 =−2√3−2.
13
Question 20
Question
Let f(x) = 2 sin(x) + √3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Amplitude
The amplitude of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by √A2+D2. In this case, A= 2 and D=√3, so the amplitude is
√22+ (√3)2=√4 + 3 = √7.
Step 2: Period
For a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C), the period
is given by 2π
|B|. In this case, since B= 1 (implicitly, since sin(x)and cos(x)
both have a coefficient of 1), the period is
2π
1= 2π.
Step 3: Phase Shift
The phase shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by C
B. Here, C= 0 (since there is no xterm inside the sine or cosine
functions), so the phase shift is 0
1= 0.
Step 4: Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is simply D. In this case, the vertical shift is
D=√3.
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
0, and the vertical shift is √3.
Question 21
Question
Find all solutions to the equation cos(2x) = sin(x)for 0≤x≤2π.
14
Solution
Step 1: We will use the double-angle formula for cosine to rewrite cos(2x)in
terms of cos(x)and sin(x).
cos(2x) = 2 cos2(x)−1
Step 2: Substitute the expression for cos(2x)into the original equation.
2 cos2(x)−1 = sin(x)
Step 3: Recall that cos2(x) = 1 −sin2(x). Substitute this into the equation.
2(1 −sin2(x)) −1 = sin(x)
Step 4: Simplify the equation.
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 5: We can treat this as a quadratic equation in terms of sin(x). Let
sin(x) = yto simplify the equation.
2y2+y−1 = 0
Step 6: Solve the quadratic equation for y.
y=−1±√12−4(2)(−1)
2(2)
y=−1±√9
4
Step 7: Solve for y:
y1=−1+3
4=2
4=1
2
y2=−1−3
4=−4
4=−1
Step 8: Recall that sin(x) = y. So, we have sin(x) = 1
2and sin(x) = −1.
Solve for xin the interval 0≤x≤2π.
For sin(x) = 1
2,x=π
6,5π
6
For sin(x) = −1,x=3π
2
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
6,5π
6,3π
2.
Question 22
Question
Solve the equation sin2(x) + cos2(x) = 5
4for 0≤x≤2π.
15
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x)=1. Therefore, we
have
sin2(x) + cos2(x) = 1 = 4
4.
Step 2: Subtracting 4
4from both sides of the given equation, we get
5
4−4
4= 1.
Step 3: Simplifying the left side of the equation gives us
1
4= 1.
Step 4: Since 1
4= 1, there are no solutions to the given equation sin2(x) +
cos2(x) = 5
4for 0≤x≤2π.
Question 23
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double angle formula for cosine: cos(2x)=1−2 sin2(x).
Therefore, we can rewrite the equation as sin(x) = 1 −2 sin2(x).
Step 2: Rearranging the equation, we get 2 sin2(x) + sin(x)−1 = 0. Let
u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 3: Solve the quadratic equation 2u2+u−1=0by factoring or us-
ing the quadratic formula. The solutions are u=−1±√12−4(2)(−1)
2(2) =−1±√9
4.
Therefore, u=−1+3
4= 1 or u=−1−3
4=−1/2.
Step 4: Since u= sin(x), the solutions are sin(x) = 1 and sin(x) = −1/2.
Step 5: For sin(x) = 1,x=π
2. For sin(x) = −1/2,x=7π
6.
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
2and x=7π
6.
Question 24
Question
Find all solutions to the equation cos(3x) = −1
2for 0◦≤x < 360◦.
16
Solution
Step 1: Recall the values of θfor which cos(θ) = −1
2. We have θ= 120◦and
θ= 240◦.
Step 2: Now, we need to solve the equation cos(3x) = −1
2. Since cos(3x) =
cos(2x+x), we can rewrite the equation as cos(2x+x) = −1
2.
Step 3: Using the sum formula for cosine, we have cos(2x) cos(x)−sin(2x) sin(x) =
−1
2.
Step 4: Recall the double angle formulas for cosine and sine: cos(2θ) =
2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ).
Step 5: Substitute these formulas into the equation from Step 3 to get
[2 cos2(x)−1] cos(x)−[2 sin(x) cos(x)] sin(x) = −1
2.
Step 6: Simplify the equation to obtain 2 cos3(x)−cos(x)−2 sin2(x) cos(x) =
−1
2.
Step 7: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1.
