MATH 122 - TRIGONOMETRY -
Trigonometric Functions
Question Bank - Set 4
Liberty University
Question 1
Question
Let f(x) = 3 sin(x) + 4 cos(x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: To find the amplitude, we need to use the formula A=√a2+b2, where
aand bare the coefficients of the sine and cosine terms, respectively. In this
case, a= 3 and b= 4, so the amplitude is
A=√32+ 42=√9 + 16 = √25 = 5.
Step 2: To find the period, we use the formula T=2π
|b|, where bis the
coefficient of the xterm. In this case, b= 1, so the period is
T=2π
|1|= 2π.
Step 3: To find the phase shift, we set sin(x) = sin(α)and cos(x) = cos(α),
where αis the phase shift. By comparing the given function f(x)with standard
forms, we know that α=−arctan (a
b). Plugging in the values a= 3 and b= 4,
we get
α=−arctan (3
4)≈ −0.6435.
Step 4: To find the vertical shift, we note that there is no vertical shift in this
case since there is no constant added/subtracted to/from the function. Thus,
the vertical shift is 0.
Therefore, the amplitude of f(x)is 5, the period is 2π, the phase shift is
approximately −0.6435 units to the right, and the vertical shift is 0.
Question 2
Question
Let f(x) = sin(x)+cos(x). Find the amplitude, period, phase shift, and vertical
shift of the function f(x).
Solution
Step 1: The amplitude of a function f(x) = asin(bx +c) + dis given by |a|.
Step 2: The amplitude of f(x) = sin(x) + cos(x)is √12+ 12=√2.
Step 3: The period of a function of the form f(x) = asin(bx +c) + dis 2π
|b|.
Step 4: The period of f(x) = sin(x) + cos(x)is 2π
1= 2π.
Step 5: The phase shift of a function f(x) = asin(bx +c) + dis given by
−c
b.
Step 6: For f(x) = sin(x) + cos(x), there is no phase shift because the
functions are both in their standard positions.
Step 7: The vertical shift of a function f(x) = asin(bx +c) + dis the value
of d.
Step 8: The vertical shift of f(x) = sin(x) + cos(x)is 0since there is no
vertical shift.
Question 3
Question
Find the general solution to the equation sin(2x) = cos(3x)in the interval
[0,2π].
Solution
Step 1: Recall the double angle identity for sine and the reflection identity for
cosine:
sin(2x) = 2 sin(x) cos(x)
cos(θ) = sin (π
2−θ)
Step 2: Substitute these identities into the equation sin(2x) = cos(3x):
2 sin(x) cos(x) = sin (π
2−3x)
Step 3: Expand the right side of the equation using the sine of difference
formula:
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
Step 4: Since sin(2x) = cos(3x)is equivalent to 2 sin(x) cos(x) = cos(3x),
we have the equation 2 sin(x) cos(x) = cos(3x).
Step 5: Now we will convert this equation into terms of sine only:
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = sin (π
2−3x)
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
Step 6: Rearrange the equation to have all terms on one side:
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
2 sin(x) cos(x)−4 sin(x) cos3(x) + 3 sin(x) cos(x) = 0
2 sin(x) cos(x) + 3 sin(x) cos(x)−4 sin(x) cos3(x) = 0
Step 7: Factor out a common sin(x)term:
sin(x)(2 cos(x) + 3 cos(x)−4 cos3(x)) = 0
sin(x)(5 cos(x)−4 cos3(x)) = 0
Step 8: Find the solutions for sin(x) = 0 and 5 cos(x)−4 cos3(x) = 0.
Step 9: For sin(x) = 0, we have x= 0 and x=π.
Step 10: For 5 cos(x)−4 cos3(x) = 0, we can factor out a cos(x):
cos(x)(5 −4 cos2(x)) = 0
cos(x)(5 −4 cos2(x)) = 0
Step 11: We find solutions for cos(x) = 0 and 5−4 cos2(x) = 0.
Step 12: For cos(x) = 0, we have x=π
2and x=3π
2.
Step 13: For 5−4 cos2(x) = 0, we solve for cos(x):
4 cos2(x) = 5
cos2(x)
3
Question 4
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can rewrite 5π
12 as the sum of two angles for which we know the
trigonometric functions. Let’s express 5π
12 as π
4+π
3.
Step 2: Using the angle addition formula for sine, we have:
sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+ cos (π
4)sin (π
3)
Step 3: Recall that sin (π
4)=√2
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and
cos (π
3)=1
2. Substituting these values into the expression, we get:
sin (5π
12 )=√2
2·1
2+√2
2·√3
2
Step 4: Simplifying further, we have:
sin (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of sin (5π
12 )is √2+√6
4.
Question 5
Question
Find the exact value of sin (5π
6·4
3).
Solution
Step 1: Recall the angle addition formula for sine: sin(α±β) = sin(α) cos(β)±
cos(α) sin(β).
Step 2: Rewrite 5π
6·4
3as 5π
6+4π
3.
Step 3: Now we apply the angle addition formula:
sin (5π
6+4π
3)= sin (5π
6)cos (4π
3)+ cos (5π
6)sin (4π
3)
=(−√3
2)(−1
2)+(1
2)(−√3
2)
=√3
4−√3
4
= 0
4
Step 4: Therefore, sin (5π
6·4
3)= 0 .
Question 6
Question
Solve the equation cos(2x) + cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We’ll use the angle addition formula for cosine, which states that
cos(a+b) = cos(a) cos(b)−sin(a) sin(b).
cos(2x) + cos(x) = 0
cos(x+x) + cos(x) = 0
cos(x) cos(x)−sin(x) sin(x) + cos(x) = 0
cos2(x)−sin2(x) + cos(x) = 0
Step 2: Remembering the Pythagorean trigonometric identity sin2(x) +
cos2(x) = 1, we can substitute cos2(x)as 1−sin2(x)in the equation.
(1 −sin2(x)) −sin2(x) + cos(x) = 0
1−2 sin2(x) + cos(x) = 0
Step 3: We can rearrange the equation in terms of sin(x)to get a quadratic
equation.
2 sin2(x)−1 + cos(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
Step 4: Now we can solve this quadratic equation for sin(x). Let’s say
y= sin(x).
2y2−1 + √1−y2= 0
2y2−1 = −√1−y2
(2y2−1)2= 1 −y2
4y4−4y2+ 1 = 1 −y2
4y4−3y2= 0
y2(4y2−3) = 0
5
Step 5: Solving the quadratic equation 4y2−3=0, we find two possible
values for sin(x).
4y2−3 = 0
4y2= 3
y2=3
4
y=±√3
2
Step 6: Since sin(x) = ±√3
2, the possible values for xare x= 60◦,x= 120◦,
x= 240◦,x= 300◦.
Step 7: We need to check the solutions in the original equation.
• For x= 60◦:cos(120◦) + cos(60◦) = −1
2+1
2= 0 (valid)
• For x= 120◦:cos(240◦) + cos(120◦) = −1
2−1
2=−1(not valid)
• For x= 240◦:cos(480◦) + cos(240◦) = −1
2+1
2= 0 (valid)
• For x= 300◦:cos(600◦) + cos(300◦) = −1
2−1
2=−1(not valid)
Step 8: Thus, the solutions for the equation cos(2x) + cos(x)=0for
0◦≤x≤
Question 7
Question
Solve for θin the interval [0,2π]:cos(2θ)−√3 sin(θ) = 0.
Solution
Step 1: Rewrite the equation using double angle identity.
cos2(θ)−sin2(θ)−√3 sin(θ) = 0
Step 2: Substitute sin2(θ) = 1 −cos2(θ)into the equation.
cos2(θ)−(1 −cos2(θ)) −√3 sin(θ) = 0
Step 3: Simplify and rearrange the equation.
2 cos2(θ) + √3 sin(θ)−1 = 0
Step 4: Rewrite cos2(θ)as 1−sin2(θ)
2(1 −sin2(θ)) + √3 sin(θ)−1 = 0
6
Step 5: Expand and rearrange the equation.
2−2 sin2(θ) + √3 sin(θ)−1 = 0
Step 6: Combine like terms.
−2 sin2(θ) + √3 sin(θ) + 1 = 0
Step 7: This is now a quadratic equation in terms of sin(θ). Let u= sin(θ).
−2u2+√3u+ 1 = 0
Step 8: Solve the quadratic equation using the quadratic formula, u=
−b±√b2−4ac
2a.
u=−√3±√(−√3)2−4(−2)(1)
2(−2)
Step 9: Simplify the expression.
u=−√3±√3+8
−4
u=−√3±√11
−4
Step 10: Solve for θby substituting back u= sin(θ).
sin(θ) = −√3±√11
−4
Step 11: Calculate the possible values of θwithin the interval [0,2π]. Since
the given equation did not specify a range for the solution, we will find all
possible solutions.
θ= arcsin (−√3 + √11
−4)
θ= arcsin (−√3−√11
−4)
Therefore, the solutions for θin the interval [0,2π]are θ= arcsin (−√3+√11
−4)
and θ= arcsin (−√3−√11
−4).
Question 8
Question
Evaluate the following expression: tan−1(2 tan(3π/4)
1−tan(3π/4) ).
7
Solution
Step 1: Recall the trigonometric identity tan(A−B) = tan A−tan B
1+tan Atan B.
Step 2: Let A= 3π/4and B= 0, so we have tan(3π/4−0) = tan 3π/4−tan 0
1+tan 3π/4 tan 0 .
Step 3: Simplify the expression to get tan(3π/4) = tan (3π
4)=tan(3π/4)−0
1+tan(3π/4)·0.
