MATH 122 - TRIGONOMETRY -
Trigonometric Functions
Question Bank - Set 2
Liberty University
Question 1
Question
Evaluate the following expression: sin2(π
3)+ cos2(2π
3).
Solution
Step 1: Recall the Pythagorean trigonometric identity: sin2(θ) + cos2(θ) = 1 for
any angle θ.
Step 2: Substitute the given angles into the expression:
sin2(π
3)+ cos2(2π
3)
Step 3: Evaluate sin (π
3)and cos (2π
3)using the unit circle or reference
angles:
sin (π
3)=√3
2,cos (2π
3)=−1
2
Step 4: Substitute the trigonometric values back into the expression:
(√3
2)2
+(−1
2)2
Step 5: Simplify the expression:
3
4+1
4
Step 6: Combine the fractions to get the final answer:
= 1
Question 2
Question
Let f(x) = 2 sin(3x)−√3 cos(3x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by R=√A2+C2. In this case, the amplitude is R=√22+ (−√3)2=
√4 + 3 = √7.
Step 2: The period of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by P=2π
|B|. In this case, the period is P=2π
3.
Step 3: To find the phase shift of the function f(x), we need to write f(x)
in the form f(x) = Rsin(B(x−h)) + k, where his the phase shift.
In this case, f(x) = √7 sin(3x−h) + k.
Comparing with f(x) = 2 sin(3x)−√3 cos(3x), we get:
•√7 sin(3x−h) = 2 sin(3x), which implies √7 = 2 and h= 0.
•k=−√3.
Therefore, the phase shift is h= 0.
Step 4: The vertical shift of the function f(x)is given by the constant kin
the form f(x) = Rsin(Bx −h) + k. In this case, the vertical shift is k=−√3.
So, the amplitude of the function is √7, the period is 2π
3, the phase shift is
0, and the vertical shift is −√3.
Question 3
Question
Solve the equation cos(2x) = √2 cos(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute this identity into the given equation to get 2 cos2(x)−1 =
√2 cos(x).
Step 3: Rearrange the equation to get 2 cos2(x)−√2 cos(x)−1 = 0.
Step 4: This is a quadratic equation in terms of cos(x). Let u= cos(x), then
we have 2u2−√2u−1 = 0.
Step 5: Solve the quadratic equation using the quadratic formula: u=
−(−√2)±√(−√2)2−4(2)(−1)
2(2) .
Step 6: Simplify the expression to find the solutions for u:u=√2±√2+8
4.
2
Step 7: Simplify further to get u=√2±√10
4.
Step 8: Since u= cos(x), the solutions for cos(x)are cos(x) = √2+√10
4and
cos(x) = √2−√10
4.
Step 9: Now, find the corresponding values of xwithin the interval [0,2π).
Step 10: From cos(x) = √2+√10
4, we have x= cos−1(√2+√10
4).
Step 11: Similarly, from cos(x) = √2−√10
4, we have x= cos−1(√2−√10
4).
Step 12: Calculate the final values of xwithin the interval [0,2π), and the
solutions to the equation are x=π
5and x=3π
5.
Question 4
Question
Find the exact value of cos (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles in the unit circle.
In this case, we rewrite 5π
12 as π
3+π
4.
Step 2: Using the angle addition formula for cosine, we have
cos (5π
12 )= cos (π
3+π
4)= cos (π
3)cos (π
4)−sin (π
3)sin (π
4).
Step 3: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2.
Step 4: Substitute these values into the formula:
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.
Question 5
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
3
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 given equation to get
2 sin(x) cos(x) = cos(x).
Step 3: Now, we can solve for sin(x):2 sin(x) cos(x) = cos(x)⇒2 sin(x) =
1⇒sin(x) = 1
2.
Step 4: The solutions for sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 6
Question
Solve the equation sin(θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Recall that cos(θ) = sin (θ+π
2). Therefore, the equation sin(θ) =
cos(θ)can be rewritten as sin(θ) = sin (θ+π
2).
Step 2: To solve sin(θ) = sin (θ+π
2), we can use the identity sin(a) = sin(b)
if and only if a=nπ + (−1)nb, where nis an integer.
Step 3: Applying this identity to sin(θ) = sin (θ+π
2), we have θ=nπ +
(−1)n(θ+π
2).
Step 4: Simplifying the above equation gives us two cases: Case 1: When n
is even, we have θ=nπ −θ−π
2, which simplifies to θ=(2n−1)π
2. Case 2:
When nis odd, we have θ=nπ +θ+π
2, which simplifies to θ=(2n+ 1)π
2.
Step 5: Now we need to find the solutions within the interval 0≤θ≤2πfor
both cases: For case 1: θ=π
2,3π
2. For case 2: θ=π
2,3π
2.
Step 6: Therefore, the solutions to the equation sin(θ) = cos(θ)for 0≤θ≤
2πare θ=π
2,3π
2.
Question 7
Question
Find the exact value of sin (5π
12 ).
4
Solution
Step 1: We will use the angle sum identity sin(A+B) = sin Acos B+cos Asin B.
Let A=π
4and B=π
3where A+B=5π
12 . Step 2: Calculate sin (π
4)and
sin (π
3):sin (π
4)=√2
2and sin (π
3)=√3
2. Step 3: Plug the values into the angle
sum identity: sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+cos (π
4)sin (π
3). Step 4:
Substitute the values we calculated: sin (5π
12 )=√2
2·1
2+√2
2·√3
2. Step 5: Simplify
the expression: sin (5π
12 )=√2
4+√6
4=√2+√6
4. Therefore, sin (5π
12 )=√2+√6
4.
Question 8
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles. Let’s write 5π
12 as
3π
12 +2π
12 , which simplifies to π
4+π
6.
Step 2: We know the sine of sum formula, which states that sin(A+B) =
sin Acos B+ cos Asin B.
Step 3: Using the sine of sum formula from Step 2, we can write sin (5π
12 )as
sin (π
4+π
6). This becomes sin (π
4)cos (π
6)+ cos (π
4)sin (π
6).
Step 4: Now, we calculate the values of sin (π
4),cos (π
4),sin (π
6), and cos (π
6).
Step 5: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2.
Step 6: Now substitute the values from Step 5 into the expression from Step
3. We get: 1
√2·√3
2+1
√2·1
2.
Step 7: Simplifying the expression from Step 6, we get √3
2√2+1
2√2.
Step 8: Further simplifying, we get √3+1
2√2.
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 9
Question
Let f(x) = 3 cos(x)−2 sin(x), defined for 0≤x≤2π. Find the maximum and
minimum values of f(x)on this interval.
Solution
Step 1: To find the maximum and minimum values of f(x), we first need to
find the critical points of f(x)where the derivative equals 0. This occurs when
5
f′(x) = 0.
Step 2: Calculate the derivative f′(x):
f′(x) = d
dx(3 cos(x)−2 sin(x)) = −3 sin(x)−2 cos(x).
Step 3: Set f′(x) = 0 to find the critical points:
−3 sin(x)−2 cos(x) = 0.
Step 4: Rearrange the equation:
−2 cos(x) = 3 sin(x) =⇒−2
3=sin(x)
cos(x).
Step 5: Recall that sin(x)
cos(x)= tan(x), so we have:
tan(x) = −2
3.
Step 6: We know that tan(x)cycles through all possible real numbers as x
ranges over 0≤x≤2π. Thus, we solve for x:
x= arctan (−2
3)≈0.588.
Step 7: There is only one critical point between 0and 2π, which is x≈0.588.
Step 8: Now, check the values of f(x)at the critical point and at the end-
points of the interval 0≤x≤2πto determine the maximum and minimum
values.
Step 9: Evaluate f(x)at x= 0, x ≈0.588,and x= 2πto find the maximum
and minimum values of f(x)on the given interval.
Step 10: f(0) = 3 cos(0)−2 sin(0) = 3,f(0.588) = 3 cos(0.588)−2 sin(0.588) ≈
3.77, and f(2π) = 3 cos(2π)−2 sin(2π) = 3.
Step 11: Therefore, the maximum value of f(x)on the interval 0≤x≤2π
is approximately 3.77, and the minimum value is 3.
Question 10
Question
Find the exact value of cos (5π
12 )using trigonometric identities.
Solution
Step 1: We can start by expressing 5π
12 as the sum of two common angles. We
have 5π
12 =π
3+π
4.
6
Step 2: Recall the following trigonometric sum identities:
cos(A+B) = cos Acos B−sin Asin B
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Apply these identities by setting A=π
3and B=π
4. We have:
cos (5π
12 )= cos (π
3+π
4)
= cos π
3cos π
4−sin π
3sin π
4
Step 4: Calculate the cosine and sine values for π
3and π
4:
cos π
3=1
2,sin π
3=√3
2
cos π
4=√2
2,sin π
4=√2
2
Step 5: Substitute these values into the expression:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4
Hence, the exact value of cos (5π
12 )is √2−√6
4.
Question 11
Question
Let f(x) = 2 sin(3x)−3 cos(3x). Find the amplitude, period, and phase shift of
the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by √A2+D2. In this case, A= 2 and D=−3, so
the amplitude is
√22+ (−3)2=√4 + 9 = √13.
7
Step 2: The period of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by 2π
|B|. In this case, B= 3, so the period is
2π
|3|=2π
3.
Step 3: To find the phase shift of the function, we need to solve Bx +C= 0
for x. In this case, B= 3 and C= 0, so the phase shift is
x=−C
B=−0
3= 0.
Therefore, the amplitude of f(x)is √13, the period is 2π
3, and the phase
shift is 0.
Question 12
Question
Solve the equation sin(2θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to rewrite sin(2θ)
in terms of sin(θ)and cos(θ).
sin(2θ) = 2 sin(θ) cos(θ) = cos(θ)
Step 2: Now we have the equation 2 sin(θ) cos(θ) = cos(θ). We can rewrite
this as 2 sin(θ) cos(θ)−cos(θ) = 0.
Step 3: Factor out cos(θ)from the left side of the equation.
cos(θ)(2 sin(θ)−1) = 0
Step 4: Set each factor equal to zero and solve for θ.
cos(θ) = 0 or 2 sin(θ)−1 = 0
Step 5: For cos(θ) = 0, we know that θ=π
2and θ=3π
2.
Step 6: For 2 sin(θ)−1 = 0, solve for sin(θ).
2 sin(θ)−1 = 0
2 sin(θ) = 1
sin(θ) = 1
2
Step 7: The solutions for sin(θ) = 1
2are π
6and 5π
6.
Step 8: Therefore, the solutions to the equation sin(2θ) = cos(θ)for 0≤θ≤
2πare π
6,π
2,5π
6, and 3π
2.
