MATH 122 - TRIGONOMETRY -
Trigonometric Functions
Question Bank - Set 5
Liberty University
Question 1
Question
Let f(x) = sin 2xand g(x) = cos x. Find the exact value of (f◦g)( π
6).
Solution
Step 1: First, we need to find g(π
6). Since g(x) = cos x, we have:
g(π
6) = cos (π
6)
Step 2: Recall that cos π
6=√3
2. Therefore, g(π
6) = cos (π
6) = √3
2.
Step 3: Next, we need to find f(√3
2). Since f(x) = sin 2x, we have:
f(√3
2) = sin (2 ·√3
2)
Step 4: We simplify the above expression to find:
f(√3
2) = sin √3
Step 5: The exact value of sin √3cannot be simplified further, so the final
answer is:
(f◦g)(π
6) = f(g(π
6)) = f(√3
2) = sin √3
Question 2
Question
Find all solutions to the equation cos(2x) = sin(x)for 0≤x≤2π.
Solution
Step 1: Use the double-angle identity to express cos(2x)in terms of cos(x)and
sin(x):
cos(2x) = 2 cos2(x)−1
Step 2: Substitute this into the given equation:
2 cos2(x)−1 = sin(x)
Step 3: Rewrite sin(x)in terms of cos(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1:
2 cos2(x)−1 = √1−cos2(x)
Step 4: Square both sides to eliminate the square root:
4 cos4(x)−4 cos2(x) + 1 = 1 −cos2(x)
Step 5: Rearrange terms to form a quadratic equation in terms of cos2(x):
4 cos4(x)−5 cos2(x) = 0
Step 6: Factor out a cos2(x):
cos2(x)(4 cos2(x)−5) = 0
Step 7: Find the solutions for cos2(x):
• Setting cos2(x) = 0: This gives cos(x) = 0. So, x=π
2,3π
2.
• Setting 4 cos2(x)−5 = 0: This gives cos(x) = ±√5
2. So, x=π
3,5π
3.
Step 8: Check the solutions in the original equation to ensure they are valid:
cos(2·π
2)= cos(π) = −1= 0 = sin(π
2)
cos(2·3π
2)= cos(3π) = −1= 0 = sin(3π
2)
cos(2·π
3)= cos(2π
3)=−1
2=−√3
2= sin(π
3)
cos(2·5π
3)= cos(10π
3)=−1
2=−√3
2= sin(5π
3)
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
3,5π
3.
2
Question 3
Question
Let f(x) = sin x
1+cos x. Determine the domain of f(x).
Solution
Step 1: The function f(x) = sin x
1+cos xwill be undefined when the denominator is
equal to 0, so we need to find the values of xthat make 1 + cos x= 0.
Step 2: Solving 1 + cos x= 0 for cos x, we get
cos x=−1
Step 3: The cosine function has a range of [−1,1], so the equation cos x=−1
has a solution only when x=π.
Step 4: Therefore, the domain of f(x) = sin x
1+cos xis all real numbers xexcept
x=π. In interval notation, the domain is (−∞, π)∪(π, ∞).
Question 4
Question
Find all solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π].
Solution
Step 1: Let u= tan(x). Then the equation becomes u2−3u−1 = 0.
Step 2: Solve the quadratic equation u2−3u−1 = 0 using the quadratic
formula:
u=−(−3) ±√(−3)2−4(1)(−1)
2(1) =3±√13
2.
Step 3: Now, we know that u= tan(x), so we have two cases to consider:
Case 1: tan(x) = 3+√13
2. To find the solutions in the interval [0,2π], we need
to consider all angles xsuch that tan(x) = 3+√13
2. These angles are π
6and 7π
6.
Case 2: tan(x) = 3−√13
2. The angles xthat satisfy this are 5π
6and 11π
6.
Hence, the solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π]are x=π
6,5π
6,7π
6,11π
6.
Question 5
Question
Solve the equation sin2(x) + sin(x)−2 = 0 for 0≤x≤2π.
3
Solution
To solve the equation sin2(x) + sin(x)−2=0, we can substitute sin(x)with a
variable y. The equation becomes y2+y−2 = 0, a quadratic equation. We can
then solve for yand then find the corresponding values of x.
Step 1: Solve the quadratic equation y2+y−2 = 0 for yby factoring or
using the quadratic formula.
The factors of −2that add up to 1are 2and −1, so we have:
y2+ 2y−y−2 = 0
y(y+ 2) −1(y+ 2) = 0
(y−1)(y+ 2) = 0
This gives us y= 1 or y=−2.
Step 2: Recall that sin(x) = y. So, we have:
sin(x) = 1 or sin(x) = −2
Step 3: Solve sin(x) = 1 for xwithin the interval [0,2π].
The solutions are x=π
2and x=3π
2.
Step 4: Solve sin(x) = −2for xwithin the interval [0,2π]. This equation
has no solutions as the range of sine function is [−1,1].
Therefore, the solutions to the equation sin2(x) + sin(x)−2 = 0 in the
interval [0,2π]are x=π
2and x=3π
2.
Question 6
Question
Find a general solution to the equation 2 sin(2x)−√3 cos(2x) = 1 for x∈[0,2π).
Solution
Step 1: Use the double angle identities to simplify the equation. Step 2: Solve
the resulting trigonometric equation. Step 3: Find the general solution within
the given interval.