Step 4: Since tan(0) = 0, the expression becomes tan(3π/4)
1.
Step 5: Therefore, tan(3π/4) = −1.
Step 6: Substitute back into the original expression: tan−1(2 tan(3π/4)
1−tan(3π/4) )=
tan−1(2(−1)
1−(−1) ).
Step 7: Simplify to get tan−1(−2).
Step 8: Finally, since tan(−π/4) = −1, the answer is −π
4.
Question 9
Question
Find the exact value of tan (4π
3).
Solution
Step 1: Recall that the tangent function is defined as tan(θ) = sin(θ)
cos(θ).
Step 2: First, let’s find the sine and cosine of 4π
3using the unit circle.
Step 3: To find the sine of 4π
3, we look at the point on the unit circle
corresponding to this angle. Since 4π
3is in the third quadrant, the y-coordinate
of the point is sin (4π
3)=−√3
2.
Step 4: To find the cosine of 4π
3, we use the definition of cosine as the x-
coordinate of the point. Since 4π
3is in the third quadrant, the x-coordinate of
the point is −1
2. Therefore, cos (4π
3)=−1
2.
Step 5: Now, we can find the value of tan (4π
3)by using the definition of
tangent and the values of sine and cosine we found earlier.
tan (4π
3)=sin (4π
3)
cos (4π
3)=−√3
2
−1
2
=√3
Therefore, tan (4π
3)=√3.
Question 10
Question
Prove the following trigonometric identity:
1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x)
8
Solution
To prove the given trigonometric identity, we will start with the right-hand side
and manipulate it to match the left-hand side.
RHS =sin(x)−cos(x)
sin(2x)
=sin(x)−cos(x)
2 sin(x) cos(x)(Using the double-angle identity for sine)
=sin(x)
2 sin(x) cos(x)−cos(x)
2 sin(x) cos(x)
=1
2 cos(x)−1
2 sin(x)(Canceling sin(x)and cos(x))
=
1
cos(x)−1
sin(x)
2
=sin(x)−cos(x)
2 sin(x) cos(x)
=sin(x)−cos(x)
sin(x) + cos(x)(Using the double-angle identity for sine)
=LHS
Therefore, we have proven that 1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x).
Question 11
Question
Prove that sin(2θ) = 2 sin(θ) cos(θ)using the angle addition formula: sin(A+B) =
sin(A) cos(B) + cos(A) sin(B).
Solution
Step 1: Let A=B=θ, so we have:
sin(2θ) = sin(θ+θ)
Step 2: Apply the angle addition formula with A=θand B=θ:
sin(θ+θ) = sin(θ) cos(θ) + cos(θ) sin(θ)
Step 3: Simplify the right-hand side of the equation:
sin(θ) cos(θ) + cos(θ) sin(θ) = 2 sin(θ) cos(θ)
Step 4: Therefore, we have shown that sin(2θ) = 2 sin(θ) cos(θ).
9
Question 12
Question
Let f(x) = 3 sin(2x)−√3 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Identify the Amplitude
The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
given by √A2+C2.
In this case, the amplitude of f(x) = 3 sin(2x)−√3 cos(2x)is √32+ (−√3)2=
√9 + 3 = √12 = 2√3.
Therefore, the amplitude of f(x)is 2√3.
Step 2: Identify the Period
The period of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is given
by 2π
|B|.
In this case, the period of f(x) = 3 sin(2x)−√3 cos(2x)is 2π
|2|=π.
Therefore, the period of f(x)is π.
Step 3: Identify the Phase Shift
To find the phase shift of a function of the form f(x) = Asin(Bx) +
Ccos(Bx), we need to find the value of xthat makes Bx equal to 0or π
2.
In this case, B= 2, so the phase shift of f(x) = 3 sin(2x)−√3 cos(2x)is 0.
Therefore, the phase shift of f(x)is 0.
Step 4: Identify the Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
simply the value of C.
In this case, the vertical shift of f(x) = 3 sin(2x)−√3 cos(2x)is −√3.
Therefore, the vertical shift of f(x)is −√3.
Question 13
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: First, let’s express 5π
12 in terms of common angles. Since π
6and π
4are
common angles, we can rewrite 5π
12 as a combination of these angles:
5π
12 =π
4+π
6
10
Step 2: Next, we will apply the sum-to-product trigonometric identity:
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Using the sum-to-product identity, we can rewrite sin (5π
12 )as:
sin (5π
12 )= sin (π
4+π
6)
= sin (π
4)cos (π
6)+ cos (π
4)sin (π
6)
Step 4: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2. Substituting these values in, we get:
sin (5π
12 )=1
√2·√3
2+1
√2·1
2
Step 5: Simplifying further, we have:
sin (5π
12 )=√3
2√2+1
2√2
=√3+1
2√2
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 14
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles. Start by dividing
the given angle by 2.
5π
12 =(4 + 1)π
12 =4π
12 +π
12 =π
3+π
12
Step 2: Next, we express π
12 in terms of common trigonometric values to
simplify the expression.
π
12 =π
6∇ · 2 = 1
2sin (π
6)=1
2(1
2)=1
4
11
Step 3: Substitute π
12 back into the expression for sin (5π
12 ).
sin (5π
12 )= sin (π
3+π
12)= sin π
3cos π
12 + cos π
3sin π
12
Step 4: Recall the values of sine and cosine for common angles.
sin π
3=√3
2,cos π
3=1
2,sin π
12 =√6−√2
4,cos π
12 =√6 + √2
4
Step 5: Substitute these values into the expression for sin (5π
12 ).
sin (5π
12 )=√3
2·√6 + √2
4+1
2·√6−√2
4
sin (5π
12 )=√18 + √6 + √6−√2
8=2√6 + √18 −√2
8=2√6+3√2
8=√6+3√2
4
Therefore, the exact value of sin (5π
12 )is √6+3√2
4.
Question 15
Question
Let f(x) = tan (1
2x). Find the period of the function f(x).
Solution
Step 1: Recall that the period of the function y= tan(ax)is π
|a|. Step 2: In
this case, a=1
2. Step 3: Therefore, the period of the function f(x) = tan (1
2x)
is π
|1
2|= 2π. Step 4: Thus, the period of the function f(x)is 2π.
Question 16
Question
Let f(x) = 2 sin(x)+3 cos(x)for 0≤x≤2π. Determine the absolute maximum
and minimum values of f(x)on this interval.
12
Solution
Step 1: Find critical points by setting the derivative of f(x)equal to 0.
d
dx[2 sin(x) + 3 cos(x)] = 2 cos(x)−3 sin(x)
2 cos(x)−3 sin(x) = 0
2 cos(x) = 3 sin(x)
2
3= tan(x)
x= arctan (2
3)
Step 2: Determine the values of f(x)at the critical point and at the end-
points of the interval.
f(0) = 2 sin(0) + 3 cos(0) = 0 + 3 = 3
f(2π) = 2 sin(2π) + 3 cos(2π) = 0 + 3 = 3
f(arctan (2
3))= 2 sin (arctan (2
3))+ 3 cos (arctan (2
3))
Step 3: Use trigonometric identities to simplify f(arctan (2
3)).
Let sin(θ) = 2
√22+ 32=2
√13 and
cos(θ) = 3
√13
f(arctan (2
3))= 2 ∗2
√13 + 3 ∗3
√13
=4+9
√13
=13
√13
=√13
Step 4: Compare the values of f(x)at the critical point and endpoints to
determine the absolute maximum and minimum values. The minimum value
of f(x)is 3 (at x= 0 and x= 2π) and the maximum value is √13 at x=
arctan (2
3).
Question 17
Question
Find the exact value of sin (5π
12 ).
13
Solution
Step 1: Begin by expressing 5π
12 as the sum or difference of two special angles
that you know the exact sine value. Step 2: Since 5π=3π
4+π
3, we can
write 5π
12 =3π
12 +4π
12 =π
4+π
3. Step 3: Now, use the sum-to-product identities
to rewrite sin (5π
12 )in terms of known values. Step 4: Applying the sum-to-
product identity, we have sin (5π
12 )= sin (π
4+π
3)= sin π
4cos π
3+ cos π
4sin π
3.
Step 5: Recall that sin π
4=1
√2and cos π
3=1
2, so sin (5π
12 )=1
√2·1
2+1
√2·√3
2.
Step 6: Simplify the expression to find the exact value of sin (5π
12 ). Step 7:
sin (5π
12 )=1
2√2+√3
2√2=1+√3
2√2. Step 8: Therefore, the exact value of sin (5π
12 )is
1+√3
2√2.
Question 18
Question
Let f(x) = 2 sin(x
2)and g(x) = cos(x
2). Find the function h(x)such that
h(x) = f(x)·g(x).
Solution
Step 1: Write out the functions f(x)and g(x):
f(x) = 2 sin (x
2)
g(x) = cos (x
2)
Step 2: Find h(x)by multiplying f(x)and g(x):
h(x) = f(x)·g(x) = 2 sin (x
2)cos (x
2)
Step 3: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to simplify
the expression:
h(x) = sin(x)
Therefore, h(x) = sin(x).
Question 19
Question
Prove the following trigonometric identity:
cos4x−sin4x= 1 −2 sin2xcos2x
14
Solution
To prove the trigonometric identity cos4x−sin4x= 1 −2 sin2xcos2x, we will
start with the left-hand side and manipulate it to match the right-hand side.