Step 8: Substitute sin2(x) = 1 −cos2(x)into the equation from Step 6 to
get 2 cos3(x)−cos(x)−2(1 −cos2(x)) cos(x) = −1
2.
Step 9: Simplify and solve the resulting cubic equation 2 cos3(x)−cos(x)−
2 cos(x) + 2 cos3(x) = −1
2.
Step 10: Combining like terms gives 4 cos3(x)−3 cos(x) + 1
2= 0.
Step 11: Use the solutions for cos(θ) = −1
2to solve the cubic equation in
Step 10: x= 120◦and x= 240◦.
Step 12: Therefore, the solutions to the equation cos(3x) = −1
2for 0◦≤x <
360◦are x= 40◦,80◦,120◦,160◦,200◦,240◦,280◦,320◦.
Question 25
Question
Find all solutions to the equation sin(2x) = cos(x)in the interval [0,2π].
Solution
Step 1: Recall the double angle identity for sine and the Pythagorean identity
for cosine.
sin(2x) = 2 sin(x) cos(x)and cos2(x) + sin2(x) = 1
Step 2: Substitute the double angle identity into the equation sin(2x) =
cos(x).
2 sin(x) cos(x) = cos(x)
Step 3: Rearrange the equation to get cos(x)terms on one side.
2 sin(x) cos(x)−cos(x) = 0
17
Step 4: Factor out a cos(x).
cos(x)(2 sin(x)−1) = 0
Step 5: Set each factor equal to zero and solve for x. For cos(x) = 0:
cos(x) = 0 ⇒x=π
2,3π
2
Step 6: For 2 sin(x)−1 = 0:
sin(x) = 1
2⇒x=π
6,5π
6
Step 7: Combine all solutions in the interval [0,2π].
x=π
6,π
2,5π
6,3π
2
18
Solution
Step 1: We are given f(x) = cos(2x)and g(x) = sin(x). The equation we
are trying to solve is f(x) = g(x). Step 2: Substitute f(x)and g(x)into the
equation:
cos(2x) = sin(x)
Step 3: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 4: Substitute the double angle identity for cos(2x):
2 cos2(x)−1 = sin(x)
Step 5: Since cos2(x) = 1 −sin2(x), substitute this into the equation:
2(1 −sin2(x)) −1 = sin(x)
Step 6: Simplify the equation:
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 7: Let u= sin(x), then the equation becomes a quadratic in u:
2u2+u−1 = 0
Step 8: Solve the quadratic equation for u:
u=−1±√12−4(2)(−1)
2(2)
u=−1±√9
4
u=−1±3
4
Step 9: Solve for u:
u1=−1+3
4=2
4=1
2
u2=−1−3
4=−4
4=−1
Step 10: Recall that u= sin(x), so x= arcsin(1
2)=π
6and x= arcsin(−1) =
−π
2. Step 11: Since we are looking for solutions in the interval [0,2π], we only
consider x=π
6. Step 12: Thus, the solution to the equation f(x) = g(x)in the
interval [0,2π]is x=π
6.
Question 3
Question
Solve the equation 2 sin2(x)−√3 sin(x)−1 = 0 for 0≤x≤2π.
2
Solution
Step 1: Let u= sin(x), then the equation becomes a quadratic equation in u:
2u2−√3u−1 = 0.
Step 2: Solve the quadratic equation by factoring or using the quadratic
formula. In this case, we will use the quadratic formula:
u=−(−√3) ±√(−√3)2−4(2)(−1)
2(2) .
u=√3±√7
4.
Step 3: Therefore, we have two possible values for u:
u1=√3 + √7
4and u2=√3−√7
4.
Step 4: Now, we recall that u= sin(x), so we need to find the angles xthat
satisfy the given values of u.
Step 5: For u1=√3+√7
4, we have x= sin−1(√3+√7
4).
Step 6: For u2=√3−√7
4, we have x= sin−1(√3−√7
4).
Step 7: Therefore, the solutions to the equation 2 sin2(x)−√3 sin(x)−1 = 0
for 0≤x≤2πare:
x= sin−1(√3 + √7
4),sin−1(√3−√7
4).
Question 4
Question
Find all solutions to the equation sin(2x) = cos(x)for 0≤x < 2π.
Solution
Step 1: Recall the double angle formula for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Rewrite the given equation using the double angle formula: 2 sin(x) cos(x) =
cos(x).
Step 3: Divide both sides by cos(x):2 sin(x) = 1.