Step 1: Using the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x)
we can rewrite the equation as:
2(2 sin(x) cos(x)) −√3(cos2(x)−sin2(x)) = 1
Step 2: Expanding the terms gives:
4 sin(x) cos(x)−√3(cos2(x)−sin2(x)) = 1
4
Using the identity cos2(x) + sin2(x) = 1 to replace cos2(x)in the equation,
we get:
4 sin(x) cos(x)−√3(1 −sin2(x)) = 1
Simplify:
4 sin(x) cos(x)−√3 + √3 sin2(x) = 1
Now, using the double angle identity for sin(2x):
2 sin(x) cos(x) = sin(2x)
we can rewrite the equation as:
2 sin(2x)−√3 + √3 sin2(x) = 1
Step 3: Now, our equation becomes:
2 sin(2x)−√3 + √3 sin2(x) = 1
Rearranging gives:
2 sin(2x) = √3−√3 sin2(x)+1
Divide by 2:
sin(2x) = √3−√3 sin2(x)+1
2
Solving for sin(x):
sin(x) = ±√1
2−1
2√7−3 sin2(x)
Thus, the general solution for x∈[0,2π)is:
x=π
12 +nπ
3or x=5π
12 +nπ
3
where nis an integer.
Question 7
Question
Compute the exact value of sin (5π
12 ).
5
Solution
Step 1: We can rewrite 5π
12 as π
12 +2π
3by common denominators.
Step 2: Using the angle addition identity sin(A+B) = sin(A) cos(B) +
cos(A) sin(B), we have:
sin (5π
12 )= sin (π
12 +2π
3).
Step 3: Recognizing that sin (π
12 )and cos (π
12 )are not familiar, we instead
use the well-known values for sin (π
6)=1
2and cos (π
6)=√3
2.
Step 4: Rewriting 2π
3as π
3+π
3and using the double angle identity sin(2A) =
2 sin(A) cos(A), we get:
sin 5π
12 = sin (π
12 +π
3+π
3).
Step 5: Apply the angle addition and double angle formulas to get:
sin 5π
12 = sin (2π
3)cos (π
12)+ cos (2π
3)sin (π
12).
Step 6: Substitute the values we know:
sin 5π
12 =(√3
2)(√3−1
2)+(−1
2)(1
2).
Step 7: Simplify the expression to get the final answer:
sin 5π
12 =3√3−√3−1
4−1
4=2√3−1
4.
Question 8
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
Solution
Step 1: We know the Pythagorean identity sin2(x) + cos2(x) = 1. We can
rewrite the given equation as sin2(x) = 1 −cos(x).
Step 2: Substitute this expression for sin2(x)into the Pythagorean identity:
(1 −cos(x))2+ cos2(x) = 1
Step 3: Expand and simplify the left-hand side of the equation:
1−2 cos(x) + cos2(x) + cos2(x) = 1
6
2 cos2(x)−2 cos(x) = 0
Step 4: Factor out a 2 cos(x)from the equation:
2 cos(x)(cos(x)−1) = 0
Step 5: Set each factor equal to zero and solve for cos(x):
2 cos(x) = 0 or cos(x)−1 = 0
Step 6: Solve the first equation 2 cos(x) = 0:
cos(x) = 0
This occurs when x=π
2or x=3π
2.
Step 7: Solve the second equation cos(x)−1 = 0:
cos(x) = 1
This occurs when x= 0.
Step 8: Thus, the solutions to the equation sin2(x) + cos(x)=1in the
interval [0,2π)are x= 0,x=π
2, and x=3π
2.
Question 9
Question
Convert the following expression to a single trigonometric function:
3 sin xcos x+ 4 sin3x
Solution
Step 1: We will first rewrite sin3xin terms of sin xand cos x. To do this,
we use the identity sin2x= 1 −cos2x, which implies sin3x= sin x·sin2x=
sin x(1 −cos2x).
Step 2: Let’s now substitute sin3xin the expression and simplify.
3 sin xcos x+ 4 sin x(1 −cos2x)
= 3 sin xcos x+ 4 sin x−4 sin xcos2x
Step 3: We can now factor out a sin xfrom the last two terms.
= sin x(3 cos x+ 4 −4 cos2x)
Step 4: To simplify further, we use the Pythagorean identity sin2x+cos2x=
1or cos2x= 1 −sin2x.
= sin x(3 cos x+ 4 −4(1 −sin2x))
7
= sin x(3 cos x+ 4 −4 + 4 sin2x)
= sin x(4 sin2x+ 3 cos x)
Step 5: We can write cos xin terms of sin xusing the Pythagorean iden-
tity sin2x+ cos2x= 1, or cos x=±√1−sin2x. Now we substitute cos x=
√1−sin2xin the expression.
= sin x(4 sin2x+ 3√1−sin2x)
Hence, the expression 3 sin xcos x+ 4 sin3xsimplifies to sin x(4 sin2x+
3√1−sin2x).
Question 10
Question
Let f(x) = ex+e−x
2and g(x) = ex−e−x
2. Find expressions for sinh(x)and cosh(x)
in terms of f(x)and g(x).
Solution
Step 1: Recall the definitions of hyperbolic sine and hyperbolic cosine:
sinh(x) = ex−e−x
2and cosh(x) = ex+e−x
2
Step 2: We can see that g(x)is equivalent to sinh(x), and f(x)is equivalent
to cosh(x). Thus, we have:
sinh(x) = g(x)and cosh(x) = f(x)
Therefore, the expressions for sinh(x)and cosh(x)in terms of f(x)and g(x)
are:
sinh(x) = g(x)and cosh(x) = f(x)
Question 11
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
8
Solution
To solve the equation sin(2x) = cos(x), we will first use the double angle identity
for sine, sin(2x) = 2 sin(x) cos(x). Then, we can substitute this into the original
equation to get 2 sin(x) cos(x) = cos(x).