Step 1: Begin with the left-hand side of the given identity: cos4x−sin4x.
cos4x−sin4x
Step 2: Rewrite cos4xin terms of sin2xusing the Pythagorean identity for
cosine: cos2x= 1 −sin2x.
(cos2x)2−sin4x= (1 −sin2x)2−sin4x
Step 3: Expand and simplify the expression.
(1 −2 sin2x+ sin4x)−sin4x= 1 −2 sin2x+ sin4x−sin4x
= 1 −2 sin2x
Step 4: Compare the simplified expression with the right-hand side of the
given identity: 1−2 sin2xcos2x. Since we have shown that cos4x−sin4x= 1−
2 sin2x, we have successfully proved the trigonometric identity cos4x−sin4x=
1−2 sin2xcos2x.
Question 20
Question
Given that sin θ=5
13 and tan θ < 0, determine the exact values of cos θ,cot θ,
and sec θ.
Solution
Step 1: Since sin θ=5
13 , we can use the Pythagorean identity to find cos θ.
cos2θ= 1 −sin2θ
cos2θ= 1 −(5
13)2
cos2θ= 1 −25
169
cos2θ=144
169
cos θ=12
13
Step 2: Since tan θ < 0and tan θ=sin θ
cos θ, we know that both sin θand cos θ
must have opposite signs. Therefore, cos θis negative.
15
Step 3: Now that we have cos θ=−12
13 , we can find cot θusing the definition
cot θ=1
tan θ.
tan θ=sin θ
cos θ=5/13
−12/13 =−5
12
cot θ=1
tan θ=1
−5
12
=−12
5
Step 4: Finally, we can find sec θusing the identity sec θ=1
cos θ.
sec θ=1
cos θ=1
−12
13
=−13
12
Therefore, the exact values of cos θ,cot θ, and sec θare −12
13 ,−12
5, and −13
12
respectively.
Question 21
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum or difference of familiar angles. Let’s
rewrite 5π
12 as a combination of angles whose trigonometric values we know.
Step 2: Notice that 5π
12 =π
3+2π
3.
Step 3: Now, we can use the sum of angles formula for sine: sin(A+B) =
sin Acos B+ cos Asin B.
Step 4: Applying the formula with A=π
3and B=2π
3, we have:
sin (5π
12 )= sin (π
3+2π
3)= sin (π
3)cos (2π
3)+ cos (π
3)sin (2π
3)
Step 5: Since sin (π
3)=√3
2,cos (π
3)=1
2,cos (2π
3)=−1
2, and sin (2π
3)=√3
2,
we can substitute these values into the equation.
Step 6: So, the calculation becomes:
sin (5π
12 )=√3
2·(−1
2) + 1
2·√3
2=−√3
4+√3
4
Step 7: Simplifying, we get:
sin (5π
12 )= 0
Therefore, sin (5π
12 )= 0.
16
Question 22
Question
Solve the following trigonometric equation for 0◦≤θ≤360◦:
2 sin2θ−3 sin θ−2 = 0
Solution
To solve the equation 2 sin2θ−3 sin θ−2=0, we can treat it as a quadratic
equation in terms of sin θ.
Step 1: Let’s denote sin θas x. The equation becomes:
2x2−3x−2 = 0
Step 2: We can factorize the quadratic equation:
(2x+ 1)(x−2) = 0
Step 3: Set each factor to zero and solve for x:
2x+ 1 = 0 =⇒x=−1
2
x−2 = 0 =⇒x= 2
Step 4: Remember that x= sin θ, so the solutions for θare:
sin θ=−1
2and sin θ= 2
Step 5: However, sin θcan only take values between -1 and 1. Therefore,
the solution sin θ= 2 is extraneous.
Step 6: Let’s find the corresponding angles for sin θ=−1
2. Using the unit
circle or reference angles, we find two possible solutions:
θ1= 210◦and θ2= 330◦
Therefore, the solutions to the trigonometric equation are θ= 210◦and
θ= 330◦.
Question 23
Question
Solve the equation tan2(x)−4 tan(x) + 4 = 0 for xin the interval [0,2π).
17
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
tan(x):
tan2(x)−4 tan(x) + 4 = 0.
Step 2: This can be factored as (tan(x)−2)2= 0.
Step 3: From the factorization, we get tan(x)−2 = 0.
Step 4: Solving for tan(x), we find tan(x) = 2.
Step 5: To find all solutions in the interval [0,2π), we need to consider the
values where tan(x) = 2.
Step 6: The value tan(x) = 2 is true in the first and third quadrants.
Step 7: In the first quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 1.1071 radians.
Step 8: In the third quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 4.2487 radians.
Step 9: Therefore, the solutions to the equation tan2(x)−4 tan(x) + 4 = 0
in the interval [0,2π)are x≈1.1071 and x≈4.2487.
Question 24
Question
If sin θ=3
5and θis in Quadrant II, determine the values of cos θ,tan θ,cot θ,
sec θ, and csc θ.
Solution
Step 1: Since sin θ=3
5and θis in Quadrant II, we can use the Pythagorean
identity to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
=16
25
cos θ=−4
5
Step 2: Now we can find tan θusing the ratios of sine and cosine.
tan θ=sin θ
cos θ=
3
5
−4
5
=−3
4
Step 3: Next, we can find cot θby taking the reciprocal of tan θ.
cot θ=1
tan θ=1
−3
4
=−4
3
18
Step 4: To find sec θ, we use the reciprocal of cos θ.
sec θ=1
cos θ=1
−4
5
=−5
4
Step 5: Finally, we can find csc θusing the reciprocal of sin θ.
csc θ=1
sin θ=1
3
5
=5
3
Therefore, the values of the trigonometric functions for θin Quadrant II are:
cos θ=−4
5,tan θ=−3
4,cot θ=−4
3,sec θ=−5
4,and csc θ=5
3
Question 25
Question
Solve the equation cos(2x) = 2 sin(x)for 0≤x≤2π.
Solution
To solve the equation cos(2x) = 2 sin(x), we will first apply the double angle
identity for cosine: cos(2x) = cos2(x)−sin2(x).
Step 1: Substitute the double angle identity into the equation.
cos2(x)−sin2(x) = 2 sin(x)
Step 2: Replace cos2(x)with 1−sin2(x).
1−sin2(x)−sin2(x) = 2 sin(x)
Step 3: Simplify the equation.
1−2 sin2(x) = 2 sin(x)
Step 4: Rearrange the equation and set it equal to zero.
2 sin2(x) + 2 sin(x)−1 = 0
Step 5: Factor the quadratic equation.
(2 sin(x)−1)(sin(x) + 1) = 0
Step 6: Set each factor to zero and solve for sin(x).
2 sin(x)−1 = 0 or sin(x) + 1 = 0
2 sin(x) = 1 or sin(x) = −1
sin(x) = 1
2or sin(x) = −1
19
Step 7: Find the corresponding values of xfor each solution. 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) = 2 sin(x)for 0≤x≤2π
are x=π
6,5π
6,3π
2.
20
Question 2
Question
Let f(x) = sin(x)+cos(x). Find the amplitude, period, phase shift, and vertical
shift of the function f(x).
Solution
Step 1: The amplitude of a function f(x) = asin(bx +c) + dis given by |a|.
Step 2: The amplitude of f(x) = sin(x) + cos(x)is √12+ 12=√2.
Step 3: The period of a function of the form f(x) = asin(bx +c) + dis 2π
|b|.
Step 4: The period of f(x) = sin(x) + cos(x)is 2π
1= 2π.
Step 5: The phase shift of a function f(x) = asin(bx +c) + dis given by
−c
b.
Step 6: For f(x) = sin(x) + cos(x), there is no phase shift because the
functions are both in their standard positions.
Step 7: The vertical shift of a function f(x) = asin(bx +c) + dis the value
of d.
Step 8: The vertical shift of f(x) = sin(x) + cos(x)is 0since there is no
vertical shift.
Question 3
Question
Find the general solution to the equation sin(2x) = cos(3x)in the interval
[0,2π].
Solution
Step 1: Recall the double angle identity for sine and the reflection identity for
cosine:
sin(2x) = 2 sin(x) cos(x)
cos(θ) = sin (π
2−θ)
Step 2: Substitute these identities into the equation sin(2x) = cos(3x):
2 sin(x) cos(x) = sin (π
2−3x)
Step 3: Expand the right side of the equation using the sine of difference
formula:
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
Step 4: Since sin(2x) = cos(3x)is equivalent to 2 sin(x) cos(x) = cos(3x),
we have the equation 2 sin(x) cos(x) = cos(3x).
Step 5: Now we will convert this equation into terms of sine only:
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = sin (π
2−3x)
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
Step 6: Rearrange the equation to have all terms on one side:
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
2 sin(x) cos(x)−4 sin(x) cos3(x) + 3 sin(x) cos(x) = 0
2 sin(x) cos(x) + 3 sin(x) cos(x)−4 sin(x) cos3(x) = 0
Step 7: Factor out a common sin(x)term:
sin(x)(2 cos(x) + 3 cos(x)−4 cos3(x)) = 0
sin(x)(5 cos(x)−4 cos3(x)) = 0
Step 8: Find the solutions for sin(x) = 0 and 5 cos(x)−4 cos3(x) = 0.
Step 9: For sin(x) = 0, we have x= 0 and x=π.
Step 10: For 5 cos(x)−4 cos3(x) = 0, we can factor out a cos(x):
cos(x)(5 −4 cos2(x)) = 0
cos(x)(5 −4 cos2(x)) = 0
Step 11: We find solutions for cos(x) = 0 and 5−4 cos2(x) = 0.