8
Question 13
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find the following expression:
f(x)·g(x)
f(x)−g(x)
Solution
Step 1: Find f(x)·g(x)
f(x)·g(x) = sin(2x)·cos(x)
Step 2: Apply the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)·cos(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos2(x)
Step 3: Find f(x)−g(x)
f(x)−g(x) = sin(2x)−cos(x)
Step 4: Appy the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)−cos(x) = 2 sin(x) cos(x)−cos(x) = cos(x)(2 sin(x)−1)
Step 5: Find the expression f(x)·g(x)
f(x)−g(x)
f(x)·g(x)
f(x)−g(x)=2 sin(x) cos2(x)
cos(x)(2 sin(x)−1)
Step 6: Simplify the expression by canceling out a factor of cos(x)
2 sin(x) cos(x)
2 sin(x)−1
Therefore, the expression f(x)·g(x)
f(x)−g(x)simplifies to 2 sin(x)
2 sin(x)−1.
Question 14
Question
Find the exact value of sin (5π
4).
9
Solution
Step 1: Determine the reference angle. To find the reference angle, we first
need to identify the co-terminal angle that lies between 0 and 2π(0 and 360◦)
that is equivalent to 5π
4. Since 5π
4is greater than π, we subtract πto find the
co-terminal angle: 5π
4−π=π
4. Therefore, the reference angle is π
4.
Step 2: Determine the quadrant. The angle 5π
4lies in the third quadrant
(3π
2<5π
4<2π).
Step 3: Calculate the sine function. In the third quadrant, the sine function
is negative. Since the reference angle π
4has a sine value of √2
2, the sine value
for 5π
4will be:
sin (5π
4)=−√2
2
Question 15
Question
Find the general solution of the equation sin(2x) = −1
2for 0≤x < 2π.
Solution
Step 1: Rewrite the equation in terms of the sine function.
sin(2x) = −1
2
Step 2: Use the double angle identity for sine.
sin(2x) = 2 sin(x) cos(x) = −1
2
Step 3: Express sin(x)in terms of cos(x).
2 sin(x) cos(x) = −1
2
sin(x) = −1
4 cos(x)
Step 4: Use the Pythagorean identity for sine and cosine.
sin2(x) + cos2(x) = 1
(−1
4 cos(x))2
+ cos2(x) = 1
Step 5: Solve for cos(x).
1
16 cos2(x)+ cos2(x) = 1
10
1 + 16 cos2(x) = 16 cos2(x)
17 cos2(x) = 1
cos2(x) = 1
17
cos(x) = ±1
√17
Step 6: Determine the possible solutions for x. Since sin(2x) = −1
2, we
must also have sin(x)<0. Therefore, sin(x) = −1
4 cos(x)will be negative when
cos(x)<0. Thus, the solutions occur in the second and third quadrants.
Step 7: Find the angles in the second and third quadrants where cos(x) =
−1
√17 .
x= cos−1(−1
√17)= 2.6779 radians
Since we need to find the general solution, the solutions in the third quadrant
will be x=π−2.6779.
Therefore, the general solution is:
x= 2.6779 + 2πn or x=π−2.6779 + 2πn
where n∈Z.
Question 16
Question
Let f(θ) = sin θ
1+cos θ. Find the values of θin the interval [0,2π]where f(θ)has a
vertical asymptote.
Solution
Step 1: Identify where f(θ)is undefined.
To find the values of θwhere f(θ)has a vertical asymptote, we need to find
where the denominator of f(θ)equals 0, since division by 0 is undefined. So we
solve 1 + cos θ= 0:
cos θ=−1
Step 2: Find the values of θthat satisfy cos θ=−1in the interval [0,2π].
The cosine function is equal to -1 at θ=πin the interval [0,2π].
Therefore, θ=πis the only value in the interval [0,2π]where f(θ)has a
vertical asymptote.
11
Question 17
Question
Given that sin(θ) = 3
5and θis in Quadrant II, find the exact values of tan(θ)
and sec(θ).
Solution
Step 1: We can use the Pythagorean identity sin2(θ) + cos2(θ) = 1 to find the
value of cos(θ).
cos2(θ) = 1 −sin2(θ)
cos2(θ) = 1 −(3
5)2
= 1 −9
25 =16
25
cos(θ) = ±4
5
Since θis in Quadrant II, cos(θ)is negative. Therefore, cos(θ) = −4
5.
Step 2: Next, we can find the values of tan(θ)and sec(θ)using the definitions
of these trigonometric functions.
tan(θ) = sin(θ)
cos(θ)=
3
5
−4
5
=−3
4
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
Therefore, the exact values of tan(θ)and sec(θ)are −3
4and −5
4, respec-
tively.
Question 18
Question
Given that sin(θ) = 3
5and θis in quadrant II, determine the exact value of
cos(2θ).
Solution
Step 1: Since we know that sin(θ) = 3
5and θis in quadrant II, we can use the
Pythagorean identity to find cos(θ).
cos(θ) = −√1−sin2(θ) = −√1−(3
5)2
=−4
5
12
Step 2: Next, we can use the double angle identity for cosine to find cos(2θ).
cos(2θ) = cos2(θ)−sin2(θ)
=(−4
5)2
−(3
5)2
=16
25 −9
25
=7
25
Therefore, the exact value of cos(2θ)is 7
25 .
Question 19
Question
Let f(θ) = 2 cos2(θ)−3 sin(θ) cos(θ), where θ∈[0,π
2]. Find the maximum
value of f(θ)in the given interval.
Solution
Step 1: Recall the double angle identities for cosine and sine:
cos(2θ) = 2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Rewrite f(θ)using the double angle identity for cosine:
f(θ) = cos(2θ)−3 sin(2θ)
Step 3: To maximize f(θ), we need to find the maximum value of cos(2θ)
and the minimum value of sin(2θ)in the interval θ∈[0,π
2].
Step 4: The maximum value of cos(2θ)in the given interval is 1, which
occurs when θ= 0.
Step 5: The minimum value of sin(2θ)in the interval θ∈[0,π
2]is -1, which
occurs when θ=π
4.
Step 6: Substituting these values into f(θ), we get the maximum value of
f(θ)in the interval:
f(θ) = 1 −3(−1) = 4
Therefore, the maximum value of f(θ)in the interval θ∈[0,π
2]is 4.
Question 20
Question
Find the exact value of sin (3π
4).
13
Solution
Step 1: We know that sin (3π
4)=−sin (π
4)since the sine function has period
2π.
Step 2: Using the angle addition identity sin(A−B) = sin Acos B−cos Asin B,
we have:
sin (3π
4)=−sin (π
4)=−(sin(π/2) cos(π/4) −cos(π/2) sin(π/4))
Step 3: We know that sin(π/2) = 1,cos(π/2) = 0,cos(π/4) = √2
2, and
sin(π/4) = √2
2. Substituting these values, we get:
sin (3π
4)=−(1·√2
2−0·√2
2)
Step 4: Simplifying further:
sin (3π
4)=−√2
2
Therefore, the exact value of sin (3π
4)is −√2
2.
Question 21
Question
Let f(x) = 3 sin(x) + 4 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), we use the formula A=√a2+b2, where
Ais the amplitude, and for f(x) = asin(x)+bcos(x),aand bare the coefficients
of sin(x)and cos(x). 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 of f(x), we know that the period of sin(x)and
cos(x)is 2π. Since there are no coefficients that affect the period, the period of
f(x)is also 2π.
Step 3: To find the phase shift of f(x), we need to set sin(x)and cos(x)
equal to their maximum/minimum values and solve for x. For sin(x), it reaches
its maximum value at x=π
2, and for cos(x), it reaches its maximum value at
x= 0. Comparing this to f(x) = 3 sin(x) + 4 cos(x), we can see that the phase
shift is π
2to the right.
14
Step 4: To find the vertical shift of f(x), we need to find the average of the
maximum and minimum values of f(x). The maximum value of f(x)is 5, and
the minimum value is −5, so the vertical shift is:
(5 + (−5))/2 = 0
Therefore, the amplitude of f(x)is 5, the period is 2π, the phase shift is π
2
to the right, and the vertical shift is 0.
Question 22
Question
Calculate the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles:
5π
12 =π
3+π
4
Step 2: Use the sum-to-product formula for sine:
sin (a
n+b
n)= sin (a
n)cos (b
n)+ cos (a
n)sin (b
n)
So we have:
sin (5π
12 )= sin (π
3+π
4)
sin (5π
12 )= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Substitute in the values of sin(π/3),cos(π/3),sin(π/4), and cos(π/4):
sin (5π
12 )=√3
2·√2
2+1
2·√2
2
sin (5π
12 )=√6
4+√2
4
sin (5π
12 )=√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
15
Question 23
Question
Solve for xin the equation sin(2x) = cos(x)for x∈[0◦,360◦].
Solution
Step 1: Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the given equation to get:
2 sin(x) cos(x) = cos(x)
Step 3: Now we have two cases to consider: Case 1: cos(x)= 0 Divide both
sides by cos(x):
2 sin(x) = 1
Step 4: Solve for sin(x):
sin(x) = 1
2
Step 5: This occurs in the first and second quadrants, so the solutions are
x= 30◦and x= 150◦.
Case 2: cos(x) = 0 This occurs when x= 90◦.
Step 6: Therefore, the solutions to the equation sin(2x) = cos(x)for x∈
[0◦,360◦]are x= 30◦,150◦,and 90◦.
Question 24
Question
Given that sin θ=3
5and θis in Quadrant II, find the exact value of cos (θ
2).
Solution
Step 1: Find cos θusing the Pythagorean identity sin2θ+ cos2θ= 1.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=±4
5
Since θis in Quadrant II, where cosine is negative, cos θ=−4
5.
16
Step 2: Use the half-angle formula for cosine to find cos (θ
2).
cos (θ
2)=±√1 + cos θ
2=±√1−4
5
2=±√1−4
5
2=±√1
10 =±1
√10 =±√10
10
Since θis in Quadrant II, where cosine is negative, cos (θ
2)=−√10
10 .
Question 25
Question
Determine all solutions to the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute the double angle identity into the given equation: 2 cos2(x)−
1 = sin(x).
Step 3: Recall the Pythagorean identity for sine and cosine: sin2(x) +
cos2(x) = 1.
Step 4: Substitute sin2(x)=1−cos2(x)into the equation from step 2:
2 cos2(x)−1 = √1−cos2(x).
Step 5: Squaring both sides of the equation gives 4 cos4(x)−4 cos2(x) + 1 =
1−cos2(x).
Step 6: Simplify the equation: 4 cos4(x)−5 cos2(x) = 0.
Step 7: Factor out a cos2(x)from the equation: cos2(x)(4 cos2(x)−5) = 0.
Step 8: Set each factor equal to zero: cos2(x) = 0 and 4 cos2(x)−5 = 0.
Step 9: Solve the first equation cos2(x) = 0 to find cos(x) = 0. This occurs
at x= 90◦and x= 270◦.
Step 10: Solve the second equation 4 cos2(x)−5=0to find cos(x) = ±√5
2.
This occurs at x= 18.19◦,x= 161.81◦,x= 198.19◦, and x= 341.81◦.
Step 11: Check all solutions in the original equation to verify their validity.