Step 4: Solve for sin(x):sin(x) = 1
2.
Step 5: The solutions for sin(x) = 1
2in the interval 0≤x < 2πare x=π
6
and x=5π
6.
Step 6: Check these solutions in the original equation: - For x=π
6,
sin(2·π
6)= sin(π
3)=√3
2and cos(π
6)=√3
2. Since they are not equal, π
6is
3
not a solution. - For x=5π
6,sin(2·5π
6)= sin(5π
3)=−√3
2and cos(5π
6)=−√3
2.
They are equal, so 5π
6is a valid solution.
Step 7: Therefore, the only solution to the equation sin(2x) = cos(x)for
0≤x < 2πis x=5π
6.
Question 5
Question
Prove the trigonometric identity:
cot(θ)−tan(θ) = −2 csc(2θ)
Solution
To prove the identity cot(θ)−tan(θ) = −2 csc(2θ), we will start by expressing
cot(θ)and tan(θ)in terms of sine and cosine functions, and csc(2θ)in terms of
sine and cosine functions.
Step 1: Expressing cot(θ)and tan(θ)
cot(θ) = cos(θ)
sin(θ)
tan(θ) = sin(θ)
cos(θ)
Step 2: Expressing csc(2θ)
csc(2θ) = 1
sin(2θ)
=1
2 sin(θ) cos(θ)
=1
2 sin(θ) cos(θ)
=1
2·1
sin(θ) cos(θ)
=1
2·2
sin(2θ)(Using double angle formula)
= csc(2θ)
Step 3: Proving the identity We will now substitute the expressions we
4
derived into the given identity:
cot(θ)−tan(θ) = −2 csc(2θ)
cos(θ)
sin(θ)−sin(θ)
cos(θ)=−2·1
2 sin(θ) cos(θ)
cos2(θ)−sin2(θ)
sin(θ) cos(θ)=−1
sin(θ) cos(θ)
cos(2θ)
sin(2θ)=−1
sin(2θ)
cos(2θ) = −1
The last step is true since cos(π) = −1. Therefore, we have proved the trigono-
metric identity cot(θ)−tan(θ) = −2 csc(2θ).
Question 6
Question
Solve the equation sin(2x)−cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We can rewrite the equation as sin(2x) = cos(x).
Step 2: Recall the double-angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 3: Substituting this identity into our equation, we get 2 sin(x) cos(x) =
cos(x).
Step 4: Dividing both sides by cos(x)(which is not zero in the given interval),
we obtain 2 sin(x) = 1.
Step 5: Solving for sin(x), we have sin(x) = 1
2.
Step 6: From the unit circle, we know that sin(30◦) = 1
2.
Step 7: So, the solutions for xare x= 30◦and x= 150◦within the given
interval.
Step 8: Therefore, the solutions to the equation sin(2x)−cos(x)=0for
0◦≤x≤360◦are x= 30◦and x= 150◦.
Question 7
Question
Let f(x) = 2 sin(x) + 3 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
5
Solution
Step 1: Amplitude The amplitude of a function of the form f(x) = asin(bx) +
ccos(dx)is given by √a2+c2. In this case, a= 2 and c= 3, so the amplitude
of f(x)is √22+ 32=√4 + 9 = √13.
Step 2: Period For a function of the form f(x) = asin(bx) + ccos(dx), the
period is given by 2π
|b|if bis the coefficient of xin the sine term, or 2π
|d|if dis
the coefficient of xin the cosine term. Since there are no coefficient before xon
either trigonometric function term in f(x), we have 2π
1= 2πas the period of
f(x).
Step 3: Phase Shift To determine the phase shift of a function f(x) =
asin(bx) + ccos(dx), we first need to express it in a different form. In this
case, f(x) = 2 sin(x) + 3 cos(x) = √13 (2
√13 sin(x) + 3
√13 cos(x)). By rewriting
2
√13 sin(x) + 3
√13 cos(x)as sin(x−α)where α= arctan(3
2), we can determine
that the phase shift of f(x)is −α=−arctan(3
2).
Step 4: Vertical Shift The vertical shift of a function is simply the constant
term in the function. In this case, the constant term is 0, so the vertical shift
of f(x)is 0.
Question 8
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: Rewrite the angle as a sum of two special angles.