Step 12: For cos(x) = 0, we have x=π
2and x=3π
2.
Step 13: For 5−4 cos2(x) = 0, we solve for cos(x):
4 cos2(x) = 5
cos2(x)
3
Question 4
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can rewrite 5π
12 as the sum of two angles for which we know the
trigonometric functions. Let’s express 5π
12 as π
4+π
3.
Step 2: Using the angle addition formula for sine, we have:
sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+ cos (π
4)sin (π
3)
Step 3: Recall that sin (π
4)=√2
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and
cos (π
3)=1
2. Substituting these values into the expression, we get:
sin (5π
12 )=√2
2·1
2+√2
2·√3
2
Step 4: Simplifying further, we have:
sin (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of sin (5π
12 )is √2+√6
4.
Question 5
Question
Find the exact value of sin (5π
6·4
3).
Solution
Step 1: Recall the angle addition formula for sine: sin(α±β) = sin(α) cos(β)±
cos(α) sin(β).
Step 2: Rewrite 5π
6·4
3as 5π
6+4π
3.
Step 3: Now we apply the angle addition formula:
sin (5π
6+4π
3)= sin (5π
6)cos (4π
3)+ cos (5π
6)sin (4π
3)
=(−√3
2)(−1
2)+(1
2)(−√3
2)
=√3
4−√3
4
= 0
4
Step 4: Therefore, sin (5π
6·4
3)= 0 .
Question 6
Question
Solve the equation cos(2x) + cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We’ll use the angle addition formula for cosine, which states that
cos(a+b) = cos(a) cos(b)−sin(a) sin(b).
cos(2x) + cos(x) = 0
cos(x+x) + cos(x) = 0
cos(x) cos(x)−sin(x) sin(x) + cos(x) = 0
cos2(x)−sin2(x) + cos(x) = 0
Step 2: Remembering the Pythagorean trigonometric identity sin2(x) +
cos2(x) = 1, we can substitute cos2(x)as 1−sin2(x)in the equation.
(1 −sin2(x)) −sin2(x) + cos(x) = 0
1−2 sin2(x) + cos(x) = 0
Step 3: We can rearrange the equation in terms of sin(x)to get a quadratic
equation.
2 sin2(x)−1 + cos(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
Step 4: Now we can solve this quadratic equation for sin(x). Let’s say
y= sin(x).
2y2−1 + √1−y2= 0
2y2−1 = −√1−y2
(2y2−1)2= 1 −y2
4y4−4y2+ 1 = 1 −y2
4y4−3y2= 0
y2(4y2−3) = 0
5
Step 5: Solving the quadratic equation 4y2−3=0, we find two possible
values for sin(x).
4y2−3 = 0
4y2= 3
y2=3
4
y=±√3
2
Step 6: Since sin(x) = ±√3
2, the possible values for xare x= 60◦,x= 120◦,
x= 240◦,x= 300◦.
Step 7: We need to check the solutions in the original equation.
• For x= 60◦:cos(120◦) + cos(60◦) = −1
2+1
2= 0 (valid)
• For x= 120◦:cos(240◦) + cos(120◦) = −1
2−1
2=−1(not valid)
• For x= 240◦:cos(480◦) + cos(240◦) = −1
2+1
2= 0 (valid)
• For x= 300◦:cos(600◦) + cos(300◦) = −1
2−1
2=−1(not valid)
Step 8: Thus, the solutions for the equation cos(2x) + cos(x)=0for
0◦≤x≤
Question 7
Question
Solve for θin the interval [0,2π]:cos(2θ)−√3 sin(θ) = 0.
Solution
Step 1: Rewrite the equation using double angle identity.
cos2(θ)−sin2(θ)−√3 sin(θ) = 0
Step 2: Substitute sin2(θ) = 1 −cos2(θ)into the equation.
cos2(θ)−(1 −cos2(θ)) −√3 sin(θ) = 0
Step 3: Simplify and rearrange the equation.
2 cos2(θ) + √3 sin(θ)−1 = 0
Step 4: Rewrite cos2(θ)as 1−sin2(θ)
2(1 −sin2(θ)) + √3 sin(θ)−1 = 0
6
Step 5: Expand and rearrange the equation.
2−2 sin2(θ) + √3 sin(θ)−1 = 0
Step 6: Combine like terms.
−2 sin2(θ) + √3 sin(θ) + 1 = 0
Step 7: This is now a quadratic equation in terms of sin(θ). Let u= sin(θ).
−2u2+√3u+ 1 = 0
Step 8: Solve the quadratic equation using the quadratic formula, u=
−b±√b2−4ac
2a.
u=−√3±√(−√3)2−4(−2)(1)
2(−2)
Step 9: Simplify the expression.
u=−√3±√3+8
−4
u=−√3±√11
−4
Step 10: Solve for θby substituting back u= sin(θ).
sin(θ) = −√3±√11
−4
Step 11: Calculate the possible values of θwithin the interval [0,2π]. Since
the given equation did not specify a range for the solution, we will find all
possible solutions.
θ= arcsin (−√3 + √11
−4)
θ= arcsin (−√3−√11
−4)
Therefore, the solutions for θin the interval [0,2π]are θ= arcsin (−√3+√11
−4)
and θ= arcsin (−√3−√11
−4).
Question 8
Question
Evaluate the following expression: tan−1(2 tan(3π/4)
1−tan(3π/4) ).
7
Solution
Step 1: Recall the trigonometric identity tan(A−B) = tan A−tan B
1+tan Atan B.
Step 2: Let A= 3π/4and B= 0, so we have tan(3π/4−0) = tan 3π/4−tan 0
1+tan 3π/4 tan 0 .
Step 3: Simplify the expression to get tan(3π/4) = tan (3π
4)=tan(3π/4)−0
1+tan(3π/4)·0.
Step 4: Since tan(0) = 0, the expression becomes tan(3π/4)
1.
Step 5: Therefore, tan(3π/4) = −1.
Step 6: Substitute back into the original expression: tan−1(2 tan(3π/4)
1−tan(3π/4) )=
tan−1(2(−1)
1−(−1) ).
Step 7: Simplify to get tan−1(−2).
Step 8: Finally, since tan(−π/4) = −1, the answer is −π
4.
Question 9
Question
Find the exact value of tan (4π
3).
Solution
Step 1: Recall that the tangent function is defined as tan(θ) = sin(θ)
cos(θ).
Step 2: First, let’s find the sine and cosine of 4π
3using the unit circle.
Step 3: To find the sine of 4π
3, we look at the point on the unit circle
corresponding to this angle. Since 4π
3is in the third quadrant, the y-coordinate
of the point is sin (4π
3)=−√3
2.
Step 4: To find the cosine of 4π
3, we use the definition of cosine as the x-
coordinate of the point. Since 4π
3is in the third quadrant, the x-coordinate of
the point is −1
2. Therefore, cos (4π
3)=−1
2.
Step 5: Now, we can find the value of tan (4π
3)by using the definition of
tangent and the values of sine and cosine we found earlier.
tan (4π
3)=sin (4π
3)
cos (4π
3)=−√3
2
−1
2
=√3
Therefore, tan (4π
3)=√3.
Question 10
Question
Prove the following trigonometric identity:
1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x)
8
Solution
To prove the given trigonometric identity, we will start with the right-hand side
and manipulate it to match the left-hand side.
RHS =sin(x)−cos(x)
sin(2x)
=sin(x)−cos(x)
2 sin(x) cos(x)(Using the double-angle identity for sine)
=sin(x)
2 sin(x) cos(x)−cos(x)
2 sin(x) cos(x)
=1
2 cos(x)−1
2 sin(x)(Canceling sin(x)and cos(x))
=
1
cos(x)−1
sin(x)
2
=sin(x)−cos(x)
2 sin(x) cos(x)
=sin(x)−cos(x)
sin(x) + cos(x)(Using the double-angle identity for sine)
=LHS
Therefore, we have proven that 1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x).
Question 11
Question
Prove that sin(2θ) = 2 sin(θ) cos(θ)using the angle addition formula: sin(A+B) =
sin(A) cos(B) + cos(A) sin(B).
Solution
Step 1: Let A=B=θ, so we have:
sin(2θ) = sin(θ+θ)
Step 2: Apply the angle addition formula with A=θand B=θ:
sin(θ+θ) = sin(θ) cos(θ) + cos(θ) sin(θ)
Step 3: Simplify the right-hand side of the equation:
sin(θ) cos(θ) + cos(θ) sin(θ) = 2 sin(θ) cos(θ)
Step 4: Therefore, we have shown that sin(2θ) = 2 sin(θ) cos(θ).
9
Question 12
Question
Let f(x) = 3 sin(2x)−√3 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Identify the Amplitude
The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
given by √A2+C2.
In this case, the amplitude of f(x) = 3 sin(2x)−√3 cos(2x)is √32+ (−√3)2=
√9 + 3 = √12 = 2√3.
Therefore, the amplitude of f(x)is 2√3.
Step 2: Identify the Period
The period of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is given
by 2π
|B|.
In this case, the period of f(x) = 3 sin(2x)−√3 cos(2x)is 2π
|2|=π.
Therefore, the period of f(x)is π.
Step 3: Identify the Phase Shift
To find the phase shift of a function of the form f(x) = Asin(Bx) +
Ccos(Bx), we need to find the value of xthat makes Bx equal to 0or π
2.
In this case, B= 2, so the phase shift of f(x) = 3 sin(2x)−√3 cos(2x)is 0.
Therefore, the phase shift of f(x)is 0.