Therefore, all the solutions to the equation cos(2x) = sin(x)for 0◦≤x≤
360◦are x= 18.19◦,90◦,161.81◦,198.19◦,270◦,341.81◦.
17
Question 2
Question
Let f(x) = 2 sin(3x)−√3 cos(3x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by R=√A2+C2. In this case, the amplitude is R=√22+ (−√3)2=
√4 + 3 = √7.
Step 2: The period of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by P=2π
|B|. In this case, the period is P=2π
3.
Step 3: To find the phase shift of the function f(x), we need to write f(x)
in the form f(x) = Rsin(B(x−h)) + k, where his the phase shift.
In this case, f(x) = √7 sin(3x−h) + k.
Comparing with f(x) = 2 sin(3x)−√3 cos(3x), we get:
•√7 sin(3x−h) = 2 sin(3x), which implies √7 = 2 and h= 0.
•k=−√3.
Therefore, the phase shift is h= 0.
Step 4: The vertical shift of the function f(x)is given by the constant kin
the form f(x) = Rsin(Bx −h) + k. In this case, the vertical shift is k=−√3.
So, the amplitude of the function is √7, the period is 2π
3, the phase shift is
0, and the vertical shift is −√3.
Question 3
Question
Solve the equation cos(2x) = √2 cos(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute this identity into the given equation to get 2 cos2(x)−1 =
√2 cos(x).
Step 3: Rearrange the equation to get 2 cos2(x)−√2 cos(x)−1 = 0.
Step 4: This is a quadratic equation in terms of cos(x). Let u= cos(x), then
we have 2u2−√2u−1 = 0.
Step 5: Solve the quadratic equation using the quadratic formula: u=
−(−√2)±√(−√2)2−4(2)(−1)
2(2) .
Step 6: Simplify the expression to find the solutions for u:u=√2±√2+8
4.
2
Step 7: Simplify further to get u=√2±√10
4.
Step 8: Since u= cos(x), the solutions for cos(x)are cos(x) = √2+√10
4and
cos(x) = √2−√10
4.
Step 9: Now, find the corresponding values of xwithin the interval [0,2π).
Step 10: From cos(x) = √2+√10
4, we have x= cos−1(√2+√10
4).
Step 11: Similarly, from cos(x) = √2−√10
4, we have x= cos−1(√2−√10
4).
Step 12: Calculate the final values of xwithin the interval [0,2π), and the
solutions to the equation are x=π
5and x=3π
5.
Question 4
Question
Find the exact value of cos (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles in the unit circle.
In this case, we rewrite 5π
12 as π
3+π
4.
Step 2: Using the angle addition formula for cosine, we have
cos (5π
12 )= cos (π
3+π
4)= cos (π
3)cos (π
4)−sin (π
3)sin (π
4).
Step 3: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2.
Step 4: Substitute these values into the formula:
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.
Question 5
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
3
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 given equation to get
2 sin(x) cos(x) = cos(x).
Step 3: Now, we can solve for sin(x):2 sin(x) cos(x) = cos(x)⇒2 sin(x) =
1⇒sin(x) = 1
2.
Step 4: The solutions for sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 6
Question
Solve the equation sin(θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Recall that cos(θ) = sin (θ+π
2). Therefore, the equation sin(θ) =
cos(θ)can be rewritten as sin(θ) = sin (θ+π
2).
Step 2: To solve sin(θ) = sin (θ+π
2), we can use the identity sin(a) = sin(b)
if and only if a=nπ + (−1)nb, where nis an integer.
Step 3: Applying this identity to sin(θ) = sin (θ+π
2), we have θ=nπ +
(−1)n(θ+π
2).
Step 4: Simplifying the above equation gives us two cases: Case 1: When n
is even, we have θ=nπ −θ−π
2, which simplifies to θ=(2n−1)π
2. Case 2:
When nis odd, we have θ=nπ +θ+π
2, which simplifies to θ=(2n+ 1)π
2.
Step 5: Now we need to find the solutions within the interval 0≤θ≤2πfor
both cases: For case 1: θ=π
2,3π
2. For case 2: θ=π
2,3π
2.
Step 6: Therefore, the solutions to the equation sin(θ) = cos(θ)for 0≤θ≤
2πare θ=π
2,3π
2.
Question 7
Question
Find the exact value of sin (5π
12 ).
4
Solution
Step 1: We will use the angle sum identity sin(A+B) = sin Acos B+cos Asin B.
Let A=π
4and B=π
3where A+B=5π
12 . Step 2: Calculate sin (π
4)and
sin (π
3):sin (π
4)=√2
2and sin (π
3)=√3
2. Step 3: Plug the values into the angle
sum identity: sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+cos (π
4)sin (π
3). Step 4:
Substitute the values we calculated: sin (5π
12 )=√2
2·1
2+√2
2·√3
2. Step 5: Simplify
the expression: sin (5π
12 )=√2
4+√6
4=√2+√6
4. Therefore, sin (5π
12 )=√2+√6
4.
Question 8
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles. Let’s write 5π
12 as
3π
12 +2π
12 , which simplifies to π
4+π
6.
Step 2: We know the sine of sum formula, which states that sin(A+B) =
sin Acos B+ cos Asin B.
Step 3: Using the sine of sum formula from Step 2, we can write sin (5π
12 )as
sin (π
4+π
6). This becomes sin (π
4)cos (π
6)+ cos (π
4)sin (π
6).
Step 4: Now, we calculate the values of sin (π
4),cos (π
4),sin (π
6), and cos (π
6).
Step 5: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2.
Step 6: Now substitute the values from Step 5 into the expression from Step
3. We get: 1
√2·√3
2+1
√2·1
2.
Step 7: Simplifying the expression from Step 6, we get √3
2√2+1
2√2.
Step 8: Further simplifying, we get √3+1
2√2.
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 9
Question
Let f(x) = 3 cos(x)−2 sin(x), defined for 0≤x≤2π. Find the maximum and
minimum values of f(x)on this interval.
Solution
Step 1: To find the maximum and minimum values of f(x), we first need to
find the critical points of f(x)where the derivative equals 0. This occurs when
5
f′(x) = 0.
Step 2: Calculate the derivative f′(x):
f′(x) = d
dx(3 cos(x)−2 sin(x)) = −3 sin(x)−2 cos(x).
Step 3: Set f′(x) = 0 to find the critical points:
−3 sin(x)−2 cos(x) = 0.
Step 4: Rearrange the equation:
−2 cos(x) = 3 sin(x) =⇒−2
3=sin(x)
cos(x).
Step 5: Recall that sin(x)
cos(x)= tan(x), so we have:
tan(x) = −2
3.
Step 6: We know that tan(x)cycles through all possible real numbers as x
ranges over 0≤x≤2π. Thus, we solve for x:
x= arctan (−2
3)≈0.588.
Step 7: There is only one critical point between 0and 2π, which is x≈0.588.
Step 8: Now, check the values of f(x)at the critical point and at the end-
points of the interval 0≤x≤2πto determine the maximum and minimum
values.
Step 9: Evaluate f(x)at x= 0, x ≈0.588,and x= 2πto find the maximum
and minimum values of f(x)on the given interval.
Step 10: f(0) = 3 cos(0)−2 sin(0) = 3,f(0.588) = 3 cos(0.588)−2 sin(0.588) ≈
3.77, and f(2π) = 3 cos(2π)−2 sin(2π) = 3.
Step 11: Therefore, the maximum value of f(x)on the interval 0≤x≤2π
is approximately 3.77, and the minimum value is 3.
Question 10
Question
Find the exact value of cos (5π
12 )using trigonometric identities.
Solution
Step 1: We can start by expressing 5π
12 as the sum of two common angles. We
have 5π
12 =π
3+π
4.
6
Step 2: Recall the following trigonometric sum identities:
cos(A+B) = cos Acos B−sin Asin B
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Apply these identities by setting A=π
3and B=π
4. We have:
cos (5π
12 )= cos (π
3+π
4)
= cos π
3cos π
4−sin π
3sin π
4
Step 4: Calculate the cosine and sine values for π
3and π
4:
cos π
3=1
2,sin π
3=√3
2
cos π
4=√2
2,sin π
4=√2
2
Step 5: Substitute these values into the expression:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4
Hence, the exact value of cos (5π
12 )is √2−√6
4.
Question 11
Question
Let f(x) = 2 sin(3x)−3 cos(3x). Find the amplitude, period, and phase shift of
the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by √A2+D2. In this case, A= 2 and D=−3, so
the amplitude is
√22+ (−3)2=√4 + 9 = √13.
7
Step 2: The period of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by 2π
|B|. In this case, B= 3, so the period is
2π
|3|=2π
3.
Step 3: To find the phase shift of the function, we need to solve Bx +C= 0
for x. In this case, B= 3 and C= 0, so the phase shift is
x=−C
B=−0
3= 0.
Therefore, the amplitude of f(x)is √13, the period is 2π
3, and the phase
shift is 0.
Question 12
Question
Solve the equation sin(2θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to rewrite sin(2θ)
in terms of sin(θ)and cos(θ).
sin(2θ) = 2 sin(θ) cos(θ) = cos(θ)
Step 2: Now we have the equation 2 sin(θ) cos(θ) = cos(θ). We can rewrite
this as 2 sin(θ) cos(θ)−cos(θ) = 0.
Step 3: Factor out cos(θ)from the left side of the equation.
cos(θ)(2 sin(θ)−1) = 0
Step 4: Set each factor equal to zero and solve for θ.
cos(θ) = 0 or 2 sin(θ)−1 = 0
Step 5: For cos(θ) = 0, we know that θ=π
2and θ=3π
2.
Step 6: For 2 sin(θ)−1 = 0, solve for sin(θ).
2 sin(θ)−1 = 0
2 sin(θ) = 1
sin(θ) = 1
2
Step 7: The solutions for sin(θ) = 1
2are π
6and 5π
6.
Step 8: Therefore, the solutions to the equation sin(2θ) = cos(θ)for 0≤θ≤
2πare π
6,π
2,5π
6, and 3π
2.
8
Question 13
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find the following expression:
f(x)·g(x)
f(x)−g(x)
Solution
Step 1: Find f(x)·g(x)
f(x)·g(x) = sin(2x)·cos(x)
Step 2: Apply the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)·cos(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos2(x)
Step 3: Find f(x)−g(x)
f(x)−g(x) = sin(2x)−cos(x)
Step 4: Appy the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)−cos(x) = 2 sin(x) cos(x)−cos(x) = cos(x)(2 sin(x)−1)
Step 5: Find the expression f(x)·g(x)
f(x)−g(x)
f(x)·g(x)
f(x)−g(x)=2 sin(x) cos2(x)
cos(x)(2 sin(x)−1)
Step 6: Simplify the expression by canceling out a factor of cos(x)
2 sin(x) cos(x)
2 sin(x)−1
Therefore, the expression f(x)·g(x)
f(x)−g(x)simplifies to 2 sin(x)
2 sin(x)−1.
Question 14
Question
Find the exact value of sin (5π
4).
