MATH 122 - TRIGONOMETRY -
Trigonometric Functions
Question Bank - Set 3
Liberty University
Question 1
Question
Given that sin(θ) = 3
5and θis in quadrant II, calculate the values of cos(θ),
tan(θ),sec(θ),csc(θ), and cot(θ).
Solution
Step 1: To find cos(θ), we can use the Pythagorean identity sin2(θ)+cos2(θ) = 1.
Since sin(θ) = 3
5, we have:
cos2(θ) = 1 −sin2(θ) = 1 −(3
5)2
= 1 −9
25 =16
25.
Taking the square root of both sides, we get cos(θ) = ±4
5. Since θis in quadrant
II, where cos(θ)<0, we have cos(θ) = −4
5.
Step 2: To find tan(θ), we can use the definition of tan(θ) = sin(θ)
cos(θ). Substi-
tuting in the given values, we have:
tan(θ) =
3
5
−4
5
=−3
4.
Step 3: To find sec(θ), we can use the reciprocal identity sec(θ) = 1
cos(θ).
Substituting in the value of cos(θ), we get:
sec(θ) = 1
−4
5
=−5
4.
Step 4: To find csc(θ), we can use the reciprocal identity csc(θ) = 1
sin(θ).
Substituting in the value of sin(θ), we have:
csc(θ) = 1
3
5
=5
3.
Step 5: To find cot(θ), we can use the definition of cot(θ) = 1
tan(θ). Substi-
tuting in the value of tan(θ), we get:
cot(θ) = 1
−3
4
=−4
3.
Therefore, the values of the trigonometric functions for θin quadrant II are:
cos(θ) = −4
5,tan(θ) = −3
4,sec(θ) = −5
4,csc(θ) = 5
3,and cot(θ) = −4
3.
Question 2
Question
If sin(θ) = −3
5and θis in Quadrant III, determine the values of the remaining
five trigonometric functions of θ.
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant III, we can determine the value
of cos(θ)using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
cos(θ) = −√16
25 =−4
5
Step 2: Now, we can determine the values of the other trigonometric func-
tions using the definitions:
tan(θ) = sin(θ)
cos(θ)=−3
5
−4
5
=3
4
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
cot(θ) = 1
tan(θ)=1
3
4
=4
3
Therefore, the remaining trigonometric functions of θare:
tan(θ) = 3
4,csc(θ) = −5
3,sec(θ) = −5
4,and cot(θ) = 4
3
2
Question 3
Question
Let f(x) = sin(3x)and g(x) = cos(2x), where xis in radians. Find the exact
value of f(π
6)·g(π
3).
Solution
Step 1: Evaluate f(π
6)= sin (3·π
6).
f(π
6)= sin (π
2)= 1
Step 2: Evaluate g(π
3)= cos (2·π
3).
g(π
3)= cos (2π
3)=−1
2
Step 3: Calculate the product f(π
6)·g(π
3).
f(π
6)·g(π
3)= 1 ·(−1
2)
=−1
2
Therefore, f(π
6)·g(π
3)=−1
2.
Question 5
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation as a quadratic equation in terms of sin(x).
sin2(x)−3 sin(x) + 2 = 0
Step 2: To simplify, let y= sin(x). The equation becomes:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
3
Step 4: Solve for y:
y= 1 or y= 2
Step 5: Since y= sin(x), we have:
sin(x) = 1 or sin(x) = 2
Step 6: Since the values of sin(x)are restricted to the interval [−1,1], the
equation sin(x) = 2 has no solutions within the given interval.
Step 7: Therefore, we only have to solve sin(x) = 1.
Step 8: The solution to sin(x) = 1 is x=π
2+ 2πn, where nis an integer.
Step 9: Hence, the solution to the equation sin2(x)−3 sin(x) + 2 = 0 for
0≤x≤2πis x=π
2.
Question 6
Question
Let f(x) = sin(x)−2 cos(x). Find the maximum and minimum values of f(x).
Solution
Step 1: We can rewrite f(x)as a single trigonometric function using the angle
addition formula for sine: sin(α−β) = sin(α) cos(β)−cos(α) sin(β).
f(x) = sin(x)−2 cos(x) = sin(x)−2 cos(x) + sin(90◦) cos(x)−cos(90◦) sin(x)
Step 2: Simplify f(x)using the angle addition formula.
f(x) = sin(x)−2 cos(x) + sin(x) cos(90◦)−cos(x) sin(90◦)
f(x) = sin(x)−2 cos(x) + sin(x)·0−cos(x)·1
f(x) = sin(x) + 2 sin(x)
f(x) = sin(x)(1 + 2)
f(x) = 3 sin(x)
Step 3: The function f(x) = 3 sin(x)has a maximum value of 3 when
sin(x)=1, and a minimum value of -3 when sin(x) = −1. Since the range
of sin(x)is [−1,1], the maximum value of f(x)is 3 and the minimum value is
-3.
Question 7
Question
Find all solutions to the equation cos(2θ) = sin(θ)for 0≤θ≤2π.
4
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Substituting this into the given equation, we have 2 cos2(θ)−1 =
sin(θ).
Step 3: Recall that sin(θ) = cos (π
2−θ).
Step 4: Substituting this identity into the equation, we get 2 cos2(θ)−1 =
cos (π
2−θ).
Step 5: Now, using the double angle identity for cosine again, we get
2 cos2(θ)−1 = 2 sin (π
2−θ)cos (π
2−θ).
Step 6: Simplifying the right side gives 2 cos2(θ)−1 = 2 cos(θ) sin(θ).
Step 7: We can rewrite this as 2 cos2(θ)−2 sin(θ) cos(θ)−1 = 0.
Step 8: Factoring the left side gives (2 cos(θ) + 1)(cos(θ)−1) = 0.
Step 9: Setting each factor to zero gives cos(θ) = −1and cos(θ) = 1.
Step 10: Solving cos(θ) = −1gives us θ=π.
Step 11: Solving cos(θ) = 1 gives us θ= 0 and θ= 2π.
Step 12: Therefore, the solutions to the equation cos(2θ) = sin(θ)for 0≤
θ≤2πare θ= 0, π, 2π.
Question 8
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan θ= cos θ−sin θ
Solution
1. Start with the left-hand side (LHS) of the equation:
1−tan2θ
1 + tan θ
2. Substitute tan θ=sin θ
cos θinto the expression:
1−(sin θ
cos θ)2
1 + sin θ
cos θ
3. Simplify the expression by squaring the term in the numerator:
1−sin2θ
cos2θ
1 + sin θ
cos θ
cos2θ−sin2θ
cos θ+ sin θ
5
4. Factor the numerator using the difference of squares formula a2−b2=
(a+b)(a−b):
(cos θ+ sin θ)(cos θ−sin θ)
cos θ+ sin θ
5. Cancel out the common term of cos θ+ sin θ:
cos θ−sin θ
6. Therefore, the left-hand side (LHS) 1−tan2θ
1+tan θsimplifies to cos θ−sin θ,
which is equal to the right-hand side (RHS) of the given identity. Hence,
the given trigonometric identity is proven.
Question 9
Question
Let f(x) = 3 sin(2x)−2 cos(2x). 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 asin(bx)+ccos(bx)
is given by √a2+c2. In this case, a= 3 and c=−2, so the amplitude is
√32+ (−2)2=√9 + 4 = √13.
Step 2: Period The period of a function of the form sin(bx)or cos(bx)is
2π
|b|. Since b= 2 in this case, the period of f(x)is
2π
|2|=π.
Step 3: Phase Shift The phase shift of a function of the form sin(bx)or
cos(bx)is given by −c
b. For the function f(x) = 3 sin(2x)−2 cos(2x), the phase
shift is −(−2)
2= 1.
Step 4: Vertical Shift The vertical shift of a function of the form asin(bx)+
ccos(bx)is simply the value of c. In this case, f(x)has a vertical shift of −2.
Question 10
Question
Find the general solution to the equation cos(2x) = sin(x).
6
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 1 −2 sin2(x).
Step 2: Substitute this identity into the equation cos(2x) = sin(x):
1−2 sin2(x) = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 3: This is a quadratic equation in terms of sin(x). Let u= sin(x)to
make it easier to solve:
2u2+u−1 = 0
Step 4: Now, solve for uusing the quadratic formula:
u=−b±√b2−4ac
2a
u=−1±√1+8
4
u=−1±√9
4
u=−1±3
4
u=2
4or u=−4
4
u=1
2or u=−1
Step 5: Remember that u= sin(x), so sin(x) = 1
2or sin(x) = −1.
Step 6: Determine the general solutions for sin(x) = 1
2and sin(x) = −1,
respectively, using your knowledge of the unit circle.
Step 7: For sin(x) = 1
2, the solutions are x=π
6+ 2πn and x=5π
6+ 2πn,
where n∈Z.
Step 8: For sin(x) = −1, the solution is x=−π
2+ 2πn, where n∈Z.
Step 9: Therefore, the general solution to the equation cos(2x) = sin(x)is
x=π
6+ 2πn,x=5π
6+ 2πn, and x=−π
2+ 2πn, where n∈Z.
Question 11
Question
Find the exact values of sin (π
12 )and cos (5π
12 )using trigonometric identities.
7
Solution
Step 1: We can use the angle addition identity for sine to find sin (π
12 ):
sin (π
12)= sin (π
6−π
4)
Step 2: Applying the angle subtraction formula for sine gives:
sin (π
12)= sin (π
6)cos (π
4)−cos (π
6)sin (π
4)
Step 3: Simplifying using known values: sin (π
6)=1
2,cos (π
6)=√3
2,sin (π
4)=
√2
2, and cos (π
4)=√2
2, we get:
sin (π
12)=1
2·√2
2−√3
2·√2
2
sin (π
12)=√2−√6
4
Step 4: Next, we use the angle addition identity for cosine to find cos (5π
12 ):
cos (5π
12 )= cos (π
3+π
4)
Step 5: Applying the angle addition formula for cosine gives:
cos (5π
12 )= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
Step 6: Using the values cos (π
3)=1
2,sin (π
3)=√3
2,cos (π
4)=√2
2, and
sin (π
4)=√2
2, we have:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
cos (5π
12 )=√2−√6
4
Question 12
Question
Solve the equation cos2x−2 cos x= 0 for xin the interval [0,2π).
8
Solution
Step 1: Let’s rewrite the equation in terms of cos x:
cos2x−2 cos x= 0
Step 2: Factor out cos x:
cos x(cos x−2) = 0
Step 3: Set each factor to zero:
cos x= 0
cos x−2 = 0
Step 4: Solve the first equation: Since cos x= 0 when x=π
2and 3π
2in the
interval [0,2π).
Step 5: Solve the second equation:
cos x−2 = 0
cos x= 2 (not possible as cos xranges from -1 to 1)
Step 6: Therefore, the solutions to the equation cos2x−2 cos x= 0 in the
interval [0,2π)are x=π
2and x=3π
2.