Step 1: Set sin(x)equal to 0or 1.
If cos(x)=0, then x=π
2,3π
2. However, x=3π
2is outside the interval
0≤x≤2π, so we reject this solution.
If sin(x) = 1, then x=π
2.
Step 2: Check the value of x=π
2in the equation sin(2x) = cos(x).
For x=π
2, we have sin(π) = cos(π
2), which is true. Therefore, the solution
to the equation is x=π
2.
Hence, the solution to the equation sin(2x) = cos(x)for 0≤x≤2πis
x=π
2.
Question 12
Question
Given that sin(θ) = −3
5and θis in Quadrant IV, find the exact values of cos(θ),
tan(θ),csc(θ),sec(θ), and cot(θ).
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant IV, we can use the Pythagorean
identity to find cos(θ).
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
Therefore, cos(θ) = ±4
5. Since θis in Quadrant IV, we have cos(θ) = 4
5.
Step 2: Next, we can find tan(θ)using the definitions of tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=−3
5
4
5
=−3
4
Step 3: To find csc(θ), we use the definition csc(θ) = 1
sin(θ).
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
Step 4: Next, we find sec(θ)using the definition sec(θ) = 1
cos(θ).
sec(θ) = 1
cos(θ)=1
4
5
=5
4
9
Step 5: Lastly, we find cot(θ)using the definition cot(θ) = 1
tan(θ).
cot(θ) = 1
tan(θ)=1
−3
4
=−4
3
Therefore, the exact values of cos(θ),tan(θ),csc(θ),sec(θ), and cot(θ)are
4
5,−3
4,−5
3,5
4, and −4
3respectively.
Question 13
Question
Suppose sin(θ) = 3
5and cos(θ)<0. Find the values of the other five trigono-
metric functions of θ.
Solution
Step 1: Since sin(θ) = 3
5and cos(θ)<0, we know that θlies in the second
quadrant. By drawing a triangle in the second quadrant, we can determine the
value of the third side using the Pythagorean theorem. Step 2: Let’s assume
the hypotenuse is 5(since the sine of θis 3
5). Using the Pythagorean theorem,
we find that the opposite side is 3. Step 3: Now, we can determine the adjacent
side using the cosine function. Since cosine is negative in the second quadrant,
the adjacent side will be negative. Therefore, the adjacent side is −4. Step 4:
Now we can find the values of the other five trigonometric functions of θ. Step
5: cos(θ) = −4
5,tan(θ) = 3
−4,cot(θ) = −4
3,sec(θ) = −5
4,csc(θ) = 5
3.
Question 14
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ)
10
Solution
Start with the left-hand side (LHS):
LHS =sin(θ)
1−cos(θ)
=sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)
=sin(θ)(1 + cos(θ))
1−cos2(θ)
=sin(θ)(1 + cos(θ))
sin2(θ)(using Pythagorean identity)
=sin(θ) + sin(θ) cos(θ)
sin2(θ)
=sin(θ)
sin(θ)+sin(θ) cos(θ)
sin(θ)
= 1 + cos(θ)
=1 + sin(θ)
sin(θ)(using Pythagorean identity)
Therefore, we have shown that sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ), which completes
the proof.
Question 15
Question
Evaluate the following trigonometric expression:
tan (5π
12 )·cot (π
12)
Solution
Step 1: First, let’s express 5π
12 in terms of π
12 :
5π
12 =2π
12 +3π
12 =π
6+π
4=3π
12 +4π
12 =3π
12 +π
3
Step 2: Since tan (π
4)= 1 and tan (π
6)=√3
3, we have:
tan (5π
12 )= tan (3π
12 +π
3)=tan (3π
12 )+ tan (π
3)
1−tan (3π
12 )·tan (π
3)
11
=tan (π
4)+ tan (π
6)
1−tan (π
4)·tan (π
6)=1 + √3
3
1−1·√3
3
=3 + √3
3−√3= 2 + √3
Step 3: Similarly, we know that cot (π
6)=√3
3, so:
cot (π
12)= cot (π
6−π
4)=cot (π
6)cot (π
4)+ 1
cot (π
6)+ cot (π
4)
=
√3
3·1+1
√3
3+ 1 =√3+3
√3+3 = 1
Step 4: Therefore, we can now evaluate the expression:
tan (5π
12 )·cot (π
12)= (2 + √3) ·1 = 2 + √3
So, tan (5π
12 )·cot (π
12 )= 2 + √3.
Question 16
Question
Solve the equation for 0≤x≤2π:2 sin(2x) + √3 = 0.
Solution
Step 1: Rewrite the equation to isolate sin(2x):
2 sin(2x) = −√3
Step 2: Divide by 2 to solve for sin(2x):
sin(2x) = −√3
2
Step 3: Recall the angle where sin(π
3)=√3
2. We can now find the solutions
for 2x:
2x=11π
6+ 2πk or 2x=7π
6+ 2πk
Step 4: Solve for x:
x=11π
12 +πk or x=7π
12 +πk
Therefore, the solutions to the equation 2 sin(2x) + √3 = 0 for 0≤x≤2π
are x=11π
12 and x=7π
12 .