Step 4: Identify the Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
simply the value of C.
In this case, the vertical shift of f(x) = 3 sin(2x)−√3 cos(2x)is −√3.
Therefore, the vertical shift of f(x)is −√3.
Question 13
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: First, let’s express 5π
12 in terms of common angles. Since π
6and π
4are
common angles, we can rewrite 5π
12 as a combination of these angles:
5π
12 =π
4+π
6
10
Step 2: Next, we will apply the sum-to-product trigonometric identity:
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Using the sum-to-product identity, we can rewrite sin (5π
12 )as:
sin (5π
12 )= sin (π
4+π
6)
= sin (π
4)cos (π
6)+ cos (π
4)sin (π
6)
Step 4: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2. Substituting these values in, we get:
sin (5π
12 )=1
√2·√3
2+1
√2·1
2
Step 5: Simplifying further, we have:
sin (5π
12 )=√3
2√2+1
2√2
=√3+1
2√2
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 14
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles. Start by dividing
the given angle by 2.
5π
12 =(4 + 1)π
12 =4π
12 +π
12 =π
3+π
12
Step 2: Next, we express π
12 in terms of common trigonometric values to
simplify the expression.
π
12 =π
6∇ · 2 = 1
2sin (π
6)=1
2(1
2)=1
4
11
Step 3: Substitute π
12 back into the expression for sin (5π
12 ).
sin (5π
12 )= sin (π
3+π
12)= sin π
3cos π
12 + cos π
3sin π
12
Step 4: Recall the values of sine and cosine for common angles.
sin π
3=√3
2,cos π
3=1
2,sin π
12 =√6−√2
4,cos π
12 =√6 + √2
4
Step 5: Substitute these values into the expression for sin (5π
12 ).
sin (5π
12 )=√3
2·√6 + √2
4+1
2·√6−√2
4
sin (5π
12 )=√18 + √6 + √6−√2
8=2√6 + √18 −√2
8=2√6+3√2
8=√6+3√2
4
Therefore, the exact value of sin (5π
12 )is √6+3√2
4.
Question 15
Question
Let f(x) = tan (1
2x). Find the period of the function f(x).
Solution
Step 1: Recall that the period of the function y= tan(ax)is π
|a|. Step 2: In
this case, a=1
2. Step 3: Therefore, the period of the function f(x) = tan (1
2x)
is π
|1
2|= 2π. Step 4: Thus, the period of the function f(x)is 2π.
Question 16
Question
Let f(x) = 2 sin(x)+3 cos(x)for 0≤x≤2π. Determine the absolute maximum
and minimum values of f(x)on this interval.
12
Solution
Step 1: Find critical points by setting the derivative of f(x)equal to 0.
d
dx[2 sin(x) + 3 cos(x)] = 2 cos(x)−3 sin(x)
2 cos(x)−3 sin(x) = 0
2 cos(x) = 3 sin(x)
2
3= tan(x)
x= arctan (2
3)
Step 2: Determine the values of f(x)at the critical point and at the end-
points of the interval.
f(0) = 2 sin(0) + 3 cos(0) = 0 + 3 = 3
f(2π) = 2 sin(2π) + 3 cos(2π) = 0 + 3 = 3
f(arctan (2
3))= 2 sin (arctan (2
3))+ 3 cos (arctan (2
3))
Step 3: Use trigonometric identities to simplify f(arctan (2
3)).
Let sin(θ) = 2
√22+ 32=2
√13 and
cos(θ) = 3
√13
f(arctan (2
3))= 2 ∗2
√13 + 3 ∗3
√13
=4+9
√13
=13
√13
=√13
Step 4: Compare the values of f(x)at the critical point and endpoints to
determine the absolute maximum and minimum values. The minimum value
of f(x)is 3 (at x= 0 and x= 2π) and the maximum value is √13 at x=
arctan (2
3).
Question 17
Question
Find the exact value of sin (5π
12 ).
13
Solution
Step 1: Begin by expressing 5π
12 as the sum or difference of two special angles
that you know the exact sine value. Step 2: Since 5π=3π
4+π
3, we can
write 5π
12 =3π
12 +4π
12 =π
4+π
3. Step 3: Now, use the sum-to-product identities
to rewrite sin (5π
12 )in terms of known values. Step 4: Applying the sum-to-
product identity, we have sin (5π
12 )= sin (π
4+π
3)= sin π
4cos π
3+ cos π
4sin π
3.
Step 5: Recall that sin π
4=1
√2and cos π
3=1
2, so sin (5π
12 )=1
√2·1
2+1
√2·√3
2.
Step 6: Simplify the expression to find the exact value of sin (5π
12 ). Step 7:
sin (5π
12 )=1
2√2+√3
2√2=1+√3
2√2. Step 8: Therefore, the exact value of sin (5π
12 )is
1+√3
2√2.
Question 18
Question
Let f(x) = 2 sin(x
2)and g(x) = cos(x
2). Find the function h(x)such that
h(x) = f(x)·g(x).
Solution
Step 1: Write out the functions f(x)and g(x):
f(x) = 2 sin (x
2)
g(x) = cos (x
2)
Step 2: Find h(x)by multiplying f(x)and g(x):
h(x) = f(x)·g(x) = 2 sin (x
2)cos (x
2)
Step 3: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to simplify
the expression:
h(x) = sin(x)
Therefore, h(x) = sin(x).
Question 19
Question
Prove the following trigonometric identity:
cos4x−sin4x= 1 −2 sin2xcos2x
14
Solution
To prove the trigonometric identity cos4x−sin4x= 1 −2 sin2xcos2x, we will
start with the left-hand side and manipulate it to match the right-hand side.
Step 1: Begin with the left-hand side of the given identity: cos4x−sin4x.
cos4x−sin4x
Step 2: Rewrite cos4xin terms of sin2xusing the Pythagorean identity for
cosine: cos2x= 1 −sin2x.
(cos2x)2−sin4x= (1 −sin2x)2−sin4x
Step 3: Expand and simplify the expression.
(1 −2 sin2x+ sin4x)−sin4x= 1 −2 sin2x+ sin4x−sin4x
= 1 −2 sin2x
Step 4: Compare the simplified expression with the right-hand side of the
given identity: 1−2 sin2xcos2x. Since we have shown that cos4x−sin4x= 1−
2 sin2x, we have successfully proved the trigonometric identity cos4x−sin4x=
1−2 sin2xcos2x.
Question 20
Question
Given that sin θ=5
13 and tan θ < 0, determine the exact values of cos θ,cot θ,
and sec θ.
Solution
Step 1: Since sin θ=5
13 , we can use the Pythagorean identity to find cos θ.
cos2θ= 1 −sin2θ
cos2θ= 1 −(5
13)2
cos2θ= 1 −25
169
cos2θ=144
169
cos θ=12
13
Step 2: Since tan θ < 0and tan θ=sin θ
cos θ, we know that both sin θand cos θ
must have opposite signs. Therefore, cos θis negative.
15
Step 3: Now that we have cos θ=−12
13 , we can find cot θusing the definition
cot θ=1
tan θ.
tan θ=sin θ
cos θ=5/13
−12/13 =−5
12
cot θ=1
tan θ=1
−5
12
=−12
5
Step 4: Finally, we can find sec θusing the identity sec θ=1
cos θ.
sec θ=1
cos θ=1
−12
13
=−13
12
Therefore, the exact values of cos θ,cot θ, and sec θare −12
13 ,−12
5, and −13
12
respectively.
Question 21
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum or difference of familiar angles. Let’s
rewrite 5π
12 as a combination of angles whose trigonometric values we know.
Step 2: Notice that 5π
12 =π
3+2π
3.
Step 3: Now, we can use the sum of angles formula for sine: sin(A+B) =
sin Acos B+ cos Asin B.
Step 4: Applying the formula with A=π
3and B=2π
3, we have:
sin (5π
12 )= sin (π
3+2π
3)= sin (π
3)cos (2π
3)+ cos (π
3)sin (2π
3)
Step 5: Since sin (π
3)=√3
2,cos (π
3)=1
2,cos (2π
3)=−1
2, and sin (2π
3)=√3
2,
we can substitute these values into the equation.
Step 6: So, the calculation becomes:
sin (5π
12 )=√3
2·(−1
2) + 1
2·√3
2=−√3
4+√3
4
Step 7: Simplifying, we get:
sin (5π
12 )= 0
Therefore, sin (5π
12 )= 0.
16
Question 22
Question
Solve the following trigonometric equation for 0◦≤θ≤360◦:
2 sin2θ−3 sin θ−2 = 0
Solution
To solve the equation 2 sin2θ−3 sin θ−2=0, we can treat it as a quadratic
equation in terms of sin θ.
Step 1: Let’s denote sin θas x. The equation becomes:
2x2−3x−2 = 0
Step 2: We can factorize the quadratic equation:
(2x+ 1)(x−2) = 0
Step 3: Set each factor to zero and solve for x:
2x+ 1 = 0 =⇒x=−1
2
x−2 = 0 =⇒x= 2
Step 4: Remember that x= sin θ, so the solutions for θare:
sin θ=−1
2and sin θ= 2
Step 5: However, sin θcan only take values between -1 and 1. Therefore,
the solution sin θ= 2 is extraneous.
Step 6: Let’s find the corresponding angles for sin θ=−1
2. Using the unit
circle or reference angles, we find two possible solutions:
θ1= 210◦and θ2= 330◦
Therefore, the solutions to the trigonometric equation are θ= 210◦and
θ= 330◦.