9
Solution
Step 1: Determine the reference angle. To find the reference angle, we first
need to identify the co-terminal angle that lies between 0 and 2π(0 and 360◦)
that is equivalent to 5π
4. Since 5π
4is greater than π, we subtract πto find the
co-terminal angle: 5π
4−π=π
4. Therefore, the reference angle is π
4.
Step 2: Determine the quadrant. The angle 5π
4lies in the third quadrant
(3π
2<5π
4<2π).
Step 3: Calculate the sine function. In the third quadrant, the sine function
is negative. Since the reference angle π
4has a sine value of √2
2, the sine value
for 5π
4will be:
sin (5π
4)=−√2
2
Question 15
Question
Find the general solution of the equation sin(2x) = −1
2for 0≤x < 2π.
Solution
Step 1: Rewrite the equation in terms of the sine function.
sin(2x) = −1
2
Step 2: Use the double angle identity for sine.
sin(2x) = 2 sin(x) cos(x) = −1
2
Step 3: Express sin(x)in terms of cos(x).
2 sin(x) cos(x) = −1
2
sin(x) = −1
4 cos(x)
Step 4: Use the Pythagorean identity for sine and cosine.
sin2(x) + cos2(x) = 1
(−1
4 cos(x))2
+ cos2(x) = 1
Step 5: Solve for cos(x).
1
16 cos2(x)+ cos2(x) = 1
10
1 + 16 cos2(x) = 16 cos2(x)
17 cos2(x) = 1
cos2(x) = 1
17
cos(x) = ±1
√17
Step 6: Determine the possible solutions for x. Since sin(2x) = −1
2, we
must also have sin(x)<0. Therefore, sin(x) = −1
4 cos(x)will be negative when
cos(x)<0. Thus, the solutions occur in the second and third quadrants.
Step 7: Find the angles in the second and third quadrants where cos(x) =
−1
√17 .
x= cos−1(−1
√17)= 2.6779 radians
Since we need to find the general solution, the solutions in the third quadrant
will be x=π−2.6779.
Therefore, the general solution is:
x= 2.6779 + 2πn or x=π−2.6779 + 2πn
where n∈Z.
Question 16
Question
Let f(θ) = sin θ
1+cos θ. Find the values of θin the interval [0,2π]where f(θ)has a
vertical asymptote.
Solution
Step 1: Identify where f(θ)is undefined.
To find the values of θwhere f(θ)has a vertical asymptote, we need to find
where the denominator of f(θ)equals 0, since division by 0 is undefined. So we
solve 1 + cos θ= 0:
cos θ=−1
Step 2: Find the values of θthat satisfy cos θ=−1in the interval [0,2π].
The cosine function is equal to -1 at θ=πin the interval [0,2π].
Therefore, θ=πis the only value in the interval [0,2π]where f(θ)has a
vertical asymptote.
11
Question 17
Question
Given that sin(θ) = 3
5and θis in Quadrant II, find the exact values of tan(θ)
and sec(θ).
Solution
Step 1: We can use the Pythagorean identity sin2(θ) + cos2(θ) = 1 to find the
value of cos(θ).
cos2(θ) = 1 −sin2(θ)
cos2(θ) = 1 −(3
5)2
= 1 −9
25 =16
25
cos(θ) = ±4
5
Since θis in Quadrant II, cos(θ)is negative. Therefore, cos(θ) = −4
5.
Step 2: Next, we can find the values of tan(θ)and sec(θ)using the definitions
of these trigonometric functions.
tan(θ) = sin(θ)
cos(θ)=
3
5
−4
5
=−3
4
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
Therefore, the exact values of tan(θ)and sec(θ)are −3
4and −5
4, respec-
tively.
Question 18
Question
Given that sin(θ) = 3
5and θis in quadrant II, determine the exact value of
cos(2θ).
Solution
Step 1: Since we know that sin(θ) = 3
5and θis in quadrant II, we can use the
Pythagorean identity to find cos(θ).
cos(θ) = −√1−sin2(θ) = −√1−(3
5)2
=−4
5
12
Step 2: Next, we can use the double angle identity for cosine to find cos(2θ).
cos(2θ) = cos2(θ)−sin2(θ)
=(−4
5)2
−(3
5)2
=16
25 −9
25
=7
25
Therefore, the exact value of cos(2θ)is 7
25 .
Question 19
Question
Let f(θ) = 2 cos2(θ)−3 sin(θ) cos(θ), where θ∈[0,π
2]. Find the maximum
value of f(θ)in the given interval.
Solution
Step 1: Recall the double angle identities for cosine and sine:
cos(2θ) = 2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Rewrite f(θ)using the double angle identity for cosine:
f(θ) = cos(2θ)−3 sin(2θ)
Step 3: To maximize f(θ), we need to find the maximum value of cos(2θ)
and the minimum value of sin(2θ)in the interval θ∈[0,π
2].
Step 4: The maximum value of cos(2θ)in the given interval is 1, which
occurs when θ= 0.
Step 5: The minimum value of sin(2θ)in the interval θ∈[0,π
2]is -1, which
occurs when θ=π
4.
Step 6: Substituting these values into f(θ), we get the maximum value of
f(θ)in the interval:
f(θ) = 1 −3(−1) = 4
Therefore, the maximum value of f(θ)in the interval θ∈[0,π
2]is 4.
Question 20
Question
Find the exact value of sin (3π
4).
13
Solution
Step 1: We know that sin (3π
4)=−sin (π
4)since the sine function has period
2π.
Step 2: Using the angle addition identity sin(A−B) = sin Acos B−cos Asin B,
we have:
sin (3π
4)=−sin (π
4)=−(sin(π/2) cos(π/4) −cos(π/2) sin(π/4))
Step 3: We know that sin(π/2) = 1,cos(π/2) = 0,cos(π/4) = √2
2, and
sin(π/4) = √2
2. Substituting these values, we get:
sin (3π
4)=−(1·√2
2−0·√2
2)
Step 4: Simplifying further:
sin (3π
4)=−√2
2
Therefore, the exact value of sin (3π
4)is −√2
2.
Question 21
Question
Let f(x) = 3 sin(x) + 4 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), we use the formula A=√a2+b2, where
Ais the amplitude, and for f(x) = asin(x)+bcos(x),aand bare the coefficients
of sin(x)and cos(x). 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 of f(x), we know that the period of sin(x)and
cos(x)is 2π. Since there are no coefficients that affect the period, the period of
f(x)is also 2π.
Step 3: To find the phase shift of f(x), we need to set sin(x)and cos(x)
equal to their maximum/minimum values and solve for x. For sin(x), it reaches
its maximum value at x=π
2, and for cos(x), it reaches its maximum value at
x= 0. Comparing this to f(x) = 3 sin(x) + 4 cos(x), we can see that the phase
shift is π
2to the right.
14
Step 4: To find the vertical shift of f(x), we need to find the average of the
maximum and minimum values of f(x). The maximum value of f(x)is 5, and
the minimum value is −5, so the vertical shift is:
(5 + (−5))/2 = 0
Therefore, the amplitude of f(x)is 5, the period is 2π, the phase shift is π
2
to the right, and the vertical shift is 0.
Question 22
Question
Calculate the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles:
5π
12 =π
3+π
4
Step 2: Use the sum-to-product formula for sine:
sin (a
n+b
n)= sin (a
n)cos (b
n)+ cos (a
n)sin (b
n)
So we have:
sin (5π
12 )= sin (π
3+π
4)
sin (5π
12 )= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Substitute in the values of sin(π/3),cos(π/3),sin(π/4), and cos(π/4):
sin (5π
12 )=√3
2·√2
2+1
2·√2
2
sin (5π
12 )=√6
4+√2
4
sin (5π
12 )=√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
15
Question 23
Question
Solve for xin the equation sin(2x) = cos(x)for x∈[0◦,360◦].
Solution
Step 1: Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the given equation to get:
2 sin(x) cos(x) = cos(x)
Step 3: Now we have two cases to consider: Case 1: cos(x)= 0 Divide both
sides by cos(x):
2 sin(x) = 1
Step 4: Solve for sin(x):
sin(x) = 1
2
Step 5: This occurs in the first and second quadrants, so the solutions are
x= 30◦and x= 150◦.
Case 2: cos(x) = 0 This occurs when x= 90◦.
Step 6: Therefore, the solutions to the equation sin(2x) = cos(x)for x∈
[0◦,360◦]are x= 30◦,150◦,and 90◦.
Question 24
Question
Given that sin θ=3
5and θis in Quadrant II, find the exact value of cos (θ
2).
Solution
Step 1: Find cos θusing the Pythagorean identity sin2θ+ cos2θ= 1.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=±4
5
Since θis in Quadrant II, where cosine is negative, cos θ=−4
5.
16
Step 2: Use the half-angle formula for cosine to find cos (θ
2).
cos (θ
2)=±√1 + cos θ
2=±√1−4
5
2=±√1−4
5
2=±√1
10 =±1
√10 =±√10
10
Since θis in Quadrant II, where cosine is negative, cos (θ
2)=−√10
10 .
Question 25
Question
Determine all solutions to the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute the double angle identity into the given equation: 2 cos2(x)−
1 = sin(x).
Step 3: Recall the Pythagorean identity for sine and cosine: sin2(x) +
cos2(x) = 1.
Step 4: Substitute sin2(x)=1−cos2(x)into the equation from step 2:
2 cos2(x)−1 = √1−cos2(x).
Step 5: Squaring both sides of the equation gives 4 cos4(x)−4 cos2(x) + 1 =
1−cos2(x).
Step 6: Simplify the equation: 4 cos4(x)−5 cos2(x) = 0.
Step 7: Factor out a cos2(x)from the equation: cos2(x)(4 cos2(x)−5) = 0.
Step 8: Set each factor equal to zero: cos2(x) = 0 and 4 cos2(x)−5 = 0.
Step 9: Solve the first equation cos2(x) = 0 to find cos(x) = 0. This occurs
at x= 90◦and x= 270◦.
Step 10: Solve the second equation 4 cos2(x)−5=0to find cos(x) = ±√5
2.
This occurs at x= 18.19◦,x= 161.81◦,x= 198.19◦, and x= 341.81◦.
Step 11: Check all solutions in the original equation to verify their validity.
Therefore, all the solutions to the equation cos(2x) = sin(x)for 0◦≤x≤
360◦are x= 18.19◦,90◦,161.81◦,198.19◦,270◦,341.81◦.
17
Question 2
Question
Let f(x) = 2 sin(3x)−√3 cos(3x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by R=√A2+C2. In this case, the amplitude is R=√22+ (−√3)2=
√4 + 3 = √7.
Step 2: The period of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by P=2π
|B|. In this case, the period is P=2π
3.
Step 3: To find the phase shift of the function f(x), we need to write f(x)
in the form f(x) = Rsin(B(x−h)) + k, where his the phase shift.
In this case, f(x) = √7 sin(3x−h) + k.
Comparing with f(x) = 2 sin(3x)−√3 cos(3x), we get:
•√7 sin(3x−h) = 2 sin(3x), which implies √7 = 2 and h= 0.