12
Question 17
Question
Find the exact value of tan (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles: π
3and π
4. Step
2: Thus, we have 5π
12 =π
3+π
4. Step 3: Using the angle addition identity for
tangent, we have
tan (5π
12 )=tan (π
3+π
4)
1−tan (π
3
tan(π
4).Step 4: Since tan(π/3) = √3and tan(π/4) = 1, we can substitute in
these values:
tan (5π
12 )=tan (π
3+π
4)
1−√3.
Step 5: Using the sum-to-product identity, we have
tan (5π
12 )=
tan(π
3)+tan(π
4)
1−tan(π
3)tan(π
4)
1−√3.
Step 6: Simplifying further gives us
tan (5π
12 )=
√3+1
1−√3
1−√3.
Step 7: Rationalizing the denominator, we get
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3).
Step 8: Simplifying the numerator gives us
tan (5π
12 )=4+2√3
−2.
Step 9: Therefore, the exact value of tan (5π
12 )is −2−√3.
Question 18
Question
Let f(x) = sin(3x)and g(x) = cos(2x), find the values of xin the interval [0,2π]
that satisfy the equation f(x) = g(x).
13
Solution
Step 1: Set up the equation f(x) = sin(3x) = g(x) = cos(2x).
Step 2: Since sin(3x) = cos (π
2−3x), rewrite the equation as cos (π
2−3x)=
cos(2x).
Step 3: In order for cos (π
2−3x)to equal cos(2x), the angles inside the cosine
function must be equal or their difference must be a multiple of 2π. Thus, we
have two cases:
• Case 1: π
2−3x= 2x
• Case 2: π
2−3x=−2x
Step 4: Solve Case 1: π
2−3x= 2x
π
2= 5x
x=π
10
Step 5: Solve Case 2: π
2−3x=−2x
π
2=x
Step 6: Check if the solutions fall within the interval [0,2π]:
0≤π
10 ≤2π
0≤π≤20π
However, πfalls outside the interval, so the only solution that satisfies x∈
[0,2π]is x=π
10 .
Question 19
Question
Solve the equation sin2(x)−cos(x) = 0 for xin the interval [0,2π].
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin2(x)−cos(x) = 0
sin2(x) = cos(x)
14
sin2(x) = √1−sin2(x)
Step 2: Square both sides of the equation to eliminate the square root.
sin2(x) = 1 −sin2(x)
2 sin2(x) = 1
sin2(x) = 1
2
Step 3: Solve for sin(x)by taking the square root of both sides, considering
the interval [0,2π].
sin(x) = ±√1
2=±√2
2
Step 4: Determine the possible values of xin the interval [0,2π]. Since sin(x)
is positive in the first and second quadrants and negative in the third and fourth
quadrants, the solutions are:
x=π
4,3π
4,5π
4,7π
4
Question 20
Question
Let θbe an angle in standard position. If sin θ=3
5and cos θ < 0, determine
the exact value of tan θ.
Solution
Step 1: Since sin θ=3
5, we can use the Pythagorean identity sin2θ+ cos2θ= 1
to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=−4
5(since cos θ < 0)
Step 2: Now, we can find tan θusing the definition tan θ=sin θ
cos θ.
tan θ=sin θ
cos θ=
3
5
−4
5
=3
5·(−5
4)=−3
4
Therefore, the exact value of tan θis −3
4.
15
Question 21
Question
Let f(x) = 3 sin(2x) + 4 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), recall that the amplitude of a function of
the form Asin(Bx)or Acos(Bx)is |A|. The amplitude of f(x)is the absolute
value of the coefficient of sin(2x)or cos(2x). In this case, the amplitude is
|3|= 3.
Step 2: The period of a function of the form sin(Bx)or cos(Bx)is 2π
|B|.
Therefore, the period of f(x)is 2π
2=π.
Step 3: To find the phase shift of f(x), set 2xequal to 0 and solve for xto
find the phase shift. We have:
2x= 0 ⇒x= 0
So, the phase shift of f(x)is 0.
Step 4: The vertical shift of a function of the form Asin(Bx) + Cor
Acos(Bx) + Cis the value of C. In this case, the vertical shift of f(x)is
0.
Question 22
Question
Solve the equation cos(2x) = sin(x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Let’s rewrite the equation using the double angle identity: 2 cos2(x)−
1 = sin(x).
Step 3: Recall that sin(x) = 2 sin(x) cos(x).
Step 4: Substitute cos(x)for 1−cos2(x)
2in the equation: 2(1−cos2(x)
2)2−1 =
sin(x).
Step 5: Simplify the equation: 2(1−2 cos2(x)+cos4(x)
4)−1 = sin(x).
Step 6: Distribute and simplify further: 1
2−cos2(x)+ 1
2cos4(x)−1 = sin(x).
Step 7: Rearrange the equation: 1
2cos4(x)−cos2(x)−sin(x)−1
2= 0.
Step 8: Let u= cos(x)and rewrite the equation: 1
2u4−u2−2u−1
2= 0.
Step 9: This is a quadratic equation in u2. Let’s solve it: u2= 2 ±√2.
16
Step 10: Since cos(x)can only take values between -1 and 1, the solutions
for u2are u2= 2 −√2and u2= 2 + √2.
Step 11: Solve for u:u=±√2−√2and u=±√2 + √2.
Step 12: Recall that cos(x) = u, we have cos(x) = √2−√2,cos(x) =
−√2−√2,cos(x) = √2 + √2, and cos(x) = −√2 + √2.