5π
12 =π
3+π
4
Step 2: Use the sum-to-product identity for sine.
sin (π
3+π
4)= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Calculate the sine and cosine values of π
3and π
4.
sin (π
3)=√3
2,cos (π
3)=1
2,sin (π
4)=√2
2,cos (π
4)=√2
2
Step 4: Substitute the values into the formula.
sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)=√3
2·√2
2+1
2·√2
2
6
Step 5: Simplify the expression.
√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
Question 9
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for (f◦g)(x)and
simplify the result.
Solution
Step 1: Find (f◦g)(x)by substituting g(x) = cos(x)into f(x) = sin(2x).
(f◦g)(x) = f(g(x)) = f(cos(x)) = sin(2 ·cos(x))
Step 2: Use the double angle identity for sine, which states that sin(2θ) =
2 sin(θ) cos(θ), to simplify sin(2 ·cos(x)).
sin(2 ·cos(x)) = 2 sin(cos(x)) cos(cos(x))
Step 3: Further simplify 2 sin(cos(x)) cos(cos(x)) since sin(cos(x)) and cos(cos(x))
are not standard trigonometric functions.
No further simplification possible
Therefore, the expression for (f◦g)(x)is 2 sin(cos(x)) cos(cos(x)).
Question 10
Question
Find the exact value of sin (7π
12 ).
Solution
Step 1: Convert 7π
12 to an angle within the unit circle.
Step 2: Since 7π
12 =π
12 +3π
4, we can rewrite this as 15◦+ 135◦in degrees.
Step 3: Identify the reference angle in the first quadrant by subtracting π
from π
12 , which gives us π
12 −π=−11π
12 .
Step 4: Use the reference angle and quadrant information to determine the
sign of sin in the second quadrant. Since sin is positive in the second quadrant,
the value of sin(7π
12 )will also be positive.
7
Step 5: Now, calculate the sine of the reference angle 11π
12 in the second
quadrant: sin (11π
12 )= sin (π−π
12 )= sin (π
12 ).
Step 6: The exact value of sin (π
12 )can be found using the angle sum identity
for sine: sin (π
12 )= sin (π
6−π
3)= sin (π
6)cos (π
3)−cos (π
6)sin (π
3).
Step 7: Substitute the values of sine and cosine for π
6and π
3:sin (π
12 )=
1
2·√3
2−√3
2·1
2=√3−√3
4=0
4= 0.
Step 8: Therefore, the exact value of sin (7π
12 )is 0.
Question 11
Question
Let f(x) = sin2(3x) + cos(3x). Find the maximum and minimum values of f(x)
for x∈[0,π
6].
Solution
Step 1: First, let’s find the critical points of f(x)in the interval [0,π
6]by taking
the derivative and solving for xwhen f′(x) = 0.
Step 2: Find f′(x):
f′(x) = 2 sin(3x) cos(3x)−3 sin(3x)
Step 3: Set f′(x) = 0 and solve for x:
2 sin(3x) cos(3x)−3 sin(3x) = 0
sin(3x)(2 cos(3x)−3) = 0
Now, we find the critical points by solving:
sin(3x) = 0 or 2 cos(3x)−3 = 0
Step 4: Solve sin(3x) = 0 for xin [0,π
6]:
sin(3x) = 0 =⇒3x= 0, π, 2π, . . .
Since x∈[0, π/6], the solution is x= 0.
Step 5: Solve 2 cos(3x)−3 = 0 for xin [0,π
6]:
2 cos(3x)−3 = 0 =⇒cos(3x) = 3
2
There are no solutions in the interval [0,π
6].
Step 6: Evaluate f(x)at the critical point and endpoints to find the maxi-
mum and minimum values:
f(0) = sin2(0) + cos(0) = 0 + 1 = 1
f(π
6)= sin2(π
2) + cos(π
2)= 1 + 0 = 1
Step 7: Therefore, the maximum value of f(x)in the interval [0,π
6]is 1, and
the minimum value is also 1.
8
Question 12
Question
Find the exact value of sin (11π
12 ).
Solution
Step 1: Recognize that 11π
12 is not one of the common angles on the unit circle,
so we need to express it in terms of more common angles.
Step 2: We can rewrite 11π
12 as 3π
4+π
6. This is because 11π
12 =9π
12 +2π
12 =3π
4+π
6.
Step 3: Now, since we know that 3π
4and π
6are common angles, we can use
the sum-to-product identities to find the exact value of sin (11π
12 ).