•k=−√3.
Therefore, the phase shift is h= 0.
Step 4: The vertical shift of the function f(x)is given by the constant kin
the form f(x) = Rsin(Bx −h) + k. In this case, the vertical shift is k=−√3.
So, the amplitude of the function is √7, the period is 2π
3, the phase shift is
0, and the vertical shift is −√3.
Question 3
Question
Solve the equation cos(2x) = √2 cos(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute this identity into the given equation to get 2 cos2(x)−1 =
√2 cos(x).
Step 3: Rearrange the equation to get 2 cos2(x)−√2 cos(x)−1 = 0.
Step 4: This is a quadratic equation in terms of cos(x). Let u= cos(x), then
we have 2u2−√2u−1 = 0.
Step 5: Solve the quadratic equation using the quadratic formula: u=
−(−√2)±√(−√2)2−4(2)(−1)
2(2) .
Step 6: Simplify the expression to find the solutions for u:u=√2±√2+8
4.
2
Step 7: Simplify further to get u=√2±√10
4.
Step 8: Since u= cos(x), the solutions for cos(x)are cos(x) = √2+√10
4and
cos(x) = √2−√10
4.
Step 9: Now, find the corresponding values of xwithin the interval [0,2π).
Step 10: From cos(x) = √2+√10
4, we have x= cos−1(√2+√10
4).
Step 11: Similarly, from cos(x) = √2−√10
4, we have x= cos−1(√2−√10
4).
Step 12: Calculate the final values of xwithin the interval [0,2π), and the
solutions to the equation are x=π
5and x=3π
5.
Question 4
Question
Find the exact value of cos (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles in the unit circle.
In this case, we rewrite 5π
12 as π
3+π
4.
Step 2: Using the angle addition formula for cosine, we have
cos (5π
12 )= cos (π
3+π
4)= cos (π
3)cos (π
4)−sin (π
3)sin (π
4).
Step 3: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2.
Step 4: Substitute these values into the formula:
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.
Question 5
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
3
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 given equation to get
2 sin(x) cos(x) = cos(x).
Step 3: Now, we can solve for sin(x):2 sin(x) cos(x) = cos(x)⇒2 sin(x) =
1⇒sin(x) = 1
2.
Step 4: The solutions for sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 6
Question
Solve the equation sin(θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Recall that cos(θ) = sin (θ+π
2). Therefore, the equation sin(θ) =
cos(θ)can be rewritten as sin(θ) = sin (θ+π
2).
Step 2: To solve sin(θ) = sin (θ+π
2), we can use the identity sin(a) = sin(b)
if and only if a=nπ + (−1)nb, where nis an integer.
Step 3: Applying this identity to sin(θ) = sin (θ+π
2), we have θ=nπ +
(−1)n(θ+π
2).
Step 4: Simplifying the above equation gives us two cases: Case 1: When n
is even, we have θ=nπ −θ−π
2, which simplifies to θ=(2n−1)π
2. Case 2:
When nis odd, we have θ=nπ +θ+π
2, which simplifies to θ=(2n+ 1)π
2.
Step 5: Now we need to find the solutions within the interval 0≤θ≤2πfor
both cases: For case 1: θ=π
2,3π
2. For case 2: θ=π
2,3π
2.
Step 6: Therefore, the solutions to the equation sin(θ) = cos(θ)for 0≤θ≤
2πare θ=π
2,3π
2.
Question 7
Question
Find the exact value of sin (5π
12 ).
4
Solution
Step 1: We will use the angle sum identity sin(A+B) = sin Acos B+cos Asin B.
Let A=π
4and B=π
3where A+B=5π
12 . Step 2: Calculate sin (π
4)and
sin (π
3):sin (π
4)=√2
2and sin (π
3)=√3
2. Step 3: Plug the values into the angle
sum identity: sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+cos (π
4)sin (π
3). Step 4:
Substitute the values we calculated: sin (5π
12 )=√2
2·1
2+√2
2·√3
2. Step 5: Simplify
the expression: sin (5π
12 )=√2
4+√6
4=√2+√6
4. Therefore, sin (5π
12 )=√2+√6
4.
Question 8
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles. Let’s write 5π
12 as
3π
12 +2π
12 , which simplifies to π
4+π
6.
Step 2: We know the sine of sum formula, which states that sin(A+B) =
sin Acos B+ cos Asin B.
Step 3: Using the sine of sum formula from Step 2, we can write sin (5π
12 )as
sin (π
4+π
6). This becomes sin (π
4)cos (π
6)+ cos (π
4)sin (π
6).
Step 4: Now, we calculate the values of sin (π
4),cos (π
4),sin (π
6), and cos (π
6).
Step 5: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2.
Step 6: Now substitute the values from Step 5 into the expression from Step
3. We get: 1
√2·√3
2+1
√2·1
2.
Step 7: Simplifying the expression from Step 6, we get √3
2√2+1
2√2.
Step 8: Further simplifying, we get √3+1
2√2.
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 9
Question
Let f(x) = 3 cos(x)−2 sin(x), defined for 0≤x≤2π. Find the maximum and
minimum values of f(x)on this interval.
Solution
Step 1: To find the maximum and minimum values of f(x), we first need to
find the critical points of f(x)where the derivative equals 0. This occurs when
5
f′(x) = 0.
Step 2: Calculate the derivative f′(x):
f′(x) = d
dx(3 cos(x)−2 sin(x)) = −3 sin(x)−2 cos(x).
Step 3: Set f′(x) = 0 to find the critical points:
−3 sin(x)−2 cos(x) = 0.
Step 4: Rearrange the equation:
−2 cos(x) = 3 sin(x) =⇒−2
3=sin(x)
cos(x).
Step 5: Recall that sin(x)
cos(x)= tan(x), so we have:
tan(x) = −2
3.
Step 6: We know that tan(x)cycles through all possible real numbers as x
ranges over 0≤x≤2π. Thus, we solve for x:
x= arctan (−2
3)≈0.588.
Step 7: There is only one critical point between 0and 2π, which is x≈0.588.
Step 8: Now, check the values of f(x)at the critical point and at the end-
points of the interval 0≤x≤2πto determine the maximum and minimum
values.
Step 9: Evaluate f(x)at x= 0, x ≈0.588,and x= 2πto find the maximum
and minimum values of f(x)on the given interval.
Step 10: f(0) = 3 cos(0)−2 sin(0) = 3,f(0.588) = 3 cos(0.588)−2 sin(0.588) ≈
3.77, and f(2π) = 3 cos(2π)−2 sin(2π) = 3.
Step 11: Therefore, the maximum value of f(x)on the interval 0≤x≤2π
is approximately 3.77, and the minimum value is 3.
Question 10
Question
Find the exact value of cos (5π
12 )using trigonometric identities.
Solution
Step 1: We can start by expressing 5π
12 as the sum of two common angles. We
have 5π
12 =π
3+π
4.
6
Step 2: Recall the following trigonometric sum identities:
cos(A+B) = cos Acos B−sin Asin B
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Apply these identities by setting A=π
3and B=π
4. We have:
cos (5π
12 )= cos (π
3+π
4)
= cos π
3cos π
4−sin π
3sin π
4
Step 4: Calculate the cosine and sine values for π
3and π
4:
cos π
3=1
2,sin π
3=√3
2
cos π
4=√2
2,sin π
4=√2
2
Step 5: Substitute these values into the expression:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4
Hence, the exact value of cos (5π
12 )is √2−√6
4.
Question 11
Question
Let f(x) = 2 sin(3x)−3 cos(3x). Find the amplitude, period, and phase shift of
the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by √A2+D2. In this case, A= 2 and D=−3, so
the amplitude is
√22+ (−3)2=√4 + 9 = √13.
7
Step 2: The period of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by 2π
|B|. In this case, B= 3, so the period is
2π
|3|=2π
3.
Step 3: To find the phase shift of the function, we need to solve Bx +C= 0
for x. In this case, B= 3 and C= 0, so the phase shift is
x=−C
B=−0
3= 0.
Therefore, the amplitude of f(x)is √13, the period is 2π
3, and the phase
shift is 0.
Question 12
Question
Solve the equation sin(2θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to rewrite sin(2θ)
in terms of sin(θ)and cos(θ).
sin(2θ) = 2 sin(θ) cos(θ) = cos(θ)
Step 2: Now we have the equation 2 sin(θ) cos(θ) = cos(θ). We can rewrite
this as 2 sin(θ) cos(θ)−cos(θ) = 0.
Step 3: Factor out cos(θ)from the left side of the equation.
cos(θ)(2 sin(θ)−1) = 0
Step 4: Set each factor equal to zero and solve for θ.
cos(θ) = 0 or 2 sin(θ)−1 = 0
Step 5: For cos(θ) = 0, we know that θ=π
2and θ=3π
2.
Step 6: For 2 sin(θ)−1 = 0, solve for sin(θ).
2 sin(θ)−1 = 0
2 sin(θ) = 1
sin(θ) = 1
2
Step 7: The solutions for sin(θ) = 1
2are π
6and 5π
6.
Step 8: Therefore, the solutions to the equation sin(2θ) = cos(θ)for 0≤θ≤
2πare π
6,π
2,5π
6, and 3π
2.
8
Question 13
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find the following expression:
f(x)·g(x)
f(x)−g(x)
Solution
Step 1: Find f(x)·g(x)
f(x)·g(x) = sin(2x)·cos(x)
Step 2: Apply the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)·cos(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos2(x)
Step 3: Find f(x)−g(x)
f(x)−g(x) = sin(2x)−cos(x)
Step 4: Appy the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)−cos(x) = 2 sin(x) cos(x)−cos(x) = cos(x)(2 sin(x)−1)
Step 5: Find the expression f(x)·g(x)
f(x)−g(x)
f(x)·g(x)
f(x)−g(x)=2 sin(x) cos2(x)
cos(x)(2 sin(x)−1)
Step 6: Simplify the expression by canceling out a factor of cos(x)
2 sin(x) cos(x)
2 sin(x)−1
Therefore, the expression f(x)·g(x)
f(x)−g(x)simplifies to 2 sin(x)
2 sin(x)−1.
Question 14
Question
Find the exact value of sin (5π
4).
9
Solution
Step 1: Determine the reference angle. To find the reference angle, we first
need to identify the co-terminal angle that lies between 0 and 2π(0 and 360◦)
that is equivalent to 5π
4. Since 5π
4is greater than π, we subtract πto find the
co-terminal angle: 5π
4−π=π
4. Therefore, the reference angle is π
4.
Step 2: Determine the quadrant. The angle 5π
4lies in the third quadrant
(3π
2<5π
4<2π).
Step 3: Calculate the sine function. In the third quadrant, the sine function
is negative. Since the reference angle π
4has a sine value of √2
2, the sine value
for 5π
4will be:
sin (5π
4)=−√2
2
Question 15
Question
Find the general solution of the equation sin(2x) = −1
2for 0≤x < 2π.