Step 13: Find the corresponding values of xusing the unit circle and the
relationships between cos(x)and x.
Step 14: The solutions to the equation are x=π
8,x=7π
8,x=3π
8, and
x=11π
8.
Question 23
Question
Let f(x) = cos(2x)and g(x) = sin(3x). Find the exact value of f(π
6)·g(π
4).
Solution
Step 1: We first find f(π
6). Since f(x) = cos(2x), we have f(π
6)= cos (2·π
6)=
cos (π
3)=1
2.
Step 2: Next, we find g(π
4). Given that g(x) = sin(3x), we have g(π
4)=
sin (3·π
4)= sin (3π
4)=−√2
2.
Step 3: Finally, we calculate the product f(π
6)·g(π
4).
f(π
6)·g(π
4)=1
2·(−√2
2)=−√2
4
Therefore, f(π
6)·g(π
4)=−√2
4.
Question 24
Question
Evaluate the exact value of cos (5π
12 ).
Solution
Step 1: We can use the sum and difference formula for cosine to rewrite the
given angle as a sum of two common angles. Step 2: In this case, we can
rewrite 5π
12 as π
3+π
4. Step 3: Using the sum formula for cosine, cos(a+b) =
cos acos b−sin asin b, we have:
cos (5π
12 )= cos (π
3+π
4)
= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
17
Step 4: Recall that cos(π/3) = 1/2,sin(π/3) = √3/2,cos(π/4) = √2/2, and
sin(π/4) = √2/2. Step 5: Substituting these values into the formula, we get:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4.
Therefore, the exact value of cos (5π
12 )is √2−√6
4.
Question 25
Question
Prove that sin(90◦−θ) = cos(θ)for any angle θ.
Solution
To prove the given trigonometric identity, we will use the sum-to-product for-
mula for sine. The sum-to-product formula states that sin(A+B) = sin Acos B+
cos Asin Bfor any angles Aand B.
Step 1: Let A= 90◦and B=−θ. Then, we have:
sin(90◦−θ) = sin 90◦cos(−θ) + cos 90◦sin(−θ)
Step 2: Recall that sin 90◦= 1 and cos 90◦= 0. We also know that
cos(−θ) = cos θand sin(−θ) = −sin θ. Substituting these values, we get:
= 1 ·cos θ+ 0 ·(−sin θ)
= cos θ+ 0
Step 3: Therefore, sin(90◦−θ) = cos θ. Hence, the identity sin(90◦−θ) =
cos θis proven.
18
Solution
Step 1: Use the double-angle identity to express cos(2x)in terms of cos(x)and
sin(x):
cos(2x) = 2 cos2(x)−1
Step 2: Substitute this into the given equation:
2 cos2(x)−1 = sin(x)
Step 3: Rewrite sin(x)in terms of cos(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1:
2 cos2(x)−1 = √1−cos2(x)
Step 4: Square both sides to eliminate the square root:
4 cos4(x)−4 cos2(x) + 1 = 1 −cos2(x)
Step 5: Rearrange terms to form a quadratic equation in terms of cos2(x):
4 cos4(x)−5 cos2(x) = 0
Step 6: Factor out a cos2(x):
cos2(x)(4 cos2(x)−5) = 0
Step 7: Find the solutions for cos2(x):
• Setting cos2(x) = 0: This gives cos(x) = 0. So, x=π
2,3π
2.
• Setting 4 cos2(x)−5 = 0: This gives cos(x) = ±√5
2. So, x=π
3,5π
3.
Step 8: Check the solutions in the original equation to ensure they are valid:
cos(2·π
2)= cos(π) = −1= 0 = sin(π
2)
cos(2·3π
2)= cos(3π) = −1= 0 = sin(3π
2)
cos(2·π
3)= cos(2π
3)=−1
2=−√3
2= sin(π
3)
cos(2·5π
3)= cos(10π
3)=−1
2=−√3
2= sin(5π
3)
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
3,5π
3.
2
Question 3
Question
Let f(x) = sin x
1+cos x. Determine the domain of f(x).
Solution
Step 1: The function f(x) = sin x
1+cos xwill be undefined when the denominator is
equal to 0, so we need to find the values of xthat make 1 + cos x= 0.
Step 2: Solving 1 + cos x= 0 for cos x, we get
cos x=−1
Step 3: The cosine function has a range of [−1,1], so the equation cos x=−1
has a solution only when x=π.
Step 4: Therefore, the domain of f(x) = sin x
1+cos xis all real numbers xexcept
x=π. In interval notation, the domain is (−∞, π)∪(π, ∞).
Question 4
Question
Find all solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π].
Solution
Step 1: Let u= tan(x). Then the equation becomes u2−3u−1 = 0.
Step 2: Solve the quadratic equation u2−3u−1 = 0 using the quadratic
formula:
u=−(−3) ±√(−3)2−4(1)(−1)
2(1) =3±√13
2.
Step 3: Now, we know that u= tan(x), so we have two cases to consider:
Case 1: tan(x) = 3+√13
2. To find the solutions in the interval [0,2π], we need
to consider all angles xsuch that tan(x) = 3+√13
2. These angles are π
6and 7π
6.
Case 2: tan(x) = 3−√13
2. The angles xthat satisfy this are 5π
6and 11π
6.
Hence, the solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π]are x=π
6,5π
6,7π
6,11π
6.