Question 13
Question
Prove the trigonometric identity:
cot(θ) = cos(θ)
sin(θ)
Solution
Step 1: Recall that cot(θ) = 1
tan(θ)and tan(θ) = sin(θ)
cos(θ). Therefore, cot(θ) =
1
sin(θ)
cos(θ)
=cos(θ)
sin(θ).
Step 2: Thus, we have shown that cot(θ) = cos(θ)
sin(θ).
Question 14
Question
Solve for xin the equation sin(x)−cos(x) = 1
2for 0◦≤x≤360◦.
9
Solution
Step 1: We can rewrite sin(x)−cos(x) = 1
2as sin(x) = cos(x) + 1
2.
Step 2: Squaring both sides of the equation, we get sin2(x) = (cos(x) + 1
2)2.
Step 3: Using the Pythagorean identity sin2(x) + cos2(x) = 1, we substitute
cos2(x) = 1−sin2(x)into the equation from step 2 to get sin2(x) = (cos(x)+ 1
2)2.
Step 4: Expanding the right side of the equation and using the Pythagorean
identity again, we have sin2(x) = cos2(x) + cos(x) + 1
4.
Step 5: Substituting 1−sin2(x)for cos2(x), we get sin2(x)=1−sin2(x) +
cos(x) + 1
4.
Step 6: Rearranging terms, we have 2 sin2(x) + cos(x) = 3
4.
Step 7: Since cos2(x) + sin2(x) = 1, we can substitute cos2(x) = 1 −sin2(x)
into the equation from step 2 to get 2 sin2(x) + √1−sin2(x) = 3
4.
Step 8: Let u= sin(x), which gives us 2u2+√1−u2=3
4.
Step 9: Solving the quadratic equation 2u2+√1−u2=3
4will give us the
values of u, from which we can determine the values of xwithin the given range.
Question 15
Question
Solve the equation sin(2x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Use the double angle formula for sine to rewrite sin(2x).
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the double angle formula back into the original equation.
2 sin(x) cos(x) + cos(x) = 0
Step 3: Factor out a cos(x)term.
cos(x)(2 sin(x) + 1) = 0
Step 4: Use the zero-product property to set each factor equal to zero.
cos(x) = 0 or 2 sin(x) + 1 = 0
Step 5: Solve cos(x) = 0. Since cos(x) = 0 when x=π
2,3π
2.
x=π
2,3π
2
Step 6: Solve 2 sin(x) + 1 = 0.
2 sin(x) = −1
10
sin(x) = −1
2
Step 7: Find the solutions for sin(x) = −1
2within the given interval 0≤
x≤2π. The solutions occur in the third and fourth quadrants, corresponding
to x=7π
6,11π
6.
Step 8: So, the solutions to the equation sin(2x) + cos(x) = 0 for 0≤x≤2π
are x=π
2,3π
2,7π
6,11π
6.
Question 16
Question
Let f(x) = 3 cos(2x) + 2 sin(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of the function can be found by using the formula A=
√a2+b2, where ais the coefficient of cos(x)and bis the coefficient of sin(x).
In this case, we have a= 3 and b= 2. Therefore, the amplitude is
A=√32+ 22=√13
.
Step 2: The period of the function can be found using the formula T=2π
|b|,
where |b|is the absolute value of the coefficient of x. In this case, |b|= 2.
Therefore, the period is
T=2π
2=π.
Step 3: The phase shift of the function can be found by setting the argument
of the cosine function equal to zero and solving for x. In this case, the argument
of cosine function is 2x. Setting 2x= 0, we find
x= 0
. Therefore, the phase shift is 0.
Step 4: The vertical shift of the function can be found by determining the
vertical translation added to the function. In this case, there is no vertical
translation added, so the vertical shift is 0.
In conclusion, the amplitude of the function is √13, the period is π, the
phase shift is 0, and the vertical shift is 0.
Question 17
Question
Let f(x) = 2 sin(3x). Find the amplitude, period, phase shift, and vertical shift
of the function f(x).
11
Solution
Step 1: Amplitude
The amplitude of a function in the form f(x) = Asin(Bx)is the absolute
value of A. In this case, A= 2, so the amplitude is |2|= 2.
Step 2: Period
The period of a function in the form f(x) = sin(Bx)is given by 2π
B. In this
case, B= 3, so the period is 2π
3.
Step 3: Phase Shift
To find the phase shift of a function in the form f(x) = sin(B(x−C)), we
set B(x−C) = 0 and solve for C. Here, 3x= 0 gives x= 0. Since the coefficient
of xin the argument of sine is positive, the phase shift is to the right. Therefore,
the phase shift is 0.
Step 4: Vertical Shift
The vertical shift of a function in the form f(x) = Asin(Bx)+Dis the value
of D. Here, there is no term added or subtracted outside the sine function, so
the vertical shift is 0.
Therefore, for the given function f(x) = 2 sin(3x), the amplitude is 2, the
period is 2π
3, the phase shift is 0, and the vertical shift is 0.
Question 18
Question
Solve the equation sin2(x) + sin(x)−6 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
sin2(x) + sin(x)−6 = 0
Step 2: Factor the quadratic equation.
To factor this quadratic equation, we need to find two numbers that multiply
to −6and add up to 1. These numbers are 3and −2.
sin2(x) + 3 sin(x)−2 sin(x)−6 = 0
sin(x)(sin(x) + 3) −2(sin(x) + 3) = 0
(sin(x)−2)(sin(x) + 3) = 0
Step 3: Set each factor to zero and solve for sin(x).
sin(x)−2 = 0 or sin(x) + 3 = 0
sin(x) = 2 or sin(x) = −3
12
Step 4: Since sin(x)must be between −1and 1, the equation sin(x) = 2 has
no solutions. So, we focus on solving sin(x) = −3.
Since sin(x) = −3has no solutions, the original trigonometric equation
sin2(x) + sin(x)−6=0does not have any solutions within the given domain
0◦≤x≤360◦.
Question 19
Question
Solve the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: We can rewrite the given equation using the double angle identity:
1−2 sin2(x) = sin(x).
Step 3: Rearrange the equation by bringing all terms to one side to get
2 sin2(x) + sin(x)−1 = 0.
Step 4: Let u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 5: Solve the quadratic equation by factoring or using the quadratic
formula: u=−b±√b2−4ac
2awhere a= 2,b= 1, and c=−1.
Step 6: Solving for ugives u=−1±√1+8
4, which simplifies to u=−1±√9
4,
leading to u=−1±3
4.
Step 7: Therefore, we find two solutions for u:u1=−1+3
4=2
4=1
2and
u2=−1−3
4=−4
4=−1.
Step 8: Since u= sin(x), we have sin(x) = 1
2and sin(x) = −1.
Step 9: Solve sin(x) = 1
2for 0◦≤x≤360◦to find x= 30◦and x= 150◦.
Step 10: Solve sin(x) = −1for 0◦≤x≤360◦to find x= 270◦.
Step 11: Therefore, the solutions to the equation cos(2x) = sin(x)for 0◦≤
x≤360◦are x= 30◦,x= 150◦, and x= 270◦.
Question 20
Question
Find all solutions to the equation cos(2x) = −1
2for 0≤x≤2π.
Solution
Step 1: Let’s start by using the double angle identity for cosine:
cos(2x) = 2 cos2(x)−1
13
Step 2: Substitute cos(2x) = −1
2into the identity:
2 cos2(x)−1 = −1
2
Step 3: Simplify the equation:
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Now, we need to find the values of xthat satisfy cos(x) = ±1
2: For
cos(x) = 1
2, we have x=π
3,5π
3. For cos(x) = −1
2, we have x=2π
3,4π
3.
Step 5: Therefore, the solutions to the given equation in the interval 0≤
x≤2πare x=π
3,2π
3,4π
3,5π
3.
Question 21
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: Express 5π
12 as the sum of two common angles.
5π
12 =π
3+π
4
Step 2: Use the angle addition identity for tangent.
tan (5π
12 )= tan (π
3+π
4)= tan (π
3
+tan (π
4)1−tan(π
3)tan(π
4)
Step 3: Find tan (π
3and tan (π
4.tan (π
3)=√3and tan (π
4)= 1
Step 4: Substitute the values into the formula for tan (5π
12 .tan (5π
12 )=√3+1
1−√3
Step 5: Rationalize the denominator to simplify.
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3) =4+2√3
1−3=4+2√3
−2=−2−√3
Therefore, tan (5π
12 )=−2−√3.
14
Question 22
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for h(x) = sin(2x) cos(x)
in terms of basic trigonometric functions.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the expression for h(x):
h(x) = sin(2x) cos(x) = (2 sin(x) cos(x)) cos(x)
Step 3: Simplify the expression by distributing the cos(x):
h(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos(x)2
Step 4: Recall the Pythagorean identity sin2(x) + cos2(x) = 1 which can be
rearranged to solve for sin2(x)as sin2(x) = 1 −cos2(x).
Step 5: Substitute sin2(x) = 1 −cos2(x)into the expression for h(x):
h(x) = 2 sin(x)(1 −cos2(x))
Step 6: Expand the expression by distributing the 2 sin(x):
h(x) = 2 sin(x)−2 sin(x) cos2(x)
Therefore, the expression for h(x) = sin(2x) cos(x)in terms of basic trigono-
metric functions is 2 sin(x)−2 sin(x) cos2(x).
Question 23
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x) + 5 cos(x) + 2 = 0
Solution
To solve the equation 2 cos2(x) + 5 cos(x) + 2 = 0, we can treat it as a quadratic
equation in terms of cos(x).
Step 1: Let’s substitute cos(x)with yto write the equation as a quadratic
equation in terms of y:2y2+ 5y+ 2 = 0.
Step 2: Now, we can factor the quadratic equation:
2y2+ 5y+ 2 = (2y+ 1)(y+ 2) = 0
15
Step 3: Set each factor to zero and solve for y:
2y+ 1 = 0 =⇒y=−1
2
y+ 2 = 0 =⇒y=−2
Step 4: Now, substitute back cos(x)for yand solve for x:
y= cos(x) = −1
2=⇒x=2π
3,4π
3
y= cos(x) = −2 (No real solutions)
Therefore, the solutions to the trigonometric equation 2 cos2(x) + 5 cos(x) +
2 = 0 in the interval [0,2π]are x=2π
3,4π
3.
Question 24
Question
Find the exact value of cos (5π
12 )using the sum or difference formula.
Solution
Step 1: Let’s express 5π
12 as the difference of two angles that we know the exact
cosine values of. In this case, let’s express it as π
3−π
4.