Solution
Step 1: Rewrite the equation in terms of the sine function.
sin(2x) = −1
2
Step 2: Use the double angle identity for sine.
sin(2x) = 2 sin(x) cos(x) = −1
2
Step 3: Express sin(x)in terms of cos(x).
2 sin(x) cos(x) = −1
2
sin(x) = −1
4 cos(x)
Step 4: Use the Pythagorean identity for sine and cosine.
sin2(x) + cos2(x) = 1
(−1
4 cos(x))2
+ cos2(x) = 1
Step 5: Solve for cos(x).
1
16 cos2(x)+ cos2(x) = 1
10
1 + 16 cos2(x) = 16 cos2(x)
17 cos2(x) = 1
cos2(x) = 1
17
cos(x) = ±1
√17
Step 6: Determine the possible solutions for x. Since sin(2x) = −1
2, we
must also have sin(x)<0. Therefore, sin(x) = −1
4 cos(x)will be negative when
cos(x)<0. Thus, the solutions occur in the second and third quadrants.
Step 7: Find the angles in the second and third quadrants where cos(x) =
−1
√17 .
x= cos−1(−1
√17)= 2.6779 radians
Since we need to find the general solution, the solutions in the third quadrant
will be x=π−2.6779.
Therefore, the general solution is:
x= 2.6779 + 2πn or x=π−2.6779 + 2πn
where n∈Z.
Question 16
Question
Let f(θ) = sin θ
1+cos θ. Find the values of θin the interval [0,2π]where f(θ)has a
vertical asymptote.
Solution
Step 1: Identify where f(θ)is undefined.
To find the values of θwhere f(θ)has a vertical asymptote, we need to find
where the denominator of f(θ)equals 0, since division by 0 is undefined. So we
solve 1 + cos θ= 0:
cos θ=−1
Step 2: Find the values of θthat satisfy cos θ=−1in the interval [0,2π].
The cosine function is equal to -1 at θ=πin the interval [0,2π].
Therefore, θ=πis the only value in the interval [0,2π]where f(θ)has a
vertical asymptote.
11
Question 17
Question
Given that sin(θ) = 3
5and θis in Quadrant II, find the exact values of tan(θ)
and sec(θ).
Solution
Step 1: We can use the Pythagorean identity sin2(θ) + cos2(θ) = 1 to find the
value of cos(θ).
cos2(θ) = 1 −sin2(θ)
cos2(θ) = 1 −(3
5)2
= 1 −9
25 =16
25
cos(θ) = ±4
5
Since θis in Quadrant II, cos(θ)is negative. Therefore, cos(θ) = −4
5.
Step 2: Next, we can find the values of tan(θ)and sec(θ)using the definitions
of these trigonometric functions.
tan(θ) = sin(θ)
cos(θ)=
3
5
−4
5
=−3
4
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
Therefore, the exact values of tan(θ)and sec(θ)are −3
4and −5
4, respec-
tively.
Question 18
Question
Given that sin(θ) = 3
5and θis in quadrant II, determine the exact value of
cos(2θ).
Solution
Step 1: Since we know that sin(θ) = 3
5and θis in quadrant II, we can use the
Pythagorean identity to find cos(θ).
cos(θ) = −√1−sin2(θ) = −√1−(3
5)2
=−4
5
12
Step 2: Next, we can use the double angle identity for cosine to find cos(2θ).
cos(2θ) = cos2(θ)−sin2(θ)
=(−4
5)2
−(3
5)2
=16
25 −9
25
=7
25
Therefore, the exact value of cos(2θ)is 7
25 .
Question 19
Question
Let f(θ) = 2 cos2(θ)−3 sin(θ) cos(θ), where θ∈[0,π
2]. Find the maximum
value of f(θ)in the given interval.
Solution
Step 1: Recall the double angle identities for cosine and sine:
cos(2θ) = 2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Rewrite f(θ)using the double angle identity for cosine:
f(θ) = cos(2θ)−3 sin(2θ)
Step 3: To maximize f(θ), we need to find the maximum value of cos(2θ)
and the minimum value of sin(2θ)in the interval θ∈[0,π
2].
Step 4: The maximum value of cos(2θ)in the given interval is 1, which
occurs when θ= 0.
Step 5: The minimum value of sin(2θ)in the interval θ∈[0,π
2]is -1, which
occurs when θ=π
4.
Step 6: Substituting these values into f(θ), we get the maximum value of
f(θ)in the interval:
f(θ) = 1 −3(−1) = 4
Therefore, the maximum value of f(θ)in the interval θ∈[0,π
2]is 4.
Question 20
Question
Find the exact value of sin (3π
4).
13
Solution
Step 1: We know that sin (3π
4)=−sin (π
4)since the sine function has period
2π.
Step 2: Using the angle addition identity sin(A−B) = sin Acos B−cos Asin B,
we have:
sin (3π
4)=−sin (π
4)=−(sin(π/2) cos(π/4) −cos(π/2) sin(π/4))
Step 3: We know that sin(π/2) = 1,cos(π/2) = 0,cos(π/4) = √2
2, and
sin(π/4) = √2
2. Substituting these values, we get:
sin (3π
4)=−(1·√2
2−0·√2
2)
Step 4: Simplifying further:
sin (3π
4)=−√2
2
Therefore, the exact value of sin (3π
4)is −√2
2.
Question 21
Question
Let f(x) = 3 sin(x) + 4 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), we use the formula A=√a2+b2, where
Ais the amplitude, and for f(x) = asin(x)+bcos(x),aand bare the coefficients
of sin(x)and cos(x). 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 of f(x), we know that the period of sin(x)and
cos(x)is 2π. Since there are no coefficients that affect the period, the period of
f(x)is also 2π.
Step 3: To find the phase shift of f(x), we need to set sin(x)and cos(x)
equal to their maximum/minimum values and solve for x. For sin(x), it reaches
its maximum value at x=π
2, and for cos(x), it reaches its maximum value at
x= 0. Comparing this to f(x) = 3 sin(x) + 4 cos(x), we can see that the phase
shift is π
2to the right.
14
Step 4: To find the vertical shift of f(x), we need to find the average of the
maximum and minimum values of f(x). The maximum value of f(x)is 5, and
the minimum value is −5, so the vertical shift is:
(5 + (−5))/2 = 0
Therefore, the amplitude of f(x)is 5, the period is 2π, the phase shift is π
2
to the right, and the vertical shift is 0.
Question 22
Question
Calculate the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles:
5π
12 =π
3+π
4
Step 2: Use the sum-to-product formula for sine:
sin (a
n+b
n)= sin (a
n)cos (b
n)+ cos (a
n)sin (b
n)
So we have:
sin (5π
12 )= sin (π
3+π
4)
sin (5π
12 )= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Substitute in the values of sin(π/3),cos(π/3),sin(π/4), and cos(π/4):
sin (5π
12 )=√3
2·√2
2+1
2·√2
2
sin (5π
12 )=√6
4+√2
4
sin (5π
12 )=√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
15
Question 23
Question
Solve for xin the equation sin(2x) = cos(x)for x∈[0◦,360◦].
Solution
Step 1: Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the given equation to get:
2 sin(x) cos(x) = cos(x)
Step 3: Now we have two cases to consider: Case 1: cos(x)= 0 Divide both
sides by cos(x):
2 sin(x) = 1
Step 4: Solve for sin(x):
sin(x) = 1
2
Step 5: This occurs in the first and second quadrants, so the solutions are
x= 30◦and x= 150◦.
Case 2: cos(x) = 0 This occurs when x= 90◦.
Step 6: Therefore, the solutions to the equation sin(2x) = cos(x)for x∈
[0◦,360◦]are x= 30◦,150◦,and 90◦.
Question 24
Question
Given that sin θ=3
5and θis in Quadrant II, find the exact value of cos (θ
2).
Solution
Step 1: Find cos θusing the Pythagorean identity sin2θ+ cos2θ= 1.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=±4
5
Since θis in Quadrant II, where cosine is negative, cos θ=−4
5.
16
Step 2: Use the half-angle formula for cosine to find cos (θ
2).
cos (θ
2)=±√1 + cos θ
2=±√1−4
5
2=±√1−4
5
2=±√1
10 =±1
√10 =±√10
10
Since θis in Quadrant II, where cosine is negative, cos (θ
2)=−√10
10 .
Question 25
Question
Determine all solutions to the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute the double angle identity into the given equation: 2 cos2(x)−
1 = sin(x).
Step 3: Recall the Pythagorean identity for sine and cosine: sin2(x) +
cos2(x) = 1.
Step 4: Substitute sin2(x)=1−cos2(x)into the equation from step 2:
2 cos2(x)−1 = √1−cos2(x).
Step 5: Squaring both sides of the equation gives 4 cos4(x)−4 cos2(x) + 1 =
1−cos2(x).
Step 6: Simplify the equation: 4 cos4(x)−5 cos2(x) = 0.
Step 7: Factor out a cos2(x)from the equation: cos2(x)(4 cos2(x)−5) = 0.
Step 8: Set each factor equal to zero: cos2(x) = 0 and 4 cos2(x)−5 = 0.
Step 9: Solve the first equation cos2(x) = 0 to find cos(x) = 0. This occurs
at x= 90◦and x= 270◦.
Step 10: Solve the second equation 4 cos2(x)−5=0to find cos(x) = ±√5
2.
This occurs at x= 18.19◦,x= 161.81◦,x= 198.19◦, and x= 341.81◦.
Step 11: Check all solutions in the original equation to verify their validity.
Therefore, all the solutions to the equation cos(2x) = sin(x)for 0◦≤x≤
360◦are x= 18.19◦,90◦,161.81◦,198.19◦,270◦,341.81◦.
17
Question 2
Question
Let f(x) = 2 sin(3x)−√3 cos(3x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by R=√A2+C2. In this case, the amplitude is R=√22+ (−√3)2=
√4 + 3 = √7.
Step 2: The period of a function of the form f(x) = Asin(Bx) + Ccos(Dx)
is given by P=2π
|B|. In this case, the period is P=2π
3.
Step 3: To find the phase shift of the function f(x), we need to write f(x)
in the form f(x) = Rsin(B(x−h)) + k, where his the phase shift.
In this case, f(x) = √7 sin(3x−h) + k.
Comparing with f(x) = 2 sin(3x)−√3 cos(3x), we get:
•√7 sin(3x−h) = 2 sin(3x), which implies √7 = 2 and h= 0.
•k=−√3.
Therefore, the phase shift is h= 0.
Step 4: The vertical shift of the function f(x)is given by the constant kin
the form f(x) = Rsin(Bx −h) + k. In this case, the vertical shift is k=−√3.
So, the amplitude of the function is √7, the period is 2π
3, the phase shift is
0, and the vertical shift is −√3.
Question 3
Question
Solve the equation cos(2x) = √2 cos(x)for xin the interval [0,2π).
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute this identity into the given equation to get 2 cos2(x)−1 =
√2 cos(x).
Step 3: Rearrange the equation to get 2 cos2(x)−√2 cos(x)−1 = 0.