Step 4: The sum-to-product identity for sine is: sin(A+B) = sin Acos B+
cos Asin B.
Step 5: Applying the sum-to-product identity, we have:
sin (11π
12 )= sin (3π
4+π
6)
= sin (3π
4)cos (π
6)+ cos (3π
4)sin (π
6)
Step 6: Using the unit circle, we find that sin (3π
4)=−√2
2,cos (π
6)=√3
2,
cos (3π
4)=−√2
2, and sin (π
6)=1
2.
Step 7: Substituting these values into our expression, we get:
sin (11π
12 )=−√2
2·√3
2−√2
2·1
2
=−√6
4−√2
4
=−√2 + √6
4
Therefore, the exact value of sin (11π
12 )is −√2+√6
4.
Question 13
Question
Solve the equation sin(3x) = cos(2x)for 0≤x≤2π.
9
Solution
Step 1: Recall the trigonometric identity sin(3x) = sin(π−3x).
Step 2: Rewrite the equation as sin(π−3x) = cos(2x).
Step 3: Since sin(π−θ) = sin(θ)and cos(π−θ) = −cos(θ), we have
sin(3x) = −cos(3x) = cos(2x).
Step 4: Set up the equation cos(2x) = −cos(3x).
Step 5: Use the double angle formula for cosine: cos(2x) = 2 cos2(x)−1.
Step 6: Substitute this into the equation: 2 cos2(x)−1 = −cos(3x).
Step 7: Use the triple angle formula for cosine: cos(3x) = 4 cos3(x)−3 cos(x).
Step 8: Substitute this into the equation: 2 cos2(x)−1 = −4 cos3(x) +
3 cos(x).
Step 9: Rearrange the equation to get 4 cos3(x)+2 cos2(x)+3 cos(x)−1 = 0.
Step 10: The equation is now a cubic equation in terms of cos(x). Solving
it may require the use of numerical methods like Newton’s method.
Question 14
Question
Determine the exact value of cos (11π
6).
Solution
Step 1: Recall the unit circle where cos (11π
6)is located. The angle 11π
6is in the
fourth quadrant, corresponding to the point (−√3
2,1
2)on the unit circle.
Step 2: Since cos (11π
6)corresponds to the x-coordinate of the point on the
unit circle, cos (11π
6)=−√3
2.
Therefore, the exact value of cos (11π
6)is −√3
2.
Question 15
Question
Solve the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤x≤2π.
Solution
Step 1: Let u= cos(x). The equation becomes 2u2+u−1 = 0.
Step 2: Now, we need to solve this quadratic equation for u. We can factor
it as (2u−1)(u+ 1) = 0.
Step 3: Set each factor equal to zero:
{2u−1 = 0
u+ 1 = 0
10
Step 4: Solve for uin each case:
{2u= 1
u=1
2
or {u=−1
Step 5: Remember, u= cos(x), so:
{cos(x) = 1
2
cos(x) = −1
Step 6: Solve each equation for x:
{x=π
3or x=5π
3
x=π
Therefore, the solutions to the equation 2 cos2(x) + cos(x)−1 = 0 for 0≤
x≤2πare x=π
3, π, 5π
3.
Question 16
Question
Let f(x) = 2 sin(x)−√3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Finding the amplitude The general form of a trigonometric function
is f(x) = Asin(Bx −C) + Dcos(Ex −F). The amplitude of f(x)is given by
A=√A2+D2. In this case, A= 2 and D=−√3. Therefore, the amplitude is
amplitude =√22+ (−√3)2=√4 + 3 = √7.
Step 2: Finding the period The period of a general trigonometric function
is given by 2π/B or 2π/E. In this case, the period of f(x)is
period =2π
1= 2π.
Step 3: Finding the phase shift To find the phase shift, we need to solve the
equations Bx −C= 0 and Ex −F= 0. In this case, the phase shift for f(x)is
phase shift =C
B=−π/6
1=−π
6.
Step 4: Finding the vertical shift The vertical shift of a trigonometric func-
tion is the value of D in the general form f(x) = Asin(Bx −C)+Dcos(Ex −F).
In this case, the vertical shift is
vertical shift =D=−√3.
11
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
−π
6, and the vertical shift is −√3.
Question 17
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double-angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: Substitute cos(2x) = 1 −2 sin2(x)into the equation sin(x) = cos(2x)
to obtain sin(x) = 1 −2 sin2(x).