Step 2: Now, we use the difference formula for cosine:
cos(A−B) = cos Acos B+ sin Asin B
Step 3: Substitute π
3for Aand π
4for B:
cos (5π
12 )= cos (π
3−π
4)= cos (π
3)cos (π
4)+ sin (π
3)sin (π
4)
Step 4: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2. Substitute these values in:
cos (5π
12 )=1
2·√2
2+√3
2·√2
2
Step 5: Simplify the expression:
cos (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of cos (5π
12 )is √2+√6
4.
16
Question 25
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the given equation:
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: To find the solutions for xin the interval [0,2π], recall the values
of sin(x)in the first and second quadrants where sin(x) = 1
2. The solutions are
x=π
6and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)for 0≤x≤2πare
x=π
6and x=5π
6.
17
Step 4: To find csc(θ), we can use the reciprocal identity csc(θ) = 1
sin(θ).
Substituting in the value of sin(θ), we have:
csc(θ) = 1
3
5
=5
3.
Step 5: To find cot(θ), we can use the definition of cot(θ) = 1
tan(θ). Substi-
tuting in the value of tan(θ), we get:
cot(θ) = 1
−3
4
=−4
3.
Therefore, the values of the trigonometric functions for θin quadrant II are:
cos(θ) = −4
5,tan(θ) = −3
4,sec(θ) = −5
4,csc(θ) = 5
3,and cot(θ) = −4
3.
Question 2
Question
If sin(θ) = −3
5and θis in Quadrant III, determine the values of the remaining
five trigonometric functions of θ.
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant III, we can determine the value
of cos(θ)using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
cos(θ) = −√16
25 =−4
5
Step 2: Now, we can determine the values of the other trigonometric func-
tions using the definitions:
tan(θ) = sin(θ)
cos(θ)=−3
5
−4
5
=3
4
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
cot(θ) = 1
tan(θ)=1
3
4
=4
3
Therefore, the remaining trigonometric functions of θare:
tan(θ) = 3
4,csc(θ) = −5
3,sec(θ) = −5
4,and cot(θ) = 4
3
2
Question 3
Question
Let f(x) = sin(3x)and g(x) = cos(2x), where xis in radians. Find the exact
value of f(π
6)·g(π
3).
Solution
Step 1: Evaluate f(π
6)= sin (3·π
6).
f(π
6)= sin (π
2)= 1
Step 2: Evaluate g(π
3)= cos (2·π
3).
g(π
3)= cos (2π
3)=−1
2
Step 3: Calculate the product f(π
6)·g(π
3).
f(π
6)·g(π
3)= 1 ·(−1
2)
=−1
2
Therefore, f(π
6)·g(π
3)=−1
2.
Question 5
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation as a quadratic equation in terms of sin(x).
sin2(x)−3 sin(x) + 2 = 0
Step 2: To simplify, let y= sin(x). The equation becomes:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
3
Step 4: Solve for y:
y= 1 or y= 2
Step 5: Since y= sin(x), we have:
sin(x) = 1 or sin(x) = 2
Step 6: Since the values of sin(x)are restricted to the interval [−1,1], the
equation sin(x) = 2 has no solutions within the given interval.
Step 7: Therefore, we only have to solve sin(x) = 1.
Step 8: The solution to sin(x) = 1 is x=π
2+ 2πn, where nis an integer.
Step 9: Hence, the solution to the equation sin2(x)−3 sin(x) + 2 = 0 for
0≤x≤2πis x=π
2.
Question 6
Question
Let f(x) = sin(x)−2 cos(x). Find the maximum and minimum values of f(x).
Solution
Step 1: We can rewrite f(x)as a single trigonometric function using the angle
addition formula for sine: sin(α−β) = sin(α) cos(β)−cos(α) sin(β).
f(x) = sin(x)−2 cos(x) = sin(x)−2 cos(x) + sin(90◦) cos(x)−cos(90◦) sin(x)
Step 2: Simplify f(x)using the angle addition formula.
f(x) = sin(x)−2 cos(x) + sin(x) cos(90◦)−cos(x) sin(90◦)
f(x) = sin(x)−2 cos(x) + sin(x)·0−cos(x)·1
f(x) = sin(x) + 2 sin(x)
f(x) = sin(x)(1 + 2)
f(x) = 3 sin(x)
Step 3: The function f(x) = 3 sin(x)has a maximum value of 3 when
sin(x)=1, and a minimum value of -3 when sin(x) = −1. Since the range
of sin(x)is [−1,1], the maximum value of f(x)is 3 and the minimum value is
-3.
Question 7
Question
Find all solutions to the equation cos(2θ) = sin(θ)for 0≤θ≤2π.
4
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Substituting this into the given equation, we have 2 cos2(θ)−1 =
sin(θ).
Step 3: Recall that sin(θ) = cos (π
2−θ).
Step 4: Substituting this identity into the equation, we get 2 cos2(θ)−1 =
cos (π
2−θ).
Step 5: Now, using the double angle identity for cosine again, we get
2 cos2(θ)−1 = 2 sin (π
2−θ)cos (π
2−θ).
Step 6: Simplifying the right side gives 2 cos2(θ)−1 = 2 cos(θ) sin(θ).
Step 7: We can rewrite this as 2 cos2(θ)−2 sin(θ) cos(θ)−1 = 0.
Step 8: Factoring the left side gives (2 cos(θ) + 1)(cos(θ)−1) = 0.
Step 9: Setting each factor to zero gives cos(θ) = −1and cos(θ) = 1.
Step 10: Solving cos(θ) = −1gives us θ=π.
Step 11: Solving cos(θ) = 1 gives us θ= 0 and θ= 2π.
Step 12: Therefore, the solutions to the equation cos(2θ) = sin(θ)for 0≤
θ≤2πare θ= 0, π, 2π.
Question 8
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan θ= cos θ−sin θ
Solution
1. Start with the left-hand side (LHS) of the equation:
1−tan2θ
1 + tan θ
2. Substitute tan θ=sin θ
cos θinto the expression:
1−(sin θ
cos θ)2
1 + sin θ
cos θ
3. Simplify the expression by squaring the term in the numerator:
1−sin2θ
cos2θ
1 + sin θ
cos θ
cos2θ−sin2θ
cos θ+ sin θ
5
4. Factor the numerator using the difference of squares formula a2−b2=
(a+b)(a−b):
(cos θ+ sin θ)(cos θ−sin θ)
cos θ+ sin θ
5. Cancel out the common term of cos θ+ sin θ:
cos θ−sin θ
6. Therefore, the left-hand side (LHS) 1−tan2θ
1+tan θsimplifies to cos θ−sin θ,
which is equal to the right-hand side (RHS) of the given identity. Hence,
the given trigonometric identity is proven.
Question 9
Question
Let f(x) = 3 sin(2x)−2 cos(2x). 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 asin(bx)+ccos(bx)
is given by √a2+c2. In this case, a= 3 and c=−2, so the amplitude is
√32+ (−2)2=√9 + 4 = √13.
Step 2: Period The period of a function of the form sin(bx)or cos(bx)is
2π
|b|. Since b= 2 in this case, the period of f(x)is
2π
|2|=π.
Step 3: Phase Shift The phase shift of a function of the form sin(bx)or
cos(bx)is given by −c
b. For the function f(x) = 3 sin(2x)−2 cos(2x), the phase
shift is −(−2)
2= 1.
Step 4: Vertical Shift The vertical shift of a function of the form asin(bx)+
ccos(bx)is simply the value of c. In this case, f(x)has a vertical shift of −2.
Question 10
Question
Find the general solution to the equation cos(2x) = sin(x).
6
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 1 −2 sin2(x).
Step 2: Substitute this identity into the equation cos(2x) = sin(x):
1−2 sin2(x) = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 3: This is a quadratic equation in terms of sin(x). Let u= sin(x)to
make it easier to solve:
2u2+u−1 = 0
Step 4: Now, solve for uusing the quadratic formula:
u=−b±√b2−4ac
2a
u=−1±√1+8
4
u=−1±√9
4
u=−1±3
4
u=2
4or u=−4
4
u=1
2or u=−1
Step 5: Remember that u= sin(x), so sin(x) = 1
2or sin(x) = −1.
Step 6: Determine the general solutions for sin(x) = 1
2and sin(x) = −1,
respectively, using your knowledge of the unit circle.
Step 7: For sin(x) = 1
2, the solutions are x=π
6+ 2πn and x=5π
6+ 2πn,
where n∈Z.
Step 8: For sin(x) = −1, the solution is x=−π
2+ 2πn, where n∈Z.
Step 9: Therefore, the general solution to the equation cos(2x) = sin(x)is
x=π
6+ 2πn,x=5π
6+ 2πn, and x=−π
2+ 2πn, where n∈Z.
Question 11
Question
Find the exact values of sin (π
12 )and cos (5π
12 )using trigonometric identities.
7
Solution
Step 1: We can use the angle addition identity for sine to find sin (π
12 ):
sin (π
12)= sin (π
6−π
4)
Step 2: Applying the angle subtraction formula for sine gives:
sin (π
12)= sin (π
6)cos (π
4)−cos (π
6)sin (π
4)
Step 3: Simplifying using known values: sin (π
6)=1
2,cos (π
6)=√3
2,sin (π
4)=
√2
2, and cos (π
4)=√2
2, we get:
sin (π
12)=1
2·√2
2−√3
2·√2
2
sin (π
12)=√2−√6
4
Step 4: Next, we use the angle addition identity for cosine to find cos (5π
12 ):
cos (5π
12 )= cos (π
3+π
4)
Step 5: Applying the angle addition formula for cosine gives:
cos (5π
12 )= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
Step 6: Using the values cos (π
3)=1
2,sin (π
3)=√3
2,cos (π
4)=√2
2, and
sin (π
4)=√2
2, we have:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
cos (5π
12 )=√2−√6
4
Question 12
Question
Solve the equation cos2x−2 cos x= 0 for xin the interval [0,2π).
8
Solution
Step 1: Let’s rewrite the equation in terms of cos x:
cos2x−2 cos x= 0
Step 2: Factor out cos x:
cos x(cos x−2) = 0
Step 3: Set each factor to zero:
cos x= 0
cos x−2 = 0
Step 4: Solve the first equation: Since cos x= 0 when x=π
2and 3π
2in the
interval [0,2π).
Step 5: Solve the second equation:
cos x−2 = 0
cos x= 2 (not possible as cos xranges from -1 to 1)
Step 6: Therefore, the solutions to the equation cos2x−2 cos x= 0 in the
interval [0,2π)are x=π
2and x=3π
2.
Question 13
Question
Prove the trigonometric identity:
cot(θ) = cos(θ)
sin(θ)
Solution
Step 1: Recall that cot(θ) = 1
tan(θ)and tan(θ) = sin(θ)
cos(θ). Therefore, cot(θ) =
1
sin(θ)
cos(θ)
=cos(θ)
sin(θ).
Step 2: Thus, we have shown that cot(θ) = cos(θ)
sin(θ).