Step 4: This is a quadratic equation in terms of cos(x). Let u= cos(x), then
we have 2u2−√2u−1 = 0.
Step 5: Solve the quadratic equation using the quadratic formula: u=
−(−√2)±√(−√2)2−4(2)(−1)
2(2) .
Step 6: Simplify the expression to find the solutions for u:u=√2±√2+8
4.
2
Step 7: Simplify further to get u=√2±√10
4.
Step 8: Since u= cos(x), the solutions for cos(x)are cos(x) = √2+√10
4and
cos(x) = √2−√10
4.
Step 9: Now, find the corresponding values of xwithin the interval [0,2π).
Step 10: From cos(x) = √2+√10
4, we have x= cos−1(√2+√10
4).
Step 11: Similarly, from cos(x) = √2−√10
4, we have x= cos−1(√2−√10
4).
Step 12: Calculate the final values of xwithin the interval [0,2π), and the
solutions to the equation are x=π
5and x=3π
5.
Question 4
Question
Find the exact value of cos (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles in the unit circle.
In this case, we rewrite 5π
12 as π
3+π
4.
Step 2: Using the angle addition formula for cosine, we have
cos (5π
12 )= cos (π
3+π
4)= cos (π
3)cos (π
4)−sin (π
3)sin (π
4).
Step 3: Recall that cos (π
3)=1
2,cos (π
4)=√2
2,sin (π
3)=√3
2, and sin (π
4)=
√2
2.
Step 4: Substitute these values into the formula:
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.
Question 5
Question
Solve the equation sin(2x) = cos(x)for xin the interval [0,2π).
3
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 given equation to get
2 sin(x) cos(x) = cos(x).
Step 3: Now, we can solve for sin(x):2 sin(x) cos(x) = cos(x)⇒2 sin(x) =
1⇒sin(x) = 1
2.
Step 4: The solutions for sin(x) = 1
2in the interval [0,2π)are x=π
6and
x=5π
6.
Therefore, the solutions to the equation sin(2x) = cos(x)in the interval
[0,2π)are x=π
6and x=5π
6.
Question 6
Question
Solve the equation sin(θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Recall that cos(θ) = sin (θ+π
2). Therefore, the equation sin(θ) =
cos(θ)can be rewritten as sin(θ) = sin (θ+π
2).
Step 2: To solve sin(θ) = sin (θ+π
2), we can use the identity sin(a) = sin(b)
if and only if a=nπ + (−1)nb, where nis an integer.
Step 3: Applying this identity to sin(θ) = sin (θ+π
2), we have θ=nπ +
(−1)n(θ+π
2).
Step 4: Simplifying the above equation gives us two cases: Case 1: When n
is even, we have θ=nπ −θ−π
2, which simplifies to θ=(2n−1)π
2. Case 2:
When nis odd, we have θ=nπ +θ+π
2, which simplifies to θ=(2n+ 1)π
2.
Step 5: Now we need to find the solutions within the interval 0≤θ≤2πfor
both cases: For case 1: θ=π
2,3π
2. For case 2: θ=π
2,3π
2.
Step 6: Therefore, the solutions to the equation sin(θ) = cos(θ)for 0≤θ≤
2πare θ=π
2,3π
2.
Question 7
Question
Find the exact value of sin (5π
12 ).
4
Solution
Step 1: We will use the angle sum identity sin(A+B) = sin Acos B+cos Asin B.
Let A=π
4and B=π
3where A+B=5π
12 . Step 2: Calculate sin (π
4)and
sin (π
3):sin (π
4)=√2
2and sin (π
3)=√3
2. Step 3: Plug the values into the angle
sum identity: sin (5π
12 )= sin (π
4+π
3)= sin (π
4)cos (π
3)+cos (π
4)sin (π
3). Step 4:
Substitute the values we calculated: sin (5π
12 )=√2
2·1
2+√2
2·√3
2. Step 5: Simplify
the expression: sin (5π
12 )=√2
4+√6
4=√2+√6
4. Therefore, sin (5π
12 )=√2+√6
4.
Question 8
Question
Determine the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles. Let’s write 5π
12 as
3π
12 +2π
12 , which simplifies to π
4+π
6.
Step 2: We know the sine of sum formula, which states that sin(A+B) =
sin Acos B+ cos Asin B.
Step 3: Using the sine of sum formula from Step 2, we can write sin (5π
12 )as
sin (π
4+π
6). This becomes sin (π
4)cos (π
6)+ cos (π
4)sin (π
6).
Step 4: Now, we calculate the values of sin (π
4),cos (π
4),sin (π
6), and cos (π
6).
Step 5: We know that sin (π
4)=1
√2,cos (π
4)=1
√2,sin (π
6)=1
2, and
cos (π
6)=√3
2.
Step 6: Now substitute the values from Step 5 into the expression from Step
3. We get: 1
√2·√3
2+1
√2·1
2.
Step 7: Simplifying the expression from Step 6, we get √3
2√2+1
2√2.
Step 8: Further simplifying, we get √3+1
2√2.
Therefore, the exact value of sin (5π
12 )is √3+1
2√2.
Question 9
Question
Let f(x) = 3 cos(x)−2 sin(x), defined for 0≤x≤2π. Find the maximum and
minimum values of f(x)on this interval.
Solution
Step 1: To find the maximum and minimum values of f(x), we first need to
find the critical points of f(x)where the derivative equals 0. This occurs when
5
f′(x) = 0.
Step 2: Calculate the derivative f′(x):
f′(x) = d
dx(3 cos(x)−2 sin(x)) = −3 sin(x)−2 cos(x).
Step 3: Set f′(x) = 0 to find the critical points:
−3 sin(x)−2 cos(x) = 0.
Step 4: Rearrange the equation:
−2 cos(x) = 3 sin(x) =⇒−2
3=sin(x)
cos(x).
Step 5: Recall that sin(x)
cos(x)= tan(x), so we have:
tan(x) = −2
3.
Step 6: We know that tan(x)cycles through all possible real numbers as x
ranges over 0≤x≤2π. Thus, we solve for x:
x= arctan (−2
3)≈0.588.
Step 7: There is only one critical point between 0and 2π, which is x≈0.588.
Step 8: Now, check the values of f(x)at the critical point and at the end-
points of the interval 0≤x≤2πto determine the maximum and minimum
values.
Step 9: Evaluate f(x)at x= 0, x ≈0.588,and x= 2πto find the maximum
and minimum values of f(x)on the given interval.
Step 10: f(0) = 3 cos(0)−2 sin(0) = 3,f(0.588) = 3 cos(0.588)−2 sin(0.588) ≈
3.77, and f(2π) = 3 cos(2π)−2 sin(2π) = 3.
Step 11: Therefore, the maximum value of f(x)on the interval 0≤x≤2π
is approximately 3.77, and the minimum value is 3.
Question 10
Question
Find the exact value of cos (5π
12 )using trigonometric identities.
Solution
Step 1: We can start by expressing 5π
12 as the sum of two common angles. We
have 5π
12 =π
3+π
4.
6
Step 2: Recall the following trigonometric sum identities:
cos(A+B) = cos Acos B−sin Asin B
sin(A+B) = sin Acos B+ cos Asin B
Step 3: Apply these identities by setting A=π
3and B=π
4. We have:
cos (5π
12 )= cos (π
3+π
4)
= cos π
3cos π
4−sin π
3sin π
4
Step 4: Calculate the cosine and sine values for π
3and π
4:
cos π
3=1
2,sin π
3=√3
2
cos π
4=√2
2,sin π
4=√2
2
Step 5: Substitute these values into the expression:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4
Hence, the exact value of cos (5π
12 )is √2−√6
4.
Question 11
Question
Let f(x) = 2 sin(3x)−3 cos(3x). Find the amplitude, period, and phase shift of
the function f(x).
Solution
Step 1: The amplitude of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by √A2+D2. In this case, A= 2 and D=−3, so
the amplitude is
√22+ (−3)2=√4 + 9 = √13.
7
Step 2: The period of a function of the form f(x) = Asin(Bx +C) +
Dcos(Bx +C)is given by 2π
|B|. In this case, B= 3, so the period is
2π
|3|=2π
3.
Step 3: To find the phase shift of the function, we need to solve Bx +C= 0
for x. In this case, B= 3 and C= 0, so the phase shift is
x=−C
B=−0
3= 0.
Therefore, the amplitude of f(x)is √13, the period is 2π
3, and the phase
shift is 0.
Question 12
Question
Solve the equation sin(2θ) = cos(θ)for 0≤θ≤2π.
Solution
Step 1: Use the double angle identity sin(2θ) = 2 sin(θ) cos(θ)to rewrite sin(2θ)
in terms of sin(θ)and cos(θ).
sin(2θ) = 2 sin(θ) cos(θ) = cos(θ)
Step 2: Now we have the equation 2 sin(θ) cos(θ) = cos(θ). We can rewrite
this as 2 sin(θ) cos(θ)−cos(θ) = 0.
Step 3: Factor out cos(θ)from the left side of the equation.
cos(θ)(2 sin(θ)−1) = 0
Step 4: Set each factor equal to zero and solve for θ.
cos(θ) = 0 or 2 sin(θ)−1 = 0
Step 5: For cos(θ) = 0, we know that θ=π
2and θ=3π
2.
Step 6: For 2 sin(θ)−1 = 0, solve for sin(θ).
2 sin(θ)−1 = 0
2 sin(θ) = 1
sin(θ) = 1
2
Step 7: The solutions for sin(θ) = 1
2are π
6and 5π
6.
Step 8: Therefore, the solutions to the equation sin(2θ) = cos(θ)for 0≤θ≤
2πare π
6,π
2,5π
6, and 3π
2.
8
Question 13
Question
Let f(x) = sin(2x)and g(x) = cos(x). Find the following expression:
f(x)·g(x)
f(x)−g(x)
Solution
Step 1: Find f(x)·g(x)
f(x)·g(x) = sin(2x)·cos(x)
Step 2: Apply the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)·cos(x) = 2 sin(x) cos(x)·cos(x) = 2 sin(x) cos2(x)
Step 3: Find f(x)−g(x)
f(x)−g(x) = sin(2x)−cos(x)
Step 4: Appy the double angle formula for sine: sin(2x) = 2 sin(x) cos(x)
sin(2x)−cos(x) = 2 sin(x) cos(x)−cos(x) = cos(x)(2 sin(x)−1)
Step 5: Find the expression f(x)·g(x)
f(x)−g(x)
f(x)·g(x)
f(x)−g(x)=2 sin(x) cos2(x)
cos(x)(2 sin(x)−1)
Step 6: Simplify the expression by canceling out a factor of cos(x)
2 sin(x) cos(x)
2 sin(x)−1
Therefore, the expression f(x)·g(x)
f(x)−g(x)simplifies to 2 sin(x)
2 sin(x)−1.
Question 14
Question
Find the exact value of sin (5π
4).