Step 3: Rearrange the equation to get 2 sin2(x) + sin(x)−1 = 0.
Step 4: Factor the quadratic equation to get (2 sin(x)−1)(sin(x) + 1) = 0.
Step 5: Set each factor to zero to find the possible solutions.
• For 2 sin(x)−1 = 0, we have sin(x) = 1
2, which gives x=π
6,5π
6in the
interval [0,2π].
• For sin(x)+1 = 0, we have sin(x) = −1, which gives x=3π
2in the interval
[0,2π].
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
6,5π
6,3π
2.
Question 18
Question
Solve the equation cos(2x) = √2
2for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute cos(2x) = √2
2into the double angle identity:
2 cos2(x)−1 = √2
2
Step 3: Add 1 to both sides:
2 cos2(x) = √2
2+ 1
12
Step 4: Simplify the right side:
2 cos2(x) = √2+2
2
Step 5: Divide by 2 to isolate cos2(x):
cos2(x) = √2+2
4
Step 6: Take the square root of both sides:
cos(x) = ±√√2+2
4
Step 7: We know that cos(x) = √2
2corresponds to the angle x=π
4in the
unit circle.
Step 8: To find the other solution, we can consider the symmetry of the
cosine function. The cosine function is positive in the first and fourth quadrants.
Hence, the other solution is x=7π
4.
Therefore, the solutions to the equation cos(2x) = √2
2for 0≤x≤2πare
x=π
4and x=7π
4.
Question 19
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: We will use the angle addition identity tan(A+B) = tan A+tan B
1−tan Atan B.
Step 2: Let A=π
3and B=π
4. Then, 5π
12 =π
3+π
4.
Step 3: We know that tan(π/3) = √3and tan(π/4) = 1.
Step 4: Substitute these values into the angle addition formula:
tan (π
3+π
4)=√3+1
1−√3·1=√3+1
1−√3
Step 5: To rationalize the denominator, multiply by the conjugate of the
denominator:
=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3)
Step 6: Simplify the expression:
=√3 + √3+1+3
1−3=−2(√3 + 1) = −2√3−2
Step 7: Therefore, tan (5π
12 =−2√3−2.
13
Question 20
Question
Let f(x) = 2 sin(x) + √3 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Amplitude
The amplitude of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by √A2+D2. In this case, A= 2 and D=√3, so the amplitude is
√22+ (√3)2=√4 + 3 = √7.
Step 2: Period
For a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C), the period
is given by 2π
|B|. In this case, since B= 1 (implicitly, since sin(x)and cos(x)
both have a coefficient of 1), the period is
2π
1= 2π.
Step 3: Phase Shift
The phase shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is given by C
B. Here, C= 0 (since there is no xterm inside the sine or cosine
functions), so the phase shift is 0
1= 0.
Step 4: Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx −C)+Dcos(Bx −C)
is simply D. In this case, the vertical shift is
D=√3.
Therefore, the amplitude of f(x)is √7, the period is 2π, the phase shift is
0, and the vertical shift is √3.
Question 21
Question
Find all solutions to the equation cos(2x) = sin(x)for 0≤x≤2π.
14
Solution
Step 1: We will use the double-angle formula for cosine to rewrite cos(2x)in
terms of cos(x)and sin(x).
cos(2x) = 2 cos2(x)−1
Step 2: Substitute the expression for cos(2x)into the original equation.
2 cos2(x)−1 = sin(x)
Step 3: Recall that cos2(x) = 1 −sin2(x). Substitute this into the equation.
2(1 −sin2(x)) −1 = sin(x)
Step 4: Simplify the equation.
2−2 sin2(x)−1 = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 5: We can treat this as a quadratic equation in terms of sin(x). Let
sin(x) = yto simplify the equation.
2y2+y−1 = 0
Step 6: Solve the quadratic equation for y.
y=−1±√12−4(2)(−1)
2(2)
y=−1±√9
4
Step 7: Solve for y:
y1=−1+3
4=2
4=1
2
y2=−1−3
4=−4
4=−1
Step 8: Recall that sin(x) = y. So, we have sin(x) = 1
2and sin(x) = −1.
Solve for xin the interval 0≤x≤2π.
For sin(x) = 1
2,x=π
6,5π
6
For sin(x) = −1,x=3π
2
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
6,5π
6,3π
2.
Question 22
Question
Solve the equation sin2(x) + cos2(x) = 5
4for 0≤x≤2π.