Question 14
Question
Solve for xin the equation sin(x)−cos(x) = 1
2for 0◦≤x≤360◦.
9
Solution
Step 1: We can rewrite sin(x)−cos(x) = 1
2as sin(x) = cos(x) + 1
2.
Step 2: Squaring both sides of the equation, we get sin2(x) = (cos(x) + 1
2)2.
Step 3: Using the Pythagorean identity sin2(x) + cos2(x) = 1, we substitute
cos2(x) = 1−sin2(x)into the equation from step 2 to get sin2(x) = (cos(x)+ 1
2)2.
Step 4: Expanding the right side of the equation and using the Pythagorean
identity again, we have sin2(x) = cos2(x) + cos(x) + 1
4.
Step 5: Substituting 1−sin2(x)for cos2(x), we get sin2(x)=1−sin2(x) +
cos(x) + 1
4.
Step 6: Rearranging terms, we have 2 sin2(x) + cos(x) = 3
4.
Step 7: Since cos2(x) + sin2(x) = 1, we can substitute cos2(x) = 1 −sin2(x)
into the equation from step 2 to get 2 sin2(x) + √1−sin2(x) = 3
4.
Step 8: Let u= sin(x), which gives us 2u2+√1−u2=3
4.
Step 9: Solving the quadratic equation 2u2+√1−u2=3
4will give us the
values of u, from which we can determine the values of xwithin the given range.
Question 15
Question
Solve the equation sin(2x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Use the double angle formula for sine to rewrite sin(2x).
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the double angle formula back into the original equation.
2 sin(x) cos(x) + cos(x) = 0
Step 3: Factor out a cos(x)term.
cos(x)(2 sin(x) + 1) = 0
Step 4: Use the zero-product property to set each factor equal to zero.
cos(x) = 0 or 2 sin(x) + 1 = 0
Step 5: Solve cos(x) = 0. Since cos(x) = 0 when x=π
2,3π
2.
x=π
2,3π
2
Step 6: Solve 2 sin(x) + 1 = 0.
2 sin(x) = −1
10
sin(x) = −1
2
Step 7: Find the solutions for sin(x) = −1
2within the given interval 0≤
x≤2π. The solutions occur in the third and fourth quadrants, corresponding
to x=7π
6,11π
6.
Step 8: So, the solutions to the equation sin(2x) + cos(x) = 0 for 0≤x≤2π
are x=π
2,3π
2,7π
6,11π
6.
Question 16
Question
Let f(x) = 3 cos(2x) + 2 sin(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of the function can be found by using the formula A=
√a2+b2, where ais the coefficient of cos(x)and bis the coefficient of sin(x).
In this case, we have a= 3 and b= 2. Therefore, the amplitude is
A=√32+ 22=√13
.
Step 2: The period of the function can be found using the formula T=2π
|b|,
where |b|is the absolute value of the coefficient of x. In this case, |b|= 2.
Therefore, the period is
T=2π
2=π.
Step 3: The phase shift of the function can be found by setting the argument
of the cosine function equal to zero and solving for x. In this case, the argument
of cosine function is 2x. Setting 2x= 0, we find
x= 0
. Therefore, the phase shift is 0.
Step 4: The vertical shift of the function can be found by determining the
vertical translation added to the function. In this case, there is no vertical
translation added, so the vertical shift is 0.
In conclusion, the amplitude of the function is √13, the period is π, the
phase shift is 0, and the vertical shift is 0.
Question 17
Question
Let f(x) = 2 sin(3x). Find the amplitude, period, phase shift, and vertical shift
of the function f(x).
11
Solution
Step 1: Amplitude
The amplitude of a function in the form f(x) = Asin(Bx)is the absolute
value of A. In this case, A= 2, so the amplitude is |2|= 2.
Step 2: Period
The period of a function in the form f(x) = sin(Bx)is given by 2π
B. In this
case, B= 3, so the period is 2π
3.
Step 3: Phase Shift
To find the phase shift of a function in the form f(x) = sin(B(x−C)), we
set B(x−C) = 0 and solve for C. Here, 3x= 0 gives x= 0. Since the coefficient
of xin the argument of sine is positive, the phase shift is to the right. Therefore,
the phase shift is 0.
Step 4: Vertical Shift
The vertical shift of a function in the form f(x) = Asin(Bx)+Dis the value
of D. Here, there is no term added or subtracted outside the sine function, so
the vertical shift is 0.
Therefore, for the given function f(x) = 2 sin(3x), the amplitude is 2, the
period is 2π
3, the phase shift is 0, and the vertical shift is 0.
Question 18
Question
Solve the equation sin2(x) + sin(x)−6 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
sin2(x) + sin(x)−6 = 0
Step 2: Factor the quadratic equation.
To factor this quadratic equation, we need to find two numbers that multiply
to −6and add up to 1. These numbers are 3and −2.
sin2(x) + 3 sin(x)−2 sin(x)−6 = 0
sin(x)(sin(x) + 3) −2(sin(x) + 3) = 0
(sin(x)−2)(sin(x) + 3) = 0
Step 3: Set each factor to zero and solve for sin(x).
sin(x)−2 = 0 or sin(x) + 3 = 0
sin(x) = 2 or sin(x) = −3
12
Step 4: Since sin(x)must be between −1and 1, the equation sin(x) = 2 has
no solutions. So, we focus on solving sin(x) = −3.
Since sin(x) = −3has no solutions, the original trigonometric equation
sin2(x) + sin(x)−6=0does not have any solutions within the given domain
0◦≤x≤360◦.
Question 19
Question
Solve the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: We can rewrite the given equation using the double angle identity:
1−2 sin2(x) = sin(x).
Step 3: Rearrange the equation by bringing all terms to one side to get
2 sin2(x) + sin(x)−1 = 0.
Step 4: Let u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 5: Solve the quadratic equation by factoring or using the quadratic
formula: u=−b±√b2−4ac
2awhere a= 2,b= 1, and c=−1.
Step 6: Solving for ugives u=−1±√1+8
4, which simplifies to u=−1±√9
4,
leading to u=−1±3
4.
Step 7: Therefore, we find two solutions for u:u1=−1+3
4=2
4=1
2and
u2=−1−3
4=−4
4=−1.
Step 8: Since u= sin(x), we have sin(x) = 1
2and sin(x) = −1.
Step 9: Solve sin(x) = 1
2for 0◦≤x≤360◦to find x= 30◦and x= 150◦.
Step 10: Solve sin(x) = −1for 0◦≤x≤360◦to find x= 270◦.
Step 11: Therefore, the solutions to the equation cos(2x) = sin(x)for 0◦≤
x≤360◦are x= 30◦,x= 150◦, and x= 270◦.
Question 20
Question
Find all solutions to the equation cos(2x) = −1
2for 0≤x≤2π.
Solution
Step 1: Let’s start by using the double angle identity for cosine:
cos(2x) = 2 cos2(x)−1
13
Step 2: Substitute cos(2x) = −1
2into the identity:
2 cos2(x)−1 = −1
2
Step 3: Simplify the equation:
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Now, we need to find the values of xthat satisfy cos(x) = ±1
2: For
cos(x) = 1
2, we have x=π
3,5π
3. For cos(x) = −1
2, we have x=2π
3,4π
3.
Step 5: Therefore, the solutions to the given equation in the interval 0≤
x≤2πare x=π
3,2π
3,4π
3,5π
3.
Question 21
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: Express 5π
12 as the sum of two common angles.
5π
12 =π
3+π
4
Step 2: Use the angle addition identity for tangent.
tan (5π
12 )= tan (π
3+π
4)= tan (π
3
+tan (π
4)1−tan(π
3)tan(π
4)
Step 3: Find tan (π
3and tan (π
4.tan (π
3)=√3and tan (π
4)= 1
Step 4: Substitute the values into the formula for tan (5π
12 .tan (5π
12 )=√3+1
1−√3
Step 5: Rationalize the denominator to simplify.
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3) =4+2√3
1−3=4+2√3
−2=−2−√3
Therefore, tan (5π
12 )=−2−√3.
14
Question 22
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for h(x) = sin(2x) cos(x)
in terms of basic trigonometric functions.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the expression for h(x):
h(x) = sin(2x) cos(x) = (2 sin(x) cos(x)) cos(x)
Step 3: Simplify the expression by distributing the cos(x):
h(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos(x)2
Step 4: Recall the Pythagorean identity sin2(x) + cos2(x) = 1 which can be
rearranged to solve for sin2(x)as sin2(x) = 1 −cos2(x).
Step 5: Substitute sin2(x) = 1 −cos2(x)into the expression for h(x):
h(x) = 2 sin(x)(1 −cos2(x))
Step 6: Expand the expression by distributing the 2 sin(x):
h(x) = 2 sin(x)−2 sin(x) cos2(x)
Therefore, the expression for h(x) = sin(2x) cos(x)in terms of basic trigono-
metric functions is 2 sin(x)−2 sin(x) cos2(x).
Question 23
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x) + 5 cos(x) + 2 = 0
Solution
To solve the equation 2 cos2(x) + 5 cos(x) + 2 = 0, we can treat it as a quadratic
equation in terms of cos(x).
Step 1: Let’s substitute cos(x)with yto write the equation as a quadratic
equation in terms of y:2y2+ 5y+ 2 = 0.
Step 2: Now, we can factor the quadratic equation:
2y2+ 5y+ 2 = (2y+ 1)(y+ 2) = 0
15
Step 3: Set each factor to zero and solve for y:
2y+ 1 = 0 =⇒y=−1
2
y+ 2 = 0 =⇒y=−2
Step 4: Now, substitute back cos(x)for yand solve for x:
y= cos(x) = −1
2=⇒x=2π
3,4π
3
y= cos(x) = −2 (No real solutions)
Therefore, the solutions to the trigonometric equation 2 cos2(x) + 5 cos(x) +
2 = 0 in the interval [0,2π]are x=2π
3,4π
3.
Question 24
Question
Find the exact value of cos (5π
12 )using the sum or difference formula.
Solution
Step 1: Let’s express 5π
12 as the difference of two angles that we know the exact
cosine values of. In this case, let’s express it as π
3−π
4.
Step 2: Now, we use the difference formula for cosine:
cos(A−B) = cos Acos B+ sin Asin B
Step 3: Substitute π
3for Aand π
4for B:
cos (5π
12 )= cos (π
3−π
4)= cos (π
3)cos (π
4)+ sin (π
3)sin (π
4)
Step 4: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2. Substitute these values in:
cos (5π
12 )=1
2·√2
2+√3
2·√2
2
Step 5: Simplify the expression:
cos (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of cos (5π
12 )is √2+√6
4.
16
Question 25
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the given equation:
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: To find the solutions for xin the interval [0,2π], recall the values
of sin(x)in the first and second quadrants where sin(x) = 1
2. The solutions are
x=π
6and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)for 0≤x≤2πare
x=π
6and x=5π
6.