9
Solution
Step 1: Determine the reference angle. To find the reference angle, we first
need to identify the co-terminal angle that lies between 0 and 2π(0 and 360◦)
that is equivalent to 5π
4. Since 5π
4is greater than π, we subtract πto find the
co-terminal angle: 5π
4−π=π
4. Therefore, the reference angle is π
4.
Step 2: Determine the quadrant. The angle 5π
4lies in the third quadrant
(3π
2<5π
4<2π).
Step 3: Calculate the sine function. In the third quadrant, the sine function
is negative. Since the reference angle π
4has a sine value of √2
2, the sine value
for 5π
4will be:
sin (5π
4)=−√2
2
Question 15
Question
Find the general solution of the equation sin(2x) = −1
2for 0≤x < 2π.
Solution
Step 1: Rewrite the equation in terms of the sine function.
sin(2x) = −1
2
Step 2: Use the double angle identity for sine.
sin(2x) = 2 sin(x) cos(x) = −1
2
Step 3: Express sin(x)in terms of cos(x).
2 sin(x) cos(x) = −1
2
sin(x) = −1
4 cos(x)
Step 4: Use the Pythagorean identity for sine and cosine.
sin2(x) + cos2(x) = 1
(−1
4 cos(x))2
+ cos2(x) = 1
Step 5: Solve for cos(x).
1
16 cos2(x)+ cos2(x) = 1
10
1 + 16 cos2(x) = 16 cos2(x)
17 cos2(x) = 1
cos2(x) = 1
17
cos(x) = ±1
√17
Step 6: Determine the possible solutions for x. Since sin(2x) = −1
2, we
must also have sin(x)<0. Therefore, sin(x) = −1
4 cos(x)will be negative when
cos(x)<0. Thus, the solutions occur in the second and third quadrants.
Step 7: Find the angles in the second and third quadrants where cos(x) =
−1
√17 .
x= cos−1(−1
√17)= 2.6779 radians
Since we need to find the general solution, the solutions in the third quadrant
will be x=π−2.6779.
Therefore, the general solution is:
x= 2.6779 + 2πn or x=π−2.6779 + 2πn
where n∈Z.
Question 16
Question
Let f(θ) = sin θ
1+cos θ. Find the values of θin the interval [0,2π]where f(θ)has a
vertical asymptote.
Solution
Step 1: Identify where f(θ)is undefined.
To find the values of θwhere f(θ)has a vertical asymptote, we need to find
where the denominator of f(θ)equals 0, since division by 0 is undefined. So we
solve 1 + cos θ= 0:
cos θ=−1
Step 2: Find the values of θthat satisfy cos θ=−1in the interval [0,2π].
The cosine function is equal to -1 at θ=πin the interval [0,2π].
Therefore, θ=πis the only value in the interval [0,2π]where f(θ)has a
vertical asymptote.
11
Question 17
Question
Given that sin(θ) = 3
5and θis in Quadrant II, find the exact values of tan(θ)
and sec(θ).
Solution
Step 1: We can use the Pythagorean identity sin2(θ) + cos2(θ) = 1 to find the
value of cos(θ).
cos2(θ) = 1 −sin2(θ)
cos2(θ) = 1 −(3
5)2
= 1 −9
25 =16
25
cos(θ) = ±4
5
Since θis in Quadrant II, cos(θ)is negative. Therefore, cos(θ) = −4
5.
Step 2: Next, we can find the values of tan(θ)and sec(θ)using the definitions
of these trigonometric functions.
tan(θ) = sin(θ)
cos(θ)=
3
5
−4
5
=−3
4
sec(θ) = 1
cos(θ)=1
−4
5
=−5
4
Therefore, the exact values of tan(θ)and sec(θ)are −3
4and −5
4, respec-
tively.
Question 18
Question
Given that sin(θ) = 3
5and θis in quadrant II, determine the exact value of
cos(2θ).
Solution
Step 1: Since we know that sin(θ) = 3
5and θis in quadrant II, we can use the
Pythagorean identity to find cos(θ).
cos(θ) = −√1−sin2(θ) = −√1−(3
5)2
=−4
5
12
Step 2: Next, we can use the double angle identity for cosine to find cos(2θ).
cos(2θ) = cos2(θ)−sin2(θ)
=(−4
5)2
−(3
5)2
=16
25 −9
25
=7
25
Therefore, the exact value of cos(2θ)is 7
25 .
Question 19
Question
Let f(θ) = 2 cos2(θ)−3 sin(θ) cos(θ), where θ∈[0,π
2]. Find the maximum
value of f(θ)in the given interval.
Solution
Step 1: Recall the double angle identities for cosine and sine:
cos(2θ) = 2 cos2(θ)−1and sin(2θ) = 2 sin(θ) cos(θ)
Step 2: Rewrite f(θ)using the double angle identity for cosine:
f(θ) = cos(2θ)−3 sin(2θ)
Step 3: To maximize f(θ), we need to find the maximum value of cos(2θ)
and the minimum value of sin(2θ)in the interval θ∈[0,π
2].
Step 4: The maximum value of cos(2θ)in the given interval is 1, which
occurs when θ= 0.
Step 5: The minimum value of sin(2θ)in the interval θ∈[0,π
2]is -1, which
occurs when θ=π
4.
Step 6: Substituting these values into f(θ), we get the maximum value of
f(θ)in the interval:
f(θ) = 1 −3(−1) = 4
Therefore, the maximum value of f(θ)in the interval θ∈[0,π
2]is 4.
Question 20
Question
Find the exact value of sin (3π
4).
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Solution
Step 1: We know that sin (3π
4)=−sin (π
4)since the sine function has period
2π.
Step 2: Using the angle addition identity sin(A−B) = sin Acos B−cos Asin B,
we have:
sin (3π
4)=−sin (π
4)=−(sin(π/2) cos(π/4) −cos(π/2) sin(π/4))
Step 3: We know that sin(π/2) = 1,cos(π/2) = 0,cos(π/4) = √2
2, and
sin(π/4) = √2
2. Substituting these values, we get:
sin (3π
4)=−(1·√2
2−0·√2
2)
Step 4: Simplifying further:
sin (3π
4)=−√2
2
Therefore, the exact value of sin (3π
4)is −√2
2.
Question 21
Question
Let f(x) = 3 sin(x) + 4 cos(x). Determine the amplitude, period, phase shift,
and vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), we use the formula A=√a2+b2, where
Ais the amplitude, and for f(x) = asin(x)+bcos(x),aand bare the coefficients
of sin(x)and cos(x). 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 of f(x), we know that the period of sin(x)and
cos(x)is 2π. Since there are no coefficients that affect the period, the period of
f(x)is also 2π.
Step 3: To find the phase shift of f(x), we need to set sin(x)and cos(x)
equal to their maximum/minimum values and solve for x. For sin(x), it reaches
its maximum value at x=π
2, and for cos(x), it reaches its maximum value at
x= 0. Comparing this to f(x) = 3 sin(x) + 4 cos(x), we can see that the phase
shift is π
2to the right.
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Step 4: To find the vertical shift of f(x), we need to find the average of the
maximum and minimum values of f(x). The maximum value of f(x)is 5, and
the minimum value is −5, so the vertical shift is:
(5 + (−5))/2 = 0
Therefore, the amplitude of f(x)is 5, the period is 2π, the phase shift is π
2
to the right, and the vertical shift is 0.
Question 22
Question
Calculate the exact value of sin (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two common angles:
5π
12 =π
3+π
4
Step 2: Use the sum-to-product formula for sine:
sin (a
n+b
n)= sin (a
n)cos (b
n)+ cos (a
n)sin (b
n)
So we have:
sin (5π
12 )= sin (π
3+π
4)
sin (5π
12 )= sin (π
3)cos (π
4)+ cos (π
3)sin (π
4)
Step 3: Substitute in the values of sin(π/3),cos(π/3),sin(π/4), and cos(π/4):
sin (5π
12 )=√3
2·√2
2+1
2·√2
2
sin (5π
12 )=√6
4+√2
4
sin (5π
12 )=√6 + √2
4
Therefore, sin (5π
12 )=√6+√2
4.
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Question 23
Question
Solve for xin the equation sin(2x) = cos(x)for x∈[0◦,360◦].
Solution
Step 1: Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
Step 2: Substitute this into the given equation to get:
2 sin(x) cos(x) = cos(x)
Step 3: Now we have two cases to consider: Case 1: cos(x)= 0 Divide both
sides by cos(x):
2 sin(x) = 1
Step 4: Solve for sin(x):
sin(x) = 1
2
Step 5: This occurs in the first and second quadrants, so the solutions are
x= 30◦and x= 150◦.
Case 2: cos(x) = 0 This occurs when x= 90◦.
Step 6: Therefore, the solutions to the equation sin(2x) = cos(x)for x∈
[0◦,360◦]are x= 30◦,150◦,and 90◦.
Question 24
Question
Given that sin θ=3
5and θis in Quadrant II, find the exact value of cos (θ
2).
Solution
Step 1: Find cos θusing the Pythagorean identity sin2θ+ cos2θ= 1.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=±4
5
Since θis in Quadrant II, where cosine is negative, cos θ=−4
5.
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Step 2: Use the half-angle formula for cosine to find cos (θ
2).
cos (θ
2)=±√1 + cos θ
2=±√1−4
5
2=±√1−4
5
2=±√1
10 =±1
√10 =±√10
10
Since θis in Quadrant II, where cosine is negative, cos (θ
2)=−√10
10 .
Question 25
Question
Determine all solutions to the equation cos(2x) = sin(x)for 0◦≤x≤360◦.
Solution
Step 1: Recall the double angle identity for cosine: cos(2x) = 2 cos2(x)−1.
Step 2: Substitute the double angle identity into the given equation: 2 cos2(x)−
1 = sin(x).
Step 3: Recall the Pythagorean identity for sine and cosine: sin2(x) +
cos2(x) = 1.
Step 4: Substitute sin2(x)=1−cos2(x)into the equation from step 2:
2 cos2(x)−1 = √1−cos2(x).
Step 5: Squaring both sides of the equation gives 4 cos4(x)−4 cos2(x) + 1 =
1−cos2(x).
Step 6: Simplify the equation: 4 cos4(x)−5 cos2(x) = 0.
Step 7: Factor out a cos2(x)from the equation: cos2(x)(4 cos2(x)−5) = 0.
Step 8: Set each factor equal to zero: cos2(x) = 0 and 4 cos2(x)−5 = 0.
Step 9: Solve the first equation cos2(x) = 0 to find cos(x) = 0. This occurs
at x= 90◦and x= 270◦.
Step 10: Solve the second equation 4 cos2(x)−5=0to find cos(x) = ±√5
2.
This occurs at x= 18.19◦,x= 161.81◦,x= 198.19◦, and x= 341.81◦.
Step 11: Check all solutions in the original equation to verify their validity.
Therefore, all the solutions to the equation cos(2x) = sin(x)for 0◦≤x≤
360◦are x= 18.19◦,90◦,161.81◦,198.19◦,270◦,341.81◦.
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