15
Solution
Step 1: Recall the Pythagorean identity: sin2(x) + cos2(x)=1. Therefore, we
have
sin2(x) + cos2(x) = 1 = 4
4.
Step 2: Subtracting 4
4from both sides of the given equation, we get
5
4−4
4= 1.
Step 3: Simplifying the left side of the equation gives us
1
4= 1.
Step 4: Since 1
4= 1, there are no solutions to the given equation sin2(x) +
cos2(x) = 5
4for 0≤x≤2π.
Question 23
Question
Find all solutions to the equation sin(x) = cos(2x)in the interval [0,2π].
Solution
Step 1: Recall the double angle formula for cosine: cos(2x)=1−2 sin2(x).
Therefore, we can rewrite the equation as sin(x) = 1 −2 sin2(x).
Step 2: Rearranging the equation, we get 2 sin2(x) + sin(x)−1 = 0. Let
u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 3: Solve the quadratic equation 2u2+u−1=0by factoring or us-
ing the quadratic formula. The solutions are u=−1±√12−4(2)(−1)
2(2) =−1±√9
4.
Therefore, u=−1+3
4= 1 or u=−1−3
4=−1/2.
Step 4: Since u= sin(x), the solutions are sin(x) = 1 and sin(x) = −1/2.
Step 5: For sin(x) = 1,x=π
2. For sin(x) = −1/2,x=7π
6.
Step 6: Therefore, the solutions to the equation sin(x) = cos(2x)in the
interval [0,2π]are x=π
2and x=7π
6.
Question 24
Question
Find all solutions to the equation cos(3x) = −1
2for 0◦≤x < 360◦.
16
Solution
Step 1: Recall the values of θfor which cos(θ) = −1
2. We have θ= 120◦and
θ= 240◦.
Step 2: Now, we need to solve the equation cos(3x) = −1
2. Since cos(3x) =
cos(2x+x), we can rewrite the equation as cos(2x+x) = −1
2.
Step 3: Using the sum formula for cosine, we have cos(2x) cos(x)−sin(2x) sin(x) =
−1
2.
Step 4: Recall the double angle formulas for cosine and sine: cos(2θ) =
2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ).
Step 5: Substitute these formulas into the equation from Step 3 to get
[2 cos2(x)−1] cos(x)−[2 sin(x) cos(x)] sin(x) = −1
2.
Step 6: Simplify the equation to obtain 2 cos3(x)−cos(x)−2 sin2(x) cos(x) =
−1
2.
Step 7: Recall the Pythagorean identity sin2(θ) + cos2(θ) = 1.
Step 8: Substitute sin2(x) = 1 −cos2(x)into the equation from Step 6 to
get 2 cos3(x)−cos(x)−2(1 −cos2(x)) cos(x) = −1
2.
Step 9: Simplify and solve the resulting cubic equation 2 cos3(x)−cos(x)−
2 cos(x) + 2 cos3(x) = −1
2.
Step 10: Combining like terms gives 4 cos3(x)−3 cos(x) + 1
2= 0.
Step 11: Use the solutions for cos(θ) = −1
2to solve the cubic equation in
Step 10: x= 120◦and x= 240◦.
Step 12: Therefore, the solutions to the equation cos(3x) = −1
2for 0◦≤x <
360◦are x= 40◦,80◦,120◦,160◦,200◦,240◦,280◦,320◦.
Question 25
Question
Find all solutions to the equation sin(2x) = cos(x)in the interval [0,2π].
Solution
Step 1: Recall the double angle identity for sine and the Pythagorean identity
for cosine.
sin(2x) = 2 sin(x) cos(x)and cos2(x) + sin2(x) = 1
Step 2: Substitute the double angle identity into the equation sin(2x) =
cos(x).
2 sin(x) cos(x) = cos(x)
Step 3: Rearrange the equation to get cos(x)terms on one side.
2 sin(x) cos(x)−cos(x) = 0
17
Step 4: Factor out a cos(x).
cos(x)(2 sin(x)−1) = 0
Step 5: Set each factor equal to zero and solve for x. For cos(x) = 0:
cos(x) = 0 ⇒x=π
2,3π
2
Step 6: For 2 sin(x)−1 = 0:
sin(x) = 1
2⇒x=π
6,5π
6
Step 7: Combine all solutions in the interval [0,2π].
x=π
6,π
2,5π
6,3π
2
18