17
Step 4: To find csc(θ), we can use the reciprocal identity csc(θ) = 1
sin(θ).
Substituting in the value of sin(θ), we have:
csc(θ) = 1
3
5
=5
3.
Step 5: To find cot(θ), we can use the definition of cot(θ) = 1
tan(θ). Substi-
tuting in the value of tan(θ), we get:
cot(θ) = 1
−3
4
=−4
3.
Therefore, the values of the trigonometric functions for θin quadrant II are:
cos(θ) = −4
5,tan(θ) = −3
4,sec(θ) = −5
4,csc(θ) = 5
3,and cot(θ) = −4
3.
Question 2
Question
If sin(θ) = −3
5and θis in Quadrant III, determine the values of the remaining
five trigonometric functions of θ.
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant III, we can determine the value
of cos(θ)using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
cos(θ) = −√16
25 =−4
5
Step 2: Now, we can determine the values of the other trigonometric func-
tions using the definitions:
tan(θ) = sin(θ)
cos(θ)=−3
5
−4
5
=3
4
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
cot(θ) = 1
tan(θ)=1
3
4
=4
3
Therefore, the remaining trigonometric functions of θare:
tan(θ) = 3
4,csc(θ) = −5
3,sec(θ) = −5
4,and cot(θ) = 4
3
2
Question 3
Question
Let f(x) = sin(3x)and g(x) = cos(2x), where xis in radians. Find the exact
value of f(π
6)·g(π
3).
Solution
Step 1: Evaluate f(π
6)= sin (3·π
6).
f(π
6)= sin (π
2)= 1
Step 2: Evaluate g(π
3)= cos (2·π
3).
g(π
3)= cos (2π
3)=−1
2
Step 3: Calculate the product f(π
6)·g(π
3).
f(π
6)·g(π
3)= 1 ·(−1
2)
=−1
2
Therefore, f(π
6)·g(π
3)=−1
2.
Question 5
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation as a quadratic equation in terms of sin(x).
sin2(x)−3 sin(x) + 2 = 0
Step 2: To simplify, let y= sin(x). The equation becomes:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
3
Step 4: Solve for y:
y= 1 or y= 2
Step 5: Since y= sin(x), we have:
sin(x) = 1 or sin(x) = 2
Step 6: Since the values of sin(x)are restricted to the interval [−1,1], the
equation sin(x) = 2 has no solutions within the given interval.
Step 7: Therefore, we only have to solve sin(x) = 1.
Step 8: The solution to sin(x) = 1 is x=π
2+ 2πn, where nis an integer.
Step 9: Hence, the solution to the equation sin2(x)−3 sin(x) + 2 = 0 for
0≤x≤2πis x=π
2.
Question 6
Question
Let f(x) = sin(x)−2 cos(x). Find the maximum and minimum values of f(x).
Solution
Step 1: We can rewrite f(x)as a single trigonometric function using the angle
addition formula for sine: sin(α−β) = sin(α) cos(β)−cos(α) sin(β).
f(x) = sin(x)−2 cos(x) = sin(x)−2 cos(x) + sin(90◦) cos(x)−cos(90◦) sin(x)
Step 2: Simplify f(x)using the angle addition formula.
f(x) = sin(x)−2 cos(x) + sin(x) cos(90◦)−cos(x) sin(90◦)
f(x) = sin(x)−2 cos(x) + sin(x)·0−cos(x)·1
f(x) = sin(x) + 2 sin(x)
f(x) = sin(x)(1 + 2)
f(x) = 3 sin(x)
Step 3: The function f(x) = 3 sin(x)has a maximum value of 3 when
sin(x)=1, and a minimum value of -3 when sin(x) = −1. Since the range
of sin(x)is [−1,1], the maximum value of f(x)is 3 and the minimum value is
-3.
Question 7
Question
Find all solutions to the equation cos(2θ) = sin(θ)for 0≤θ≤2π.
4
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Substituting this into the given equation, we have 2 cos2(θ)−1 =
sin(θ).
Step 3: Recall that sin(θ) = cos (π
2−θ).
Step 4: Substituting this identity into the equation, we get 2 cos2(θ)−1 =
cos (π
2−θ).
Step 5: Now, using the double angle identity for cosine again, we get
2 cos2(θ)−1 = 2 sin (π
2−θ)cos (π
2−θ).
Step 6: Simplifying the right side gives 2 cos2(θ)−1 = 2 cos(θ) sin(θ).
Step 7: We can rewrite this as 2 cos2(θ)−2 sin(θ) cos(θ)−1 = 0.
Step 8: Factoring the left side gives (2 cos(θ) + 1)(cos(θ)−1) = 0.
Step 9: Setting each factor to zero gives cos(θ) = −1and cos(θ) = 1.
Step 10: Solving cos(θ) = −1gives us θ=π.
Step 11: Solving cos(θ) = 1 gives us θ= 0 and θ= 2π.
Step 12: Therefore, the solutions to the equation cos(2θ) = sin(θ)for 0≤
θ≤2πare θ= 0, π, 2π.
Question 8
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan θ= cos θ−sin θ
Solution
1. Start with the left-hand side (LHS) of the equation:
1−tan2θ
1 + tan θ
2. Substitute tan θ=sin θ
cos θinto the expression:
1−(sin θ
cos θ)2
1 + sin θ
cos θ
3. Simplify the expression by squaring the term in the numerator:
1−sin2θ
cos2θ
1 + sin θ
cos θ
cos2θ−sin2θ
cos θ+ sin θ
5
4. Factor the numerator using the difference of squares formula a2−b2=
(a+b)(a−b):
(cos θ+ sin θ)(cos θ−sin θ)
cos θ+ sin θ
5. Cancel out the common term of cos θ+ sin θ:
cos θ−sin θ
6. Therefore, the left-hand side (LHS) 1−tan2θ
1+tan θsimplifies to cos θ−sin θ,
which is equal to the right-hand side (RHS) of the given identity. Hence,
the given trigonometric identity is proven.
Question 9
Question
Let f(x) = 3 sin(2x)−2 cos(2x). 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 asin(bx)+ccos(bx)
is given by √a2+c2. In this case, a= 3 and c=−2, so the amplitude is
√32+ (−2)2=√9 + 4 = √13.
Step 2: Period The period of a function of the form sin(bx)or cos(bx)is
2π
|b|. Since b= 2 in this case, the period of f(x)is
2π
|2|=π.
Step 3: Phase Shift The phase shift of a function of the form sin(bx)or
cos(bx)is given by −c
b. For the function f(x) = 3 sin(2x)−2 cos(2x), the phase
shift is −(−2)
2= 1.
Step 4: Vertical Shift The vertical shift of a function of the form asin(bx)+
ccos(bx)is simply the value of c. In this case, f(x)has a vertical shift of −2.
Question 10
Question
Find the general solution to the equation cos(2x) = sin(x).
6
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 1 −2 sin2(x).
Step 2: Substitute this identity into the equation cos(2x) = sin(x):
1−2 sin2(x) = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 3: This is a quadratic equation in terms of sin(x). Let u= sin(x)to
make it easier to solve:
2u2+u−1 = 0
Step 4: Now, solve for uusing the quadratic formula:
u=−b±√b2−4ac
2a
u=−1±√1+8
4
u=−1±√9
4
u=−1±3
4
u=2
4or u=−4
4
u=1
2or u=−1
Step 5: Remember that u= sin(x), so sin(x) = 1
2or sin(x) = −1.
Step 6: Determine the general solutions for sin(x) = 1
2and sin(x) = −1,
respectively, using your knowledge of the unit circle.
Step 7: For sin(x) = 1
2, the solutions are x=π
6+ 2πn and x=5π
6+ 2πn,
where n∈Z.
Step 8: For sin(x) = −1, the solution is x=−π
2+ 2πn, where n∈Z.
Step 9: Therefore, the general solution to the equation cos(2x) = sin(x)is
x=π
6+ 2πn,x=5π
6+ 2πn, and x=−π
2+ 2πn, where n∈Z.
Question 11
Question
Find the exact values of sin (π
12 )and cos (5π
12 )using trigonometric identities.
7
Solution
Step 1: We can use the angle addition identity for sine to find sin (π
12 ):
sin (π
12)= sin (π
6−π
4)
Step 2: Applying the angle subtraction formula for sine gives:
sin (π
12)= sin (π
6)cos (π
4)−cos (π
6)sin (π
4)
Step 3: Simplifying using known values: sin (π
6)=1
2,cos (π
6)=√3
2,sin (π
4)=
√2
2, and cos (π
4)=√2
2, we get:
sin (π
12)=1
2·√2
2−√3
2·√2
2
sin (π
12)=√2−√6
4
Step 4: Next, we use the angle addition identity for cosine to find cos (5π
12 ):
cos (5π
12 )= cos (π
3+π
4)
Step 5: Applying the angle addition formula for cosine gives:
cos (5π
12 )= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
Step 6: Using the values cos (π
3)=1
2,sin (π
3)=√3
2,cos (π
4)=√2
2, and
sin (π
4)=√2
2, we have:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
cos (5π
12 )=√2−√6
4
Question 12
Question
Solve the equation cos2x−2 cos x= 0 for xin the interval [0,2π).
8
Solution
Step 1: Let’s rewrite the equation in terms of cos x:
cos2x−2 cos x= 0
Step 2: Factor out cos x:
cos x(cos x−2) = 0
Step 3: Set each factor to zero:
cos x= 0
cos x−2 = 0
Step 4: Solve the first equation: Since cos x= 0 when x=π
2and 3π
2in the
interval [0,2π).
Step 5: Solve the second equation:
cos x−2 = 0
cos x= 2 (not possible as cos xranges from -1 to 1)
Step 6: Therefore, the solutions to the equation cos2x−2 cos x= 0 in the
interval [0,2π)are x=π
2and x=3π
2.
Question 13
Question
Prove the trigonometric identity:
cot(θ) = cos(θ)
sin(θ)
Solution
Step 1: Recall that cot(θ) = 1
tan(θ)and tan(θ) = sin(θ)
cos(θ). Therefore, cot(θ) =
1
sin(θ)
cos(θ)
=cos(θ)
sin(θ).
Step 2: Thus, we have shown that cot(θ) = cos(θ)
sin(θ).
Question 14
Question
Solve for xin the equation sin(x)−cos(x) = 1
2for 0◦≤x≤360◦.
9
Solution
Step 1: We can rewrite sin(x)−cos(x) = 1
2as sin(x) = cos(x) + 1
2.
Step 2: Squaring both sides of the equation, we get sin2(x) = (cos(x) + 1
2)2.
Step 3: Using the Pythagorean identity sin2(x) + cos2(x) = 1, we substitute
cos2(x) = 1−sin2(x)into the equation from step 2 to get sin2(x) = (cos(x)+ 1
2)2.