Question 23
Question
Solve the equation tan2(x)−4 tan(x) + 4 = 0 for xin the interval [0,2π).
17
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
tan(x):
tan2(x)−4 tan(x) + 4 = 0.
Step 2: This can be factored as (tan(x)−2)2= 0.
Step 3: From the factorization, we get tan(x)−2 = 0.
Step 4: Solving for tan(x), we find tan(x) = 2.
Step 5: To find all solutions in the interval [0,2π), we need to consider the
values where tan(x) = 2.
Step 6: The value tan(x) = 2 is true in the first and third quadrants.
Step 7: In the first quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 1.1071 radians.
Step 8: In the third quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 4.2487 radians.
Step 9: Therefore, the solutions to the equation tan2(x)−4 tan(x) + 4 = 0
in the interval [0,2π)are x≈1.1071 and x≈4.2487.
Question 24
Question
If sin θ=3
5and θis in Quadrant II, determine the values of cos θ,tan θ,cot θ,
sec θ, and csc θ.
Solution
Step 1: Since sin θ=3
5and θis in Quadrant II, we can use the Pythagorean
identity to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
=16
25
cos θ=−4
5
Step 2: Now we can find tan θusing the ratios of sine and cosine.
tan θ=sin θ
cos θ=
3
5
−4
5
=−3
4
Step 3: Next, we can find cot θby taking the reciprocal of tan θ.
cot θ=1
tan θ=1
−3
4
=−4
3
18
Step 4: To find sec θ, we use the reciprocal of cos θ.
sec θ=1
cos θ=1
−4
5
=−5
4
Step 5: Finally, we can find csc θusing the reciprocal of sin θ.
csc θ=1
sin θ=1
3
5
=5
3
Therefore, the values of the trigonometric functions for θin Quadrant II are:
cos θ=−4
5,tan θ=−3
4,cot θ=−4
3,sec θ=−5
4,and csc θ=5
3
Question 25
Question
Solve the equation cos(2x) = 2 sin(x)for 0≤x≤2π.
Solution
To solve the equation cos(2x) = 2 sin(x), we will first apply the double angle
identity for cosine: cos(2x) = cos2(x)−sin2(x).
Step 1: Substitute the double angle identity into the equation.
cos2(x)−sin2(x) = 2 sin(x)
Step 2: Replace cos2(x)with 1−sin2(x).
1−sin2(x)−sin2(x) = 2 sin(x)
Step 3: Simplify the equation.
1−2 sin2(x) = 2 sin(x)
Step 4: Rearrange the equation and set it equal to zero.
2 sin2(x) + 2 sin(x)−1 = 0
Step 5: Factor the quadratic equation.
(2 sin(x)−1)(sin(x) + 1) = 0
Step 6: Set each factor to zero and solve for sin(x).
2 sin(x)−1 = 0 or sin(x) + 1 = 0
2 sin(x) = 1 or sin(x) = −1
sin(x) = 1
2or sin(x) = −1
19
Step 7: Find the corresponding values of xfor each solution. 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) = 2 sin(x)for 0≤x≤2π
are x=π
6,5π
6,3π
2.
20
Question 2
Question
Let f(x) = sin(x)+cos(x). Find the amplitude, period, phase shift, and vertical
shift of the function f(x).
Solution
Step 1: The amplitude of a function f(x) = asin(bx +c) + dis given by |a|.
Step 2: The amplitude of f(x) = sin(x) + cos(x)is √12+ 12=√2.
Step 3: The period of a function of the form f(x) = asin(bx +c) + dis 2π
|b|.
Step 4: The period of f(x) = sin(x) + cos(x)is 2π
1= 2π.
Step 5: The phase shift of a function f(x) = asin(bx +c) + dis given by
−c
b.
Step 6: For f(x) = sin(x) + cos(x), there is no phase shift because the
functions are both in their standard positions.
Step 7: The vertical shift of a function f(x) = asin(bx +c) + dis the value
of d.
Step 8: The vertical shift of f(x) = sin(x) + cos(x)is 0since there is no
vertical shift.
Question 3
Question
Find the general solution to the equation sin(2x) = cos(3x)in the interval
[0,2π].
Solution
Step 1: Recall the double angle identity for sine and the reflection identity for
cosine:
sin(2x) = 2 sin(x) cos(x)
cos(θ) = sin (π
2−θ)
Step 2: Substitute these identities into the equation sin(2x) = cos(3x):
2 sin(x) cos(x) = sin (π
2−3x)
Step 3: Expand the right side of the equation using the sine of difference
formula:
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
Step 4: Since sin(2x) = cos(3x)is equivalent to 2 sin(x) cos(x) = cos(3x),
we have the equation 2 sin(x) cos(x) = cos(3x).
Step 5: Now we will convert this equation into terms of sine only:
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = sin (π
2−3x)
2 sin(x) cos(x) = sin (π
2)cos(3x)−cos (π
2)sin(3x)
2 sin(x) cos(x) = 1 ·cos(3x)−0·sin(3x)
2 sin(x) cos(x) = cos(3x)
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
Step 6: Rearrange the equation to have all terms on one side:
2 sin(x) cos(x) = 4 sin(x) cos3(x)−3 sin(x) cos(x)
2 sin(x) cos(x)−4 sin(x) cos3(x) + 3 sin(x) cos(x) = 0
2 sin(x) cos(x) + 3 sin(x) cos(x)−4 sin(x) cos3(x) = 0
Step 7: Factor out a common sin(x)term:
sin(x)(2 cos(x) + 3 cos(x)−4 cos3(x)) = 0
sin(x)(5 cos(x)−4 cos3(x)) = 0
Step 8: Find the solutions for sin(x) = 0 and 5 cos(x)−4 cos3(x) = 0.
Step 9: For sin(x) = 0, we have x= 0 and x=π.
Step 10: For 5 cos(x)−4 cos3(x) = 0, we can factor out a cos(x):
cos(x)(5 −4 cos2(x)) = 0
cos(x)(5 −4 cos2(x)) = 0
Step 11: We find solutions for cos(x) = 0 and 5−4 cos2(x) = 0.
Step 12: For cos(x) = 0, we have x=π
2and x=3π
2.
Step 13: For 5−4 cos2(x) = 0, we solve for cos(x):
4 cos2(x) = 5
cos2(x)
3
Question 4
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can rewrite 5π
12 as the sum of two angles for which we know the
trigonometric functions. Let’s express 5π
12 as π
4+π
3.
Step 2: Using the angle addition formula for sine, we have:
sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+ cos (π
4)sin (π
3)
Step 3: Recall that sin (π
4)=√2
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and
cos (π
3)=1
2. Substituting these values into the expression, we get:
sin (5π
12 )=√2
2·1
2+√2
2·√3
2
Step 4: Simplifying further, we have:
sin (5π
12 )=√2
4+√6
4=√2 + √6
4
Therefore, the exact value of sin (5π
12 )is √2+√6
4.
Question 5
Question
Find the exact value of sin (5π
6·4
3).
Solution
Step 1: Recall the angle addition formula for sine: sin(α±β) = sin(α) cos(β)±
cos(α) sin(β).
Step 2: Rewrite 5π
6·4
3as 5π
6+4π
3.
Step 3: Now we apply the angle addition formula:
sin (5π
6+4π
3)= sin (5π
6)cos (4π
3)+ cos (5π
6)sin (4π
3)
=(−√3
2)(−1
2)+(1
2)(−√3
2)
=√3
4−√3
4
= 0
4
Step 4: Therefore, sin (5π
6·4
3)= 0 .
Question 6
Question
Solve the equation cos(2x) + cos(x) = 0 for 0◦≤x≤360◦.
Solution
Step 1: We’ll use the angle addition formula for cosine, which states that
cos(a+b) = cos(a) cos(b)−sin(a) sin(b).
cos(2x) + cos(x) = 0
cos(x+x) + cos(x) = 0
cos(x) cos(x)−sin(x) sin(x) + cos(x) = 0
cos2(x)−sin2(x) + cos(x) = 0
Step 2: Remembering the Pythagorean trigonometric identity sin2(x) +
cos2(x) = 1, we can substitute cos2(x)as 1−sin2(x)in the equation.
(1 −sin2(x)) −sin2(x) + cos(x) = 0
1−2 sin2(x) + cos(x) = 0
Step 3: We can rearrange the equation in terms of sin(x)to get a quadratic
equation.
2 sin2(x)−1 + cos(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
2 sin2(x)−1 + √1−sin2(x) = 0
Step 4: Now we can solve this quadratic equation for sin(x). Let’s say
y= sin(x).
2y2−1 + √1−y2= 0
2y2−1 = −√1−y2
(2y2−1)2= 1 −y2
4y4−4y2+ 1 = 1 −y2
4y4−3y2= 0
y2(4y2−3) = 0
5
Step 5: Solving the quadratic equation 4y2−3=0, we find two possible
values for sin(x).
4y2−3 = 0
4y2= 3
y2=3
4
y=±√3
2
Step 6: Since sin(x) = ±√3
2, the possible values for xare x= 60◦,x= 120◦,
x= 240◦,x= 300◦.
Step 7: We need to check the solutions in the original equation.
• For x= 60◦:cos(120◦) + cos(60◦) = −1
2+1
2= 0 (valid)
• For x= 120◦:cos(240◦) + cos(120◦) = −1
2−1
2=−1(not valid)
• For x= 240◦:cos(480◦) + cos(240◦) = −1
2+1
2= 0 (valid)
• For x= 300◦:cos(600◦) + cos(300◦) = −1
2−1
2=−1(not valid)
Step 8: Thus, the solutions for the equation cos(2x) + cos(x)=0for
0◦≤x≤
Question 7
Question
Solve for θin the interval [0,2π]:cos(2θ)−√3 sin(θ) = 0.