Question 5
Question
Solve the equation sin2(x) + sin(x)−2 = 0 for 0≤x≤2π.
3
Solution
To solve the equation sin2(x) + sin(x)−2=0, we can substitute sin(x)with a
variable y. The equation becomes y2+y−2 = 0, a quadratic equation. We can
then solve for yand then find the corresponding values of x.
Step 1: Solve the quadratic equation y2+y−2 = 0 for yby factoring or
using the quadratic formula.
The factors of −2that add up to 1are 2and −1, so we have:
y2+ 2y−y−2 = 0
y(y+ 2) −1(y+ 2) = 0
(y−1)(y+ 2) = 0
This gives us y= 1 or y=−2.
Step 2: Recall that sin(x) = y. So, we have:
sin(x) = 1 or sin(x) = −2
Step 3: Solve sin(x) = 1 for xwithin the interval [0,2π].
The solutions are x=π
2and x=3π
2.
Step 4: Solve sin(x) = −2for xwithin the interval [0,2π]. This equation
has no solutions as the range of sine function is [−1,1].
Therefore, the solutions to the equation sin2(x) + sin(x)−2 = 0 in the
interval [0,2π]are x=π
2and x=3π
2.
Question 6
Question
Find a general solution to the equation 2 sin(2x)−√3 cos(2x) = 1 for x∈[0,2π).
Solution
Step 1: Use the double angle identities to simplify the equation. Step 2: Solve
the resulting trigonometric equation. Step 3: Find the general solution within
the given interval.
Step 1: Using the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x)
we can rewrite the equation as:
2(2 sin(x) cos(x)) −√3(cos2(x)−sin2(x)) = 1
Step 2: Expanding the terms gives:
4 sin(x) cos(x)−√3(cos2(x)−sin2(x)) = 1
4
Using the identity cos2(x) + sin2(x) = 1 to replace cos2(x)in the equation,
we get:
4 sin(x) cos(x)−√3(1 −sin2(x)) = 1
Simplify:
4 sin(x) cos(x)−√3 + √3 sin2(x) = 1
Now, using the double angle identity for sin(2x):
2 sin(x) cos(x) = sin(2x)
we can rewrite the equation as:
2 sin(2x)−√3 + √3 sin2(x) = 1
Step 3: Now, our equation becomes:
2 sin(2x)−√3 + √3 sin2(x) = 1
Rearranging gives:
2 sin(2x) = √3−√3 sin2(x)+1
Divide by 2:
sin(2x) = √3−√3 sin2(x)+1
2
Solving for sin(x):
sin(x) = ±√1
2−1
2√7−3 sin2(x)
Thus, the general solution for x∈[0,2π)is:
x=π
12 +nπ
3or x=5π
12 +nπ
3
where nis an integer.
Question 7
Question
Compute the exact value of sin (5π
12 ).
5
Solution
Step 1: We can rewrite 5π
12 as π
12 +2π
3by common denominators.
Step 2: Using the angle addition identity sin(A+B) = sin(A) cos(B) +
cos(A) sin(B), we have:
sin (5π
12 )= sin (π
12 +2π
3).
Step 3: Recognizing that sin (π
12 )and cos (π
12 )are not familiar, we instead
use the well-known values for sin (π
6)=1
2and cos (π
6)=√3
2.
Step 4: Rewriting 2π
3as π
3+π
3and using the double angle identity sin(2A) =
2 sin(A) cos(A), we get:
sin 5π
12 = sin (π
12 +π
3+π
3).
Step 5: Apply the angle addition and double angle formulas to get:
sin 5π
12 = sin (2π
3)cos (π
12)+ cos (2π
3)sin (π
12).
Step 6: Substitute the values we know:
sin 5π
12 =(√3
2)(√3−1
2)+(−1
2)(1
2).
Step 7: Simplify the expression to get the final answer:
sin 5π
12 =3√3−√3−1
4−1
4=2√3−1
4.
Question 8
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
Solution
Step 1: We know the Pythagorean identity sin2(x) + cos2(x) = 1. We can
rewrite the given equation as sin2(x) = 1 −cos(x).
Step 2: Substitute this expression for sin2(x)into the Pythagorean identity:
(1 −cos(x))2+ cos2(x) = 1
Step 3: Expand and simplify the left-hand side of the equation:
1−2 cos(x) + cos2(x) + cos2(x) = 1
6
2 cos2(x)−2 cos(x) = 0
Step 4: Factor out a 2 cos(x)from the equation:
2 cos(x)(cos(x)−1) = 0
Step 5: Set each factor equal to zero and solve for cos(x):
2 cos(x) = 0 or cos(x)−1 = 0
Step 6: Solve the first equation 2 cos(x) = 0:
cos(x) = 0
This occurs when x=π
2or x=3π
2.
Step 7: Solve the second equation cos(x)−1 = 0:
cos(x) = 1
This occurs when x= 0.
Step 8: Thus, the solutions to the equation sin2(x) + cos(x)=1in the
interval [0,2π)are x= 0,x=π
2, and x=3π
2.
Question 9
Question
Convert the following expression to a single trigonometric function:
3 sin xcos x+ 4 sin3x
Solution
Step 1: We will first rewrite sin3xin terms of sin xand cos x. To do this,
we use the identity sin2x= 1 −cos2x, which implies sin3x= sin x·sin2x=
sin x(1 −cos2x).
Step 2: Let’s now substitute sin3xin the expression and simplify.