Step 4: Expanding the right side of the equation and using the Pythagorean
identity again, we have sin2(x) = cos2(x) + cos(x) + 1
4.
Step 5: Substituting 1−sin2(x)for cos2(x), we get sin2(x)=1−sin2(x) +
cos(x) + 1
4.
Step 6: Rearranging terms, we have 2 sin2(x) + cos(x) = 3
4.
Step 7: Since cos2(x) + sin2(x) = 1, we can substitute cos2(x) = 1 −sin2(x)
into the equation from step 2 to get 2 sin2(x) + √1−sin2(x) = 3
4.
Step 8: Let u= sin(x), which gives us 2u2+√1−u2=3
4.
Step 9: Solving the quadratic equation 2u2+√1−u2=3
4will give us the
values of u, from which we can determine the values of xwithin the given range.
Question 15
Question
Solve the equation sin(2x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Use the double angle formula for sine to rewrite sin(2x).
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the double angle formula back into the original equation.
2 sin(x) cos(x) + cos(x) = 0
Step 3: Factor out a cos(x)term.
cos(x)(2 sin(x) + 1) = 0
Step 4: Use the zero-product property to set each factor equal to zero.
cos(x) = 0 or 2 sin(x) + 1 = 0
Step 5: Solve cos(x) = 0. Since cos(x) = 0 when x=π
2,3π
2.
x=π
2,3π
2
Step 6: Solve 2 sin(x) + 1 = 0.
2 sin(x) = −1
10
sin(x) = −1
2
Step 7: Find the solutions for sin(x) = −1
2within the given interval 0≤
x≤2π. The solutions occur in the third and fourth quadrants, corresponding
to x=7π
6,11π
6.
Step 8: So, the solutions to the equation sin(2x) + cos(x) = 0 for 0≤x≤2π
are x=π
2,3π
2,7π
6,11π
6.
Question 16
Question
Let f(x) = 3 cos(2x) + 2 sin(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of the function can be found by using the formula A=
√a2+b2, where ais the coefficient of cos(x)and bis the coefficient of sin(x).
In this case, we have a= 3 and b= 2. Therefore, the amplitude is
A=√32+ 22=√13
.
Step 2: The period of the function can be found using the formula T=2π
|b|,
where |b|is the absolute value of the coefficient of x. In this case, |b|= 2.
Therefore, the period is
T=2π
2=π.
Step 3: The phase shift of the function can be found by setting the argument
of the cosine function equal to zero and solving for x. In this case, the argument
of cosine function is 2x. Setting 2x= 0, we find
x= 0
. Therefore, the phase shift is 0.
Step 4: The vertical shift of the function can be found by determining the
vertical translation added to the function. In this case, there is no vertical
translation added, so the vertical shift is 0.
In conclusion, the amplitude of the function is √13, the period is π, the
phase shift is 0, and the vertical shift is 0.
Question 17
Question
Let f(x) = 2 sin(3x). Find the amplitude, period, phase shift, and vertical shift
of the function f(x).
11
Solution
Step 1: Amplitude
The amplitude of a function in the form f(x) = Asin(Bx)is the absolute
value of A. In this case, A= 2, so the amplitude is |2|= 2.
Step 2: Period
The period of a function in the form f(x) = sin(Bx)is given by 2π
B. In this
case, B= 3, so the period is 2π
3.
Step 3: Phase Shift
To find the phase shift of a function in the form f(x) = sin(B(x−C)), we
set B(x−C) = 0 and solve for C. Here, 3x= 0 gives x= 0. Since the coefficient
of xin the argument of sine is positive, the phase shift is to the right. Therefore,
the phase shift is 0.
Step 4: Vertical Shift
The vertical shift of a function in the form f(x) = Asin(Bx)+Dis the value
of D. Here, there is no term added or subtracted outside the sine function, so
the vertical shift is 0.
Therefore, for the given function f(x) = 2 sin(3x), the amplitude is 2, the
period is 2π
3, the phase shift is 0, and the vertical shift is 0.
Question 18
Question
Solve the equation sin2(x) + sin(x)−6 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
sin2(x) + sin(x)−6 = 0
Step 2: Factor the quadratic equation.
To factor this quadratic equation, we need to find two numbers that multiply
to −6and add up to 1. These numbers are 3and −2.
sin2(x) + 3 sin(x)−2 sin(x)−6 = 0
sin(x)(sin(x) + 3) −2(sin(x) + 3) = 0
(sin(x)−2)(sin(x) + 3) = 0
Step 3: Set each factor to zero and solve for sin(x).
sin(x)−2 = 0 or sin(x) + 3 = 0
sin(x) = 2 or sin(x) = −3
12
Step 4: Since sin(x)must be between −1and 1, the equation sin(x) = 2 has
no solutions. So, we focus on solving sin(x) = −3.
Since sin(x) = −3has no solutions, the original trigonometric equation
sin2(x) + sin(x)−6=0does not have any solutions within the given domain
0◦≤x≤360◦.
Question 19
Question
Solve the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: We can rewrite the given equation using the double angle identity:
1−2 sin2(x) = sin(x).
Step 3: Rearrange the equation by bringing all terms to one side to get
2 sin2(x) + sin(x)−1 = 0.
Step 4: Let u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 5: Solve the quadratic equation by factoring or using the quadratic
formula: u=−b±√b2−4ac
2awhere a= 2,b= 1, and c=−1.
Step 6: Solving for ugives u=−1±√1+8
4, which simplifies to u=−1±√9
4,
leading to u=−1±3
4.
Step 7: Therefore, we find two solutions for u:u1=−1+3
4=2
4=1
2and
u2=−1−3
4=−4
4=−1.
Step 8: Since u= sin(x), we have sin(x) = 1
2and sin(x) = −1.
Step 9: Solve sin(x) = 1
2for 0◦≤x≤360◦to find x= 30◦and x= 150◦.
Step 10: Solve sin(x) = −1for 0◦≤x≤360◦to find x= 270◦.
Step 11: Therefore, the solutions to the equation cos(2x) = sin(x)for 0◦≤
x≤360◦are x= 30◦,x= 150◦, and x= 270◦.
Question 20
Question
Find all solutions to the equation cos(2x) = −1
2for 0≤x≤2π.
Solution
Step 1: Let’s start by using the double angle identity for cosine:
cos(2x) = 2 cos2(x)−1
13
Step 2: Substitute cos(2x) = −1
2into the identity:
2 cos2(x)−1 = −1
2
Step 3: Simplify the equation:
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Now, we need to find the values of xthat satisfy cos(x) = ±1
2: For
cos(x) = 1
2, we have x=π
3,5π
3. For cos(x) = −1
2, we have x=2π
3,4π
3.
Step 5: Therefore, the solutions to the given equation in the interval 0≤
x≤2πare x=π
3,2π
3,4π
3,5π
3.
Question 21
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: Express 5π
12 as the sum of two common angles.
5π
12 =π
3+π
4
Step 2: Use the angle addition identity for tangent.
tan (5π
12 )= tan (π
3+π
4)= tan (π
3
+tan (π
4)1−tan(π
3)tan(π
4)
Step 3: Find tan (π
3and tan (π
4.tan (π
3)=√3and tan (π
4)= 1
Step 4: Substitute the values into the formula for tan (5π
12 .tan (5π
12 )=√3+1
1−√3
Step 5: Rationalize the denominator to simplify.
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3) =4+2√3
1−3=4+2√3
−2=−2−√3
Therefore, tan (5π
12 )=−2−√3.
14
Question 22
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for h(x) = sin(2x) cos(x)
in terms of basic trigonometric functions.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the expression for h(x):
h(x) = sin(2x) cos(x) = (2 sin(x) cos(x)) cos(x)
Step 3: Simplify the expression by distributing the cos(x):
h(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos(x)2
Step 4: Recall the Pythagorean identity sin2(x) + cos2(x) = 1 which can be
rearranged to solve for sin2(x)as sin2(x) = 1 −cos2(x).
Step 5: Substitute sin2(x) = 1 −cos2(x)into the expression for h(x):
h(x) = 2 sin(x)(1 −cos2(x))
Step 6: Expand the expression by distributing the 2 sin(x):
h(x) = 2 sin(x)−2 sin(x) cos2(x)
Therefore, the expression for h(x) = sin(2x) cos(x)in terms of basic trigono-
metric functions is 2 sin(x)−2 sin(x) cos2(x).
Question 23
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x) + 5 cos(x) + 2 = 0
Solution
To solve the equation 2 cos2(x) + 5 cos(x) + 2 = 0, we can treat it as a quadratic
equation in terms of cos(x).
Step 1: Let’s substitute cos(x)with yto write the equation as a quadratic
equation in terms of y:2y2+ 5y+ 2 = 0.
Step 2: Now, we can factor the quadratic equation:
2y2+ 5y+ 2 = (2y+ 1)(y+ 2) = 0
15
Step 3: Set each factor to zero and solve for y:
2y+ 1 = 0 =⇒y=−1
2
y+ 2 = 0 =⇒y=−2
Step 4: Now, substitute back cos(x)for yand solve for x:
y= cos(x) = −1
2=⇒x=2π
3,4π
3
y= cos(x) = −2 (No real solutions)
Therefore, the solutions to the trigonometric equation 2 cos2(x) + 5 cos(x) +
2 = 0 in the interval [0,2π]are x=2π
3,4π
3.
Question 24
Question
Find the exact value of cos (5π
12 )using the sum or difference formula.
Solution
Step 1: Let’s express 5π
12 as the difference of two angles that we know the exact
cosine values of. In this case, let’s express it as π
3−π
4.
Step 2: Now, we use the difference formula for cosine:
cos(A−B) = cos Acos B+ sin Asin B
Step 3: Substitute π
3for Aand π
4for B:
cos (5π
12 )= cos (π
3−π
4)= cos (π
3)cos (π
4)+ sin (π
3)sin (π
4)
Step 4: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2. Substitute these values in:
cos (5π
12 )=1
2·√2
2+√3
2·√2
2
Step 5: Simplify the expression:
cos (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of cos (5π
12 )is √2+√6
4.
16
Question 25
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the given equation:
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: To find the solutions for xin the interval [0,2π], recall the values
of sin(x)in the first and second quadrants where sin(x) = 1
2. The solutions are
x=π
6and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)for 0≤x≤2πare
x=π
6and x=5π
6.
17
Step 4: To find csc(θ), we can use the reciprocal identity csc(θ) = 1
sin(θ).
Substituting in the value of sin(θ), we have:
csc(θ) = 1
3
5
=5
3.
Step 5: To find cot(θ), we can use the definition of cot(θ) = 1
tan(θ). Substi-
tuting in the value of tan(θ), we get:
cot(θ) = 1
−3
4
=−4
3.
Therefore, the values of the trigonometric functions for θin quadrant II are:
cos(θ) = −4
5,tan(θ) = −3
4,sec(θ) = −5
4,csc(θ) = 5
3,and cot(θ) = −4
3.