Solution
Step 1: Rewrite the equation using double angle identity.
cos2(θ)−sin2(θ)−√3 sin(θ) = 0
Step 2: Substitute sin2(θ) = 1 −cos2(θ)into the equation.
cos2(θ)−(1 −cos2(θ)) −√3 sin(θ) = 0
Step 3: Simplify and rearrange the equation.
2 cos2(θ) + √3 sin(θ)−1 = 0
Step 4: Rewrite cos2(θ)as 1−sin2(θ)
2(1 −sin2(θ)) + √3 sin(θ)−1 = 0
6
Step 5: Expand and rearrange the equation.
2−2 sin2(θ) + √3 sin(θ)−1 = 0
Step 6: Combine like terms.
−2 sin2(θ) + √3 sin(θ) + 1 = 0
Step 7: This is now a quadratic equation in terms of sin(θ). Let u= sin(θ).
−2u2+√3u+ 1 = 0
Step 8: Solve the quadratic equation using the quadratic formula, u=
−b±√b2−4ac
2a.
u=−√3±√(−√3)2−4(−2)(1)
2(−2)
Step 9: Simplify the expression.
u=−√3±√3+8
−4
u=−√3±√11
−4
Step 10: Solve for θby substituting back u= sin(θ).
sin(θ) = −√3±√11
−4
Step 11: Calculate the possible values of θwithin the interval [0,2π]. Since
the given equation did not specify a range for the solution, we will find all
possible solutions.
θ= arcsin (−√3 + √11
−4)
θ= arcsin (−√3−√11
−4)
Therefore, the solutions for θin the interval [0,2π]are θ= arcsin (−√3+√11
−4)
and θ= arcsin (−√3−√11
−4).
Question 8
Question
Evaluate the following expression: tan−1(2 tan(3π/4)
1−tan(3π/4) ).
7
Solution
Step 1: Recall the trigonometric identity tan(A−B) = tan A−tan B
1+tan Atan B.
Step 2: Let A= 3π/4and B= 0, so we have tan(3π/4−0) = tan 3π/4−tan 0
1+tan 3π/4 tan 0 .
Step 3: Simplify the expression to get tan(3π/4) = tan (3π
4)=tan(3π/4)−0
1+tan(3π/4)·0.
Step 4: Since tan(0) = 0, the expression becomes tan(3π/4)
1.
Step 5: Therefore, tan(3π/4) = −1.
Step 6: Substitute back into the original expression: tan−1(2 tan(3π/4)
1−tan(3π/4) )=
tan−1(2(−1)
1−(−1) ).
Step 7: Simplify to get tan−1(−2).
Step 8: Finally, since tan(−π/4) = −1, the answer is −π
4.
Question 9
Question
Find the exact value of tan (4π
3).
Solution
Step 1: Recall that the tangent function is defined as tan(θ) = sin(θ)
cos(θ).
Step 2: First, let’s find the sine and cosine of 4π
3using the unit circle.
Step 3: To find the sine of 4π
3, we look at the point on the unit circle
corresponding to this angle. Since 4π
3is in the third quadrant, the y-coordinate
of the point is sin (4π
3)=−√3
2.
Step 4: To find the cosine of 4π
3, we use the definition of cosine as the x-
coordinate of the point. Since 4π
3is in the third quadrant, the x-coordinate of
the point is −1
2. Therefore, cos (4π
3)=−1
2.
Step 5: Now, we can find the value of tan (4π
3)by using the definition of
tangent and the values of sine and cosine we found earlier.
tan (4π
3)=sin (4π
3)
cos (4π
3)=−√3
2
−1
2
=√3
Therefore, tan (4π
3)=√3.
Question 10
Question
Prove the following trigonometric identity:
1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x)
8
Solution
To prove the given trigonometric identity, we will start with the right-hand side
and manipulate it to match the left-hand side.
RHS =sin(x)−cos(x)
sin(2x)
=sin(x)−cos(x)
2 sin(x) cos(x)(Using the double-angle identity for sine)
=sin(x)
2 sin(x) cos(x)−cos(x)
2 sin(x) cos(x)
=1
2 cos(x)−1
2 sin(x)(Canceling sin(x)and cos(x))
=
1
cos(x)−1
sin(x)
2
=sin(x)−cos(x)
2 sin(x) cos(x)
=sin(x)−cos(x)
sin(x) + cos(x)(Using the double-angle identity for sine)
=LHS
Therefore, we have proven that 1
sin(x) + cos(x)=sin(x)−cos(x)
sin(2x).
Question 11
Question
Prove that sin(2θ) = 2 sin(θ) cos(θ)using the angle addition formula: sin(A+B) =
sin(A) cos(B) + cos(A) sin(B).
Solution
Step 1: Let A=B=θ, so we have:
sin(2θ) = sin(θ+θ)
Step 2: Apply the angle addition formula with A=θand B=θ:
sin(θ+θ) = sin(θ) cos(θ) + cos(θ) sin(θ)
Step 3: Simplify the right-hand side of the equation:
sin(θ) cos(θ) + cos(θ) sin(θ) = 2 sin(θ) cos(θ)
Step 4: Therefore, we have shown that sin(2θ) = 2 sin(θ) cos(θ).
9
Question 12
Question
Let f(x) = 3 sin(2x)−√3 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: Identify the Amplitude
The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
given by √A2+C2.
In this case, the amplitude of f(x) = 3 sin(2x)−√3 cos(2x)is √32+ (−√3)2=
√9 + 3 = √12 = 2√3.
Therefore, the amplitude of f(x)is 2√3.
Step 2: Identify the Period
The period of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is given
by 2π
|B|.
In this case, the period of f(x) = 3 sin(2x)−√3 cos(2x)is 2π
|2|=π.
Therefore, the period of f(x)is π.
Step 3: Identify the Phase Shift
To find the phase shift of a function of the form f(x) = Asin(Bx) +
Ccos(Bx), we need to find the value of xthat makes Bx equal to 0or π
2.
In this case, B= 2, so the phase shift of f(x) = 3 sin(2x)−√3 cos(2x)is 0.
Therefore, the phase shift of f(x)is 0.
Step 4: Identify the Vertical Shift
The vertical shift of a function of the form f(x) = Asin(Bx) + Ccos(Bx)is
simply the value of C.
In this case, the vertical shift of f(x) = 3 sin(2x)−√3 cos(2x)is −√3.
Therefore, the vertical shift of f(x)is −√3.
Question 13
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: First, let’s express 5π
12 in terms of common angles. Since π
6and π
4are
common angles, we can rewrite 5π
12 as a combination of these angles:
5π
12 =π
4+π
6
10
Step 2: Next, we will apply the sum-to-product trigonometric identity:
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Using the sum-to-product identity, we can rewrite sin (5π
12 )as:
sin (5π
12 )= sin (π
4+π
6)
= sin (π
4)cos (π
6)+ cos (π
4)sin (π
6)
Step 4: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2. Substituting these values in, we get:
sin (5π
12 )=1
√2·√3
2+1
√2·1
2
Step 5: Simplifying further, we have:
sin (5π
12 )=√3
2√2+1
2√2
=√3+1
2√2
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 14
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles. Start by dividing
the given angle by 2.
5π
12 =(4 + 1)π
12 =4π
12 +π
12 =π
3+π
12
Step 2: Next, we express π
12 in terms of common trigonometric values to
simplify the expression.
π
12 =π
6∇ · 2 = 1
2sin (π
6)=1
2(1
2)=1
4
11
Step 3: Substitute π
12 back into the expression for sin (5π
12 ).
sin (5π
12 )= sin (π
3+π
12)= sin π
3cos π
12 + cos π
3sin π
12
Step 4: Recall the values of sine and cosine for common angles.
sin π
3=√3
2,cos π
3=1
2,sin π
12 =√6−√2
4,cos π
12 =√6 + √2
4
Step 5: Substitute these values into the expression for sin (5π
12 ).
sin (5π
12 )=√3
2·√6 + √2
4+1
2·√6−√2
4
sin (5π
12 )=√18 + √6 + √6−√2
8=2√6 + √18 −√2
8=2√6+3√2
8=√6+3√2
4
Therefore, the exact value of sin (5π
12 )is √6+3√2
4.
Question 15
Question
Let f(x) = tan (1
2x). Find the period of the function f(x).
Solution
Step 1: Recall that the period of the function y= tan(ax)is π
|a|. Step 2: In
this case, a=1
2. Step 3: Therefore, the period of the function f(x) = tan (1
2x)
is π
|1
2|= 2π. Step 4: Thus, the period of the function f(x)is 2π.
Question 16
Question
Let f(x) = 2 sin(x)+3 cos(x)for 0≤x≤2π. Determine the absolute maximum
and minimum values of f(x)on this interval.
12
Solution
Step 1: Find critical points by setting the derivative of f(x)equal to 0.
d
dx[2 sin(x) + 3 cos(x)] = 2 cos(x)−3 sin(x)
2 cos(x)−3 sin(x) = 0
2 cos(x) = 3 sin(x)
2
3= tan(x)
x= arctan (2
3)
Step 2: Determine the values of f(x)at the critical point and at the end-
points of the interval.
f(0) = 2 sin(0) + 3 cos(0) = 0 + 3 = 3
f(2π) = 2 sin(2π) + 3 cos(2π) = 0 + 3 = 3
f(arctan (2
3))= 2 sin (arctan (2
3))+ 3 cos (arctan (2
3))
Step 3: Use trigonometric identities to simplify f(arctan (2
3)).