3 sin xcos x+ 4 sin x(1 −cos2x)
= 3 sin xcos x+ 4 sin x−4 sin xcos2x
Step 3: We can now factor out a sin xfrom the last two terms.
= sin x(3 cos x+ 4 −4 cos2x)
Step 4: To simplify further, we use the Pythagorean identity sin2x+cos2x=
1or cos2x= 1 −sin2x.
= sin x(3 cos x+ 4 −4(1 −sin2x))
7
= sin x(3 cos x+ 4 −4 + 4 sin2x)
= sin x(4 sin2x+ 3 cos x)
Step 5: We can write cos xin terms of sin xusing the Pythagorean iden-
tity sin2x+ cos2x= 1, or cos x=±√1−sin2x. Now we substitute cos x=
√1−sin2xin the expression.
= sin x(4 sin2x+ 3√1−sin2x)
Hence, the expression 3 sin xcos x+ 4 sin3xsimplifies to sin x(4 sin2x+
3√1−sin2x).
Question 10
Question
Let f(x) = ex+e−x
2and g(x) = ex−e−x
2. Find expressions for sinh(x)and cosh(x)
in terms of f(x)and g(x).
Solution
Step 1: Recall the definitions of hyperbolic sine and hyperbolic cosine:
sinh(x) = ex−e−x
2and cosh(x) = ex+e−x
2
Step 2: We can see that g(x)is equivalent to sinh(x), and f(x)is equivalent
to cosh(x). Thus, we have:
sinh(x) = g(x)and cosh(x) = f(x)
Therefore, the expressions for sinh(x)and cosh(x)in terms of f(x)and g(x)
are:
sinh(x) = g(x)and cosh(x) = f(x)
Question 11
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
8
Solution
To solve the equation sin(2x) = cos(x), we will first use the double angle identity
for sine, sin(2x) = 2 sin(x) cos(x). Then, we can substitute this into the original
equation to get 2 sin(x) cos(x) = cos(x).
Step 1: Set sin(x)equal to 0or 1.
If cos(x)=0, then x=π
2,3π
2. However, x=3π
2is outside the interval
0≤x≤2π, so we reject this solution.
If sin(x) = 1, then x=π
2.
Step 2: Check the value of x=π
2in the equation sin(2x) = cos(x).
For x=π
2, we have sin(π) = cos(π
2), which is true. Therefore, the solution
to the equation is x=π
2.
Hence, the solution to the equation sin(2x) = cos(x)for 0≤x≤2πis
x=π
2.
Question 12
Question
Given that sin(θ) = −3
5and θis in Quadrant IV, find the exact values of cos(θ),
tan(θ),csc(θ),sec(θ), and cot(θ).
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant IV, we can use the Pythagorean
identity to find cos(θ).
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
Therefore, cos(θ) = ±4
5. Since θis in Quadrant IV, we have cos(θ) = 4
5.
Step 2: Next, we can find tan(θ)using the definitions of tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=−3
5
4
5
=−3
4
Step 3: To find csc(θ), we use the definition csc(θ) = 1
sin(θ).
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
Step 4: Next, we find sec(θ)using the definition sec(θ) = 1
cos(θ).
sec(θ) = 1
cos(θ)=1
4
5
=5
4
9
Step 5: Lastly, we find cot(θ)using the definition cot(θ) = 1
tan(θ).
cot(θ) = 1
tan(θ)=1
−3
4
=−4
3
Therefore, the exact values of cos(θ),tan(θ),csc(θ),sec(θ), and cot(θ)are
4
5,−3
4,−5
3,5
4, and −4
3respectively.
Question 13
Question
Suppose sin(θ) = 3
5and cos(θ)<0. Find the values of the other five trigono-
metric functions of θ.
Solution
Step 1: Since sin(θ) = 3
5and cos(θ)<0, we know that θlies in the second
quadrant. By drawing a triangle in the second quadrant, we can determine the
value of the third side using the Pythagorean theorem. Step 2: Let’s assume
the hypotenuse is 5(since the sine of θis 3
5). Using the Pythagorean theorem,
we find that the opposite side is 3. Step 3: Now, we can determine the adjacent
side using the cosine function. Since cosine is negative in the second quadrant,
the adjacent side will be negative. Therefore, the adjacent side is −4. Step 4:
Now we can find the values of the other five trigonometric functions of θ. Step
5: cos(θ) = −4
5,tan(θ) = 3
−4,cot(θ) = −4
3,sec(θ) = −5
4,csc(θ) = 5
3.
Question 14
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ)
10
Solution
Start with the left-hand side (LHS):
LHS =sin(θ)
1−cos(θ)
=sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)
=sin(θ)(1 + cos(θ))
1−cos2(θ)
=sin(θ)(1 + cos(θ))
sin2(θ)(using Pythagorean identity)
=sin(θ) + sin(θ) cos(θ)
sin2(θ)
=sin(θ)
sin(θ)+sin(θ) cos(θ)
sin(θ)
= 1 + cos(θ)
=1 + sin(θ)
sin(θ)(using Pythagorean identity)
Therefore, we have shown that sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ), which completes
the proof.