Question 2
Question
If sin(θ) = −3
5and θis in Quadrant III, determine the values of the remaining
five trigonometric functions of θ.
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant III, we can determine the value
of cos(θ)using the Pythagorean identity sin2(θ) + cos2(θ) = 1.
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
cos(θ) = −√16
25 =−4
5
Step 2: Now, we can determine the values of the other trigonometric func-
tions using the definitions:
tan(θ) = sin(θ)
cos(θ)=−3
5
−4
5
=3
4
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
cot(θ) = 1
tan(θ)=1
3
4
=4
3
Therefore, the remaining trigonometric functions of θare:
tan(θ) = 3
4,csc(θ) = −5
3,sec(θ) = −5
4,and cot(θ) = 4
3
2
Question 3
Question
Let f(x) = sin(3x)and g(x) = cos(2x), where xis in radians. Find the exact
value of f(π
6)·g(π
3).
Solution
Step 1: Evaluate f(π
6)= sin (3·π
6).
f(π
6)= sin (π
2)= 1
Step 2: Evaluate g(π
3)= cos (2·π
3).
g(π
3)= cos (2π
3)=−1
2
Step 3: Calculate the product f(π
6)·g(π
3).
f(π
6)·g(π
3)= 1 ·(−1
2)
=−1
2
Therefore, f(π
6)·g(π
3)=−1
2.
Question 5
Question
Solve the equation sin2(x)−3 sin(x) + 2 = 0 for 0≤x≤2π.
Solution
Step 1: Let’s rewrite the equation as a quadratic equation in terms of sin(x).
sin2(x)−3 sin(x) + 2 = 0
Step 2: To simplify, let y= sin(x). The equation becomes:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
3
Step 4: Solve for y:
y= 1 or y= 2
Step 5: Since y= sin(x), we have:
sin(x) = 1 or sin(x) = 2
Step 6: Since the values of sin(x)are restricted to the interval [−1,1], the
equation sin(x) = 2 has no solutions within the given interval.
Step 7: Therefore, we only have to solve sin(x) = 1.
Step 8: The solution to sin(x) = 1 is x=π
2+ 2πn, where nis an integer.
Step 9: Hence, the solution to the equation sin2(x)−3 sin(x) + 2 = 0 for
0≤x≤2πis x=π
2.
Question 6
Question
Let f(x) = sin(x)−2 cos(x). Find the maximum and minimum values of f(x).
Solution
Step 1: We can rewrite f(x)as a single trigonometric function using the angle
addition formula for sine: sin(α−β) = sin(α) cos(β)−cos(α) sin(β).
f(x) = sin(x)−2 cos(x) = sin(x)−2 cos(x) + sin(90◦) cos(x)−cos(90◦) sin(x)
Step 2: Simplify f(x)using the angle addition formula.
f(x) = sin(x)−2 cos(x) + sin(x) cos(90◦)−cos(x) sin(90◦)
f(x) = sin(x)−2 cos(x) + sin(x)·0−cos(x)·1
f(x) = sin(x) + 2 sin(x)
f(x) = sin(x)(1 + 2)
f(x) = 3 sin(x)
Step 3: The function f(x) = 3 sin(x)has a maximum value of 3 when
sin(x)=1, and a minimum value of -3 when sin(x) = −1. Since the range
of sin(x)is [−1,1], the maximum value of f(x)is 3 and the minimum value is
-3.
Question 7
Question
Find all solutions to the equation cos(2θ) = sin(θ)for 0≤θ≤2π.
4
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Substituting this into the given equation, we have 2 cos2(θ)−1 =
sin(θ).
Step 3: Recall that sin(θ) = cos (π
2−θ).
Step 4: Substituting this identity into the equation, we get 2 cos2(θ)−1 =
cos (π
2−θ).
Step 5: Now, using the double angle identity for cosine again, we get
2 cos2(θ)−1 = 2 sin (π
2−θ)cos (π
2−θ).
Step 6: Simplifying the right side gives 2 cos2(θ)−1 = 2 cos(θ) sin(θ).
Step 7: We can rewrite this as 2 cos2(θ)−2 sin(θ) cos(θ)−1 = 0.
Step 8: Factoring the left side gives (2 cos(θ) + 1)(cos(θ)−1) = 0.
Step 9: Setting each factor to zero gives cos(θ) = −1and cos(θ) = 1.
Step 10: Solving cos(θ) = −1gives us θ=π.
Step 11: Solving cos(θ) = 1 gives us θ= 0 and θ= 2π.
Step 12: Therefore, the solutions to the equation cos(2θ) = sin(θ)for 0≤
θ≤2πare θ= 0, π, 2π.
Question 8
Question
Prove the following trigonometric identity:
1−tan2θ
1 + tan θ= cos θ−sin θ
Solution
1. Start with the left-hand side (LHS) of the equation:
1−tan2θ
1 + tan θ
2. Substitute tan θ=sin θ
cos θinto the expression:
1−(sin θ
cos θ)2
1 + sin θ
cos θ
3. Simplify the expression by squaring the term in the numerator:
1−sin2θ
cos2θ
1 + sin θ
cos θ
cos2θ−sin2θ
cos θ+ sin θ
5
4. Factor the numerator using the difference of squares formula a2−b2=
(a+b)(a−b):
(cos θ+ sin θ)(cos θ−sin θ)
cos θ+ sin θ
5. Cancel out the common term of cos θ+ sin θ:
cos θ−sin θ
6. Therefore, the left-hand side (LHS) 1−tan2θ
1+tan θsimplifies to cos θ−sin θ,
which is equal to the right-hand side (RHS) of the given identity. Hence,
the given trigonometric identity is proven.
Question 9
Question
Let f(x) = 3 sin(2x)−2 cos(2x). 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 asin(bx)+ccos(bx)
is given by √a2+c2. In this case, a= 3 and c=−2, so the amplitude is
√32+ (−2)2=√9 + 4 = √13.
Step 2: Period The period of a function of the form sin(bx)or cos(bx)is
2π
|b|. Since b= 2 in this case, the period of f(x)is
2π
|2|=π.
Step 3: Phase Shift The phase shift of a function of the form sin(bx)or
cos(bx)is given by −c
b. For the function f(x) = 3 sin(2x)−2 cos(2x), the phase
shift is −(−2)
2= 1.
Step 4: Vertical Shift The vertical shift of a function of the form asin(bx)+
ccos(bx)is simply the value of c. In this case, f(x)has a vertical shift of −2.
Question 10
Question
Find the general solution to the equation cos(2x) = sin(x).
6
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 1 −2 sin2(x).
Step 2: Substitute this identity into the equation cos(2x) = sin(x):
1−2 sin2(x) = sin(x)
2 sin2(x) + sin(x)−1 = 0
Step 3: This is a quadratic equation in terms of sin(x). Let u= sin(x)to
make it easier to solve:
2u2+u−1 = 0
Step 4: Now, solve for uusing the quadratic formula:
u=−b±√b2−4ac
2a
u=−1±√1+8
4
u=−1±√9
4
u=−1±3
4
u=2
4or u=−4
4
u=1
2or u=−1
Step 5: Remember that u= sin(x), so sin(x) = 1
2or sin(x) = −1.
Step 6: Determine the general solutions for sin(x) = 1
2and sin(x) = −1,
respectively, using your knowledge of the unit circle.
Step 7: For sin(x) = 1
2, the solutions are x=π
6+ 2πn and x=5π
6+ 2πn,
where n∈Z.
Step 8: For sin(x) = −1, the solution is x=−π
2+ 2πn, where n∈Z.
Step 9: Therefore, the general solution to the equation cos(2x) = sin(x)is
x=π
6+ 2πn,x=5π
6+ 2πn, and x=−π
2+ 2πn, where n∈Z.
Question 11
Question
Find the exact values of sin (π
12 )and cos (5π
12 )using trigonometric identities.
7
Solution
Step 1: We can use the angle addition identity for sine to find sin (π
12 ):
sin (π
12)= sin (π
6−π
4)
Step 2: Applying the angle subtraction formula for sine gives:
sin (π
12)= sin (π
6)cos (π
4)−cos (π
6)sin (π
4)
Step 3: Simplifying using known values: sin (π
6)=1
2,cos (π
6)=√3
2,sin (π
4)=
√2
2, and cos (π
4)=√2
2, we get:
sin (π
12)=1
2·√2
2−√3
2·√2
2
sin (π
12)=√2−√6
4
Step 4: Next, we use the angle addition identity for cosine to find cos (5π
12 ):
cos (5π
12 )= cos (π
3+π
4)
Step 5: Applying the angle addition formula for cosine gives:
cos (5π
12 )= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
Step 6: Using the values cos (π
3)=1
2,sin (π
3)=√3
2,cos (π
4)=√2
2, and
sin (π
4)=√2
2, we have:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
cos (5π
12 )=√2−√6
4
Question 12
Question
Solve the equation cos2x−2 cos x= 0 for xin the interval [0,2π).
8
Solution
Step 1: Let’s rewrite the equation in terms of cos x:
cos2x−2 cos x= 0
Step 2: Factor out cos x:
cos x(cos x−2) = 0
Step 3: Set each factor to zero:
cos x= 0
cos x−2 = 0
Step 4: Solve the first equation: Since cos x= 0 when x=π
2and 3π
2in the
interval [0,2π).
Step 5: Solve the second equation:
cos x−2 = 0
cos x= 2 (not possible as cos xranges from -1 to 1)
Step 6: Therefore, the solutions to the equation cos2x−2 cos x= 0 in the
interval [0,2π)are x=π
2and x=3π
2.
Question 13
Question
Prove the trigonometric identity:
cot(θ) = cos(θ)
sin(θ)
Solution
Step 1: Recall that cot(θ) = 1
tan(θ)and tan(θ) = sin(θ)
cos(θ). Therefore, cot(θ) =
1
sin(θ)
cos(θ)
=cos(θ)
sin(θ).
Step 2: Thus, we have shown that cot(θ) = cos(θ)
sin(θ).
Question 14
Question
Solve for xin the equation sin(x)−cos(x) = 1
2for 0◦≤x≤360◦.
9
Solution
Step 1: We can rewrite sin(x)−cos(x) = 1
2as sin(x) = cos(x) + 1
2.
Step 2: Squaring both sides of the equation, we get sin2(x) = (cos(x) + 1
2)2.
Step 3: Using the Pythagorean identity sin2(x) + cos2(x) = 1, we substitute
cos2(x) = 1−sin2(x)into the equation from step 2 to get sin2(x) = (cos(x)+ 1
2)2.
Step 4: Expanding the right side of the equation and using the Pythagorean
identity again, we have sin2(x) = cos2(x) + cos(x) + 1
4.
Step 5: Substituting 1−sin2(x)for cos2(x), we get sin2(x)=1−sin2(x) +
cos(x) + 1
4.