Let sin(θ) = 2
√22+ 32=2
√13 and
cos(θ) = 3
√13
f(arctan (2
3))= 2 ∗2
√13 + 3 ∗3
√13
=4+9
√13
=13
√13
=√13
Step 4: Compare the values of f(x)at the critical point and endpoints to
determine the absolute maximum and minimum values. The minimum value
of f(x)is 3 (at x= 0 and x= 2π) and the maximum value is √13 at x=
arctan (2
3).
Question 17
Question
Find the exact value of sin (5π
12 ).
13
Solution
Step 1: Begin by expressing 5π
12 as the sum or difference of two special angles
that you know the exact sine value. Step 2: Since 5π=3π
4+π
3, we can
write 5π
12 =3π
12 +4π
12 =π
4+π
3. Step 3: Now, use the sum-to-product identities
to rewrite sin (5π
12 )in terms of known values. Step 4: Applying the sum-to-
product identity, we have sin (5π
12 )= sin (π
4+π
3)= sin π
4cos π
3+ cos π
4sin π
3.
Step 5: Recall that sin π
4=1
√2and cos π
3=1
2, so sin (5π
12 )=1
√2·1
2+1
√2·√3
2.
Step 6: Simplify the expression to find the exact value of sin (5π
12 ). Step 7:
sin (5π
12 )=1
2√2+√3
2√2=1+√3
2√2. Step 8: Therefore, the exact value of sin (5π
12 )is
1+√3
2√2.
Question 18
Question
Let f(x) = 2 sin(x
2)and g(x) = cos(x
2). Find the function h(x)such that
h(x) = f(x)·g(x).
Solution
Step 1: Write out the functions f(x)and g(x):
f(x) = 2 sin (x
2)
g(x) = cos (x
2)
Step 2: Find h(x)by multiplying f(x)and g(x):
h(x) = f(x)·g(x) = 2 sin (x
2)cos (x
2)
Step 3: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to simplify
the expression:
h(x) = sin(x)
Therefore, h(x) = sin(x).
Question 19
Question
Prove the following trigonometric identity:
cos4x−sin4x= 1 −2 sin2xcos2x
14
Solution
To prove the trigonometric identity cos4x−sin4x= 1 −2 sin2xcos2x, we will
start with the left-hand side and manipulate it to match the right-hand side.
Step 1: Begin with the left-hand side of the given identity: cos4x−sin4x.
cos4x−sin4x
Step 2: Rewrite cos4xin terms of sin2xusing the Pythagorean identity for
cosine: cos2x= 1 −sin2x.
(cos2x)2−sin4x= (1 −sin2x)2−sin4x
Step 3: Expand and simplify the expression.
(1 −2 sin2x+ sin4x)−sin4x= 1 −2 sin2x+ sin4x−sin4x
= 1 −2 sin2x
Step 4: Compare the simplified expression with the right-hand side of the
given identity: 1−2 sin2xcos2x. Since we have shown that cos4x−sin4x= 1−
2 sin2x, we have successfully proved the trigonometric identity cos4x−sin4x=
1−2 sin2xcos2x.
Question 20
Question
Given that sin θ=5
13 and tan θ < 0, determine the exact values of cos θ,cot θ,
and sec θ.
Solution
Step 1: Since sin θ=5
13 , we can use the Pythagorean identity to find cos θ.
cos2θ= 1 −sin2θ
cos2θ= 1 −(5
13)2
cos2θ= 1 −25
169
cos2θ=144
169
cos θ=12
13
Step 2: Since tan θ < 0and tan θ=sin θ
cos θ, we know that both sin θand cos θ
must have opposite signs. Therefore, cos θis negative.
15
Step 3: Now that we have cos θ=−12
13 , we can find cot θusing the definition
cot θ=1
tan θ.
tan θ=sin θ
cos θ=5/13
−12/13 =−5
12
cot θ=1
tan θ=1
−5
12
=−12
5
Step 4: Finally, we can find sec θusing the identity sec θ=1
cos θ.
sec θ=1
cos θ=1
−12
13
=−13
12
Therefore, the exact values of cos θ,cot θ, and sec θare −12
13 ,−12
5, and −13
12
respectively.
Question 21
Question
Find the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum or difference of familiar angles. Let’s
rewrite 5π
12 as a combination of angles whose trigonometric values we know.
Step 2: Notice that 5π
12 =π
3+2π
3.
Step 3: Now, we can use the sum of angles formula for sine: sin(A+B) =
sin Acos B+ cos Asin B.
Step 4: Applying the formula with A=π
3and B=2π
3, we have:
sin (5π
12 )= sin (π
3+2π
3)= sin (π
3)cos (2π
3)+ cos (π
3)sin (2π
3)
Step 5: Since sin (π
3)=√3
2,cos (π
3)=1
2,cos (2π
3)=−1
2, and sin (2π
3)=√3
2,
we can substitute these values into the equation.
Step 6: So, the calculation becomes:
sin (5π
12 )=√3
2·(−1
2) + 1
2·√3
2=−√3
4+√3
4
Step 7: Simplifying, we get:
sin (5π
12 )= 0
Therefore, sin (5π
12 )= 0.
16
Question 22
Question
Solve the following trigonometric equation for 0◦≤θ≤360◦:
2 sin2θ−3 sin θ−2 = 0
Solution
To solve the equation 2 sin2θ−3 sin θ−2=0, we can treat it as a quadratic
equation in terms of sin θ.
Step 1: Let’s denote sin θas x. The equation becomes:
2x2−3x−2 = 0
Step 2: We can factorize the quadratic equation:
(2x+ 1)(x−2) = 0
Step 3: Set each factor to zero and solve for x:
2x+ 1 = 0 =⇒x=−1
2
x−2 = 0 =⇒x= 2
Step 4: Remember that x= sin θ, so the solutions for θare:
sin θ=−1
2and sin θ= 2
Step 5: However, sin θcan only take values between -1 and 1. Therefore,
the solution sin θ= 2 is extraneous.
Step 6: Let’s find the corresponding angles for sin θ=−1
2. Using the unit
circle or reference angles, we find two possible solutions:
θ1= 210◦and θ2= 330◦
Therefore, the solutions to the trigonometric equation are θ= 210◦and
θ= 330◦.
Question 23
Question
Solve the equation tan2(x)−4 tan(x) + 4 = 0 for xin the interval [0,2π).
17
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation in terms of
tan(x):
tan2(x)−4 tan(x) + 4 = 0.
Step 2: This can be factored as (tan(x)−2)2= 0.
Step 3: From the factorization, we get tan(x)−2 = 0.
Step 4: Solving for tan(x), we find tan(x) = 2.
Step 5: To find all solutions in the interval [0,2π), we need to consider the
values where tan(x) = 2.
Step 6: The value tan(x) = 2 is true in the first and third quadrants.
Step 7: In the first quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 1.1071 radians.
Step 8: In the third quadrant, tan(x) = 2 corresponds to an angle measuring
approximately 4.2487 radians.
Step 9: Therefore, the solutions to the equation tan2(x)−4 tan(x) + 4 = 0
in the interval [0,2π)are x≈1.1071 and x≈4.2487.
Question 24
Question
If sin θ=3
5and θis in Quadrant II, determine the values of cos θ,tan θ,cot θ,
sec θ, and csc θ.
Solution
Step 1: Since sin θ=3
5and θis in Quadrant II, we can use the Pythagorean
identity to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
=16
25
cos θ=−4
5
Step 2: Now we can find tan θusing the ratios of sine and cosine.
tan θ=sin θ
cos θ=
3
5
−4
5
=−3
4
Step 3: Next, we can find cot θby taking the reciprocal of tan θ.
cot θ=1
tan θ=1
−3
4
=−4
3
18
Step 4: To find sec θ, we use the reciprocal of cos θ.
sec θ=1
cos θ=1
−4
5
=−5
4
Step 5: Finally, we can find csc θusing the reciprocal of sin θ.
csc θ=1
sin θ=1
3
5
=5
3
Therefore, the values of the trigonometric functions for θin Quadrant II are:
cos θ=−4
5,tan θ=−3
4,cot θ=−4
3,sec θ=−5
4,and csc θ=5
3
Question 25
Question
Solve the equation cos(2x) = 2 sin(x)for 0≤x≤2π.
Solution
To solve the equation cos(2x) = 2 sin(x), we will first apply the double angle
identity for cosine: cos(2x) = cos2(x)−sin2(x).
Step 1: Substitute the double angle identity into the equation.
cos2(x)−sin2(x) = 2 sin(x)
Step 2: Replace cos2(x)with 1−sin2(x).
1−sin2(x)−sin2(x) = 2 sin(x)
Step 3: Simplify the equation.
1−2 sin2(x) = 2 sin(x)
Step 4: Rearrange the equation and set it equal to zero.
2 sin2(x) + 2 sin(x)−1 = 0
Step 5: Factor the quadratic equation.
(2 sin(x)−1)(sin(x) + 1) = 0
Step 6: Set each factor to zero and solve for sin(x).
2 sin(x)−1 = 0 or sin(x) + 1 = 0
2 sin(x) = 1 or sin(x) = −1
sin(x) = 1
2or sin(x) = −1
19
Step 7: Find the corresponding values of xfor each solution. 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) = 2 sin(x)for 0≤x≤2π
are x=π
6,5π
6,3π
2.
20