Step 6: Rearranging terms, we have 2 sin2(x) + cos(x) = 3
4.
Step 7: Since cos2(x) + sin2(x) = 1, we can substitute cos2(x) = 1 −sin2(x)
into the equation from step 2 to get 2 sin2(x) + √1−sin2(x) = 3
4.
Step 8: Let u= sin(x), which gives us 2u2+√1−u2=3
4.
Step 9: Solving the quadratic equation 2u2+√1−u2=3
4will give us the
values of u, from which we can determine the values of xwithin the given range.
Question 15
Question
Solve the equation sin(2x) + cos(x) = 0 for 0≤x≤2π.
Solution
Step 1: Use the double angle formula for sine to rewrite sin(2x).
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute the double angle formula back into the original equation.
2 sin(x) cos(x) + cos(x) = 0
Step 3: Factor out a cos(x)term.
cos(x)(2 sin(x) + 1) = 0
Step 4: Use the zero-product property to set each factor equal to zero.
cos(x) = 0 or 2 sin(x) + 1 = 0
Step 5: Solve cos(x) = 0. Since cos(x) = 0 when x=π
2,3π
2.
x=π
2,3π
2
Step 6: Solve 2 sin(x) + 1 = 0.
2 sin(x) = −1
10
sin(x) = −1
2
Step 7: Find the solutions for sin(x) = −1
2within the given interval 0≤
x≤2π. The solutions occur in the third and fourth quadrants, corresponding
to x=7π
6,11π
6.
Step 8: So, the solutions to the equation sin(2x) + cos(x) = 0 for 0≤x≤2π
are x=π
2,3π
2,7π
6,11π
6.
Question 16
Question
Let f(x) = 3 cos(2x) + 2 sin(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of the function can be found by using the formula A=
√a2+b2, where ais the coefficient of cos(x)and bis the coefficient of sin(x).
In this case, we have a= 3 and b= 2. Therefore, the amplitude is
A=√32+ 22=√13
.
Step 2: The period of the function can be found using the formula T=2π
|b|,
where |b|is the absolute value of the coefficient of x. In this case, |b|= 2.
Therefore, the period is
T=2π
2=π.
Step 3: The phase shift of the function can be found by setting the argument
of the cosine function equal to zero and solving for x. In this case, the argument
of cosine function is 2x. Setting 2x= 0, we find
x= 0
. Therefore, the phase shift is 0.
Step 4: The vertical shift of the function can be found by determining the
vertical translation added to the function. In this case, there is no vertical
translation added, so the vertical shift is 0.
In conclusion, the amplitude of the function is √13, the period is π, the
phase shift is 0, and the vertical shift is 0.
Question 17
Question
Let f(x) = 2 sin(3x). Find the amplitude, period, phase shift, and vertical shift
of the function f(x).
11
Solution
Step 1: Amplitude
The amplitude of a function in the form f(x) = Asin(Bx)is the absolute
value of A. In this case, A= 2, so the amplitude is |2|= 2.
Step 2: Period
The period of a function in the form f(x) = sin(Bx)is given by 2π
B. In this
case, B= 3, so the period is 2π
3.
Step 3: Phase Shift
To find the phase shift of a function in the form f(x) = sin(B(x−C)), we
set B(x−C) = 0 and solve for C. Here, 3x= 0 gives x= 0. Since the coefficient
of xin the argument of sine is positive, the phase shift is to the right. Therefore,
the phase shift is 0.
Step 4: Vertical Shift
The vertical shift of a function in the form f(x) = Asin(Bx)+Dis the value
of D. Here, there is no term added or subtracted outside the sine function, so
the vertical shift is 0.
Therefore, for the given function f(x) = 2 sin(3x), the amplitude is 2, the
period is 2π
3, the phase shift is 0, and the vertical shift is 0.
Question 18
Question
Solve the equation sin2(x) + sin(x)−6 = 0 for 0◦≤x≤360◦.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
sin(x).
sin2(x) + sin(x)−6 = 0
Step 2: Factor the quadratic equation.
To factor this quadratic equation, we need to find two numbers that multiply
to −6and add up to 1. These numbers are 3and −2.
sin2(x) + 3 sin(x)−2 sin(x)−6 = 0
sin(x)(sin(x) + 3) −2(sin(x) + 3) = 0
(sin(x)−2)(sin(x) + 3) = 0
Step 3: Set each factor to zero and solve for sin(x).
sin(x)−2 = 0 or sin(x) + 3 = 0
sin(x) = 2 or sin(x) = −3
12
Step 4: Since sin(x)must be between −1and 1, the equation sin(x) = 2 has
no solutions. So, we focus on solving sin(x) = −3.
Since sin(x) = −3has no solutions, the original trigonometric equation
sin2(x) + sin(x)−6=0does not have any solutions within the given domain
0◦≤x≤360◦.
Question 19
Question
Solve the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 1 −2 sin2(θ).
Step 2: We can rewrite the given equation using the double angle identity:
1−2 sin2(x) = sin(x).
Step 3: Rearrange the equation by bringing all terms to one side to get
2 sin2(x) + sin(x)−1 = 0.
Step 4: Let u= sin(x), then the equation becomes 2u2+u−1 = 0.
Step 5: Solve the quadratic equation by factoring or using the quadratic
formula: u=−b±√b2−4ac
2awhere a= 2,b= 1, and c=−1.
Step 6: Solving for ugives u=−1±√1+8
4, which simplifies to u=−1±√9
4,
leading to u=−1±3
4.
Step 7: Therefore, we find two solutions for u:u1=−1+3
4=2
4=1
2and
u2=−1−3
4=−4
4=−1.
Step 8: Since u= sin(x), we have sin(x) = 1
2and sin(x) = −1.
Step 9: Solve sin(x) = 1
2for 0◦≤x≤360◦to find x= 30◦and x= 150◦.
Step 10: Solve sin(x) = −1for 0◦≤x≤360◦to find x= 270◦.
Step 11: Therefore, the solutions to the equation cos(2x) = sin(x)for 0◦≤
x≤360◦are x= 30◦,x= 150◦, and x= 270◦.
Question 20
Question
Find all solutions to the equation cos(2x) = −1
2for 0≤x≤2π.
Solution
Step 1: Let’s start by using the double angle identity for cosine:
cos(2x) = 2 cos2(x)−1
13
Step 2: Substitute cos(2x) = −1
2into the identity:
2 cos2(x)−1 = −1
2
Step 3: Simplify the equation:
2 cos2(x) = 1
2
cos2(x) = 1
4
cos(x) = ±1
2
Step 4: Now, we need to find the values of xthat satisfy cos(x) = ±1
2: For
cos(x) = 1
2, we have x=π
3,5π
3. For cos(x) = −1
2, we have x=2π
3,4π
3.
Step 5: Therefore, the solutions to the given equation in the interval 0≤
x≤2πare x=π
3,2π
3,4π
3,5π
3.
Question 21
Question
Find the exact value of tan (5π
12 .
Solution
Step 1: Express 5π
12 as the sum of two common angles.
5π
12 =π
3+π
4
Step 2: Use the angle addition identity for tangent.
tan (5π
12 )= tan (π
3+π
4)= tan (π
3
+tan (π
4)1−tan(π
3)tan(π
4)
Step 3: Find tan (π
3and tan (π
4.tan (π
3)=√3and tan (π
4)= 1
Step 4: Substitute the values into the formula for tan (5π
12 .tan (5π
12 )=√3+1
1−√3
Step 5: Rationalize the denominator to simplify.
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3) =4+2√3
1−3=4+2√3
−2=−2−√3
Therefore, tan (5π
12 )=−2−√3.
14
Question 22
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find an expression for h(x) = sin(2x) cos(x)
in terms of basic trigonometric functions.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute sin(2x) = 2 sin(x) cos(x)into the expression for h(x):
h(x) = sin(2x) cos(x) = (2 sin(x) cos(x)) cos(x)
Step 3: Simplify the expression by distributing the cos(x):
h(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos(x)2
Step 4: Recall the Pythagorean identity sin2(x) + cos2(x) = 1 which can be
rearranged to solve for sin2(x)as sin2(x) = 1 −cos2(x).
Step 5: Substitute sin2(x) = 1 −cos2(x)into the expression for h(x):
h(x) = 2 sin(x)(1 −cos2(x))
Step 6: Expand the expression by distributing the 2 sin(x):
h(x) = 2 sin(x)−2 sin(x) cos2(x)
Therefore, the expression for h(x) = sin(2x) cos(x)in terms of basic trigono-
metric functions is 2 sin(x)−2 sin(x) cos2(x).
Question 23
Question
Solve the trigonometric equation for xin the interval [0,2π]:
2 cos2(x) + 5 cos(x) + 2 = 0
Solution
To solve the equation 2 cos2(x) + 5 cos(x) + 2 = 0, we can treat it as a quadratic
equation in terms of cos(x).
Step 1: Let’s substitute cos(x)with yto write the equation as a quadratic
equation in terms of y:2y2+ 5y+ 2 = 0.
Step 2: Now, we can factor the quadratic equation:
2y2+ 5y+ 2 = (2y+ 1)(y+ 2) = 0
15
Step 3: Set each factor to zero and solve for y:
2y+ 1 = 0 =⇒y=−1
2
y+ 2 = 0 =⇒y=−2
Step 4: Now, substitute back cos(x)for yand solve for x:
y= cos(x) = −1
2=⇒x=2π
3,4π
3
y= cos(x) = −2 (No real solutions)
Therefore, the solutions to the trigonometric equation 2 cos2(x) + 5 cos(x) +
2 = 0 in the interval [0,2π]are x=2π
3,4π
3.
Question 24
Question
Find the exact value of cos (5π
12 )using the sum or difference formula.
Solution
Step 1: Let’s express 5π
12 as the difference of two angles that we know the exact
cosine values of. In this case, let’s express it as π
3−π
4.
Step 2: Now, we use the difference formula for cosine:
cos(A−B) = cos Acos B+ sin Asin B
Step 3: Substitute π
3for Aand π
4for B:
cos (5π
12 )= cos (π
3−π
4)= cos (π
3)cos (π
4)+ sin (π
3)sin (π
4)
Step 4: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2. Substitute these values in:
cos (5π
12 )=1
2·√2
2+√3
2·√2
2
Step 5: Simplify the expression:
cos (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of cos (5π
12 )is √2+√6
4.
16
Question 25
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
Solution
Step 1: Recall the double angle identity for sine: sin(2x) = 2 sin(x) cos(x).
Step 2: Substitute the double angle identity into the given equation:
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: To find the solutions for xin the interval [0,2π], recall the values
of sin(x)in the first and second quadrants where sin(x) = 1
2. The solutions are
x=π
6and x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)for 0≤x≤2πare
x=π
6and x=5π
6.
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