Question 15
Question
Evaluate the following trigonometric expression:
tan (5π
12 )·cot (π
12)
Solution
Step 1: First, let’s express 5π
12 in terms of π
12 :
5π
12 =2π
12 +3π
12 =π
6+π
4=3π
12 +4π
12 =3π
12 +π
3
Step 2: Since tan (π
4)= 1 and tan (π
6)=√3
3, we have:
tan (5π
12 )= tan (3π
12 +π
3)=tan (3π
12 )+ tan (π
3)
1−tan (3π
12 )·tan (π
3)
11
=tan (π
4)+ tan (π
6)
1−tan (π
4)·tan (π
6)=1 + √3
3
1−1·√3
3
=3 + √3
3−√3= 2 + √3
Step 3: Similarly, we know that cot (π
6)=√3
3, so:
cot (π
12)= cot (π
6−π
4)=cot (π
6)cot (π
4)+ 1
cot (π
6)+ cot (π
4)
=
√3
3·1+1
√3
3+ 1 =√3+3
√3+3 = 1
Step 4: Therefore, we can now evaluate the expression:
tan (5π
12 )·cot (π
12)= (2 + √3) ·1 = 2 + √3
So, tan (5π
12 )·cot (π
12 )= 2 + √3.
Question 16
Question
Solve the equation for 0≤x≤2π:2 sin(2x) + √3 = 0.
Solution
Step 1: Rewrite the equation to isolate sin(2x):
2 sin(2x) = −√3
Step 2: Divide by 2 to solve for sin(2x):
sin(2x) = −√3
2
Step 3: Recall the angle where sin(π
3)=√3
2. We can now find the solutions
for 2x:
2x=11π
6+ 2πk or 2x=7π
6+ 2πk
Step 4: Solve for x:
x=11π
12 +πk or x=7π
12 +πk
Therefore, the solutions to the equation 2 sin(2x) + √3 = 0 for 0≤x≤2π
are x=11π
12 and x=7π
12 .
12
Question 17
Question
Find the exact value of tan (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles: π
3and π
4. Step
2: Thus, we have 5π
12 =π
3+π
4. Step 3: Using the angle addition identity for
tangent, we have
tan (5π
12 )=tan (π
3+π
4)
1−tan (π
3
tan(π
4).Step 4: Since tan(π/3) = √3and tan(π/4) = 1, we can substitute in
these values:
tan (5π
12 )=tan (π
3+π
4)
1−√3.
Step 5: Using the sum-to-product identity, we have
tan (5π
12 )=
tan(π
3)+tan(π
4)
1−tan(π
3)tan(π
4)
1−√3.
Step 6: Simplifying further gives us
tan (5π
12 )=
√3+1
1−√3
1−√3.
Step 7: Rationalizing the denominator, we get
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3).
Step 8: Simplifying the numerator gives us
tan (5π
12 )=4+2√3
−2.
Step 9: Therefore, the exact value of tan (5π
12 )is −2−√3.
Question 18
Question
Let f(x) = sin(3x)and g(x) = cos(2x), find the values of xin the interval [0,2π]
that satisfy the equation f(x) = g(x).
13
Solution
Step 1: Set up the equation f(x) = sin(3x) = g(x) = cos(2x).
Step 2: Since sin(3x) = cos (π
2−3x), rewrite the equation as cos (π
2−3x)=
cos(2x).
Step 3: In order for cos (π
2−3x)to equal cos(2x), the angles inside the cosine
function must be equal or their difference must be a multiple of 2π. Thus, we
have two cases:
• Case 1: π
2−3x= 2x
• Case 2: π
2−3x=−2x
Step 4: Solve Case 1: π
2−3x= 2x
π
2= 5x
x=π
10
Step 5: Solve Case 2: π
2−3x=−2x
π
2=x
Step 6: Check if the solutions fall within the interval [0,2π]:
0≤π
10 ≤2π
0≤π≤20π
However, πfalls outside the interval, so the only solution that satisfies x∈
[0,2π]is x=π
10 .
Question 19
Question
Solve the equation sin2(x)−cos(x) = 0 for xin the interval [0,2π].
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin2(x)−cos(x) = 0
sin2(x) = cos(x)
14
sin2(x) = √1−sin2(x)
Step 2: Square both sides of the equation to eliminate the square root.
sin2(x) = 1 −sin2(x)
2 sin2(x) = 1
sin2(x) = 1
2
Step 3: Solve for sin(x)by taking the square root of both sides, considering
the interval [0,2π].
sin(x) = ±√1
2=±√2
2
Step 4: Determine the possible values of xin the interval [0,2π]. Since sin(x)
is positive in the first and second quadrants and negative in the third and fourth
quadrants, the solutions are:
x=π
4,3π
4,5π
4,7π
4
Question 20
Question
Let θbe an angle in standard position. If sin θ=3
5and cos θ < 0, determine
the exact value of tan θ.
Solution
Step 1: Since sin θ=3
5, we can use the Pythagorean identity sin2θ+ cos2θ= 1
to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=−4
5(since cos θ < 0)
Step 2: Now, we can find tan θusing the definition tan θ=sin θ
cos θ.
tan θ=sin θ
cos θ=
3
5
−4
5
=3
5·(−5
4)=−3
4
Therefore, the exact value of tan θis −3
4.
15
Question 21
Question
Let f(x) = 3 sin(2x) + 4 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), recall that the amplitude of a function of
the form Asin(Bx)or Acos(Bx)is |A|. The amplitude of f(x)is the absolute
value of the coefficient of sin(2x)or cos(2x). In this case, the amplitude is
|3|= 3.
Step 2: The period of a function of the form sin(Bx)or cos(Bx)is 2π
|B|.
Therefore, the period of f(x)is 2π
2=π.
Step 3: To find the phase shift of f(x), set 2xequal to 0 and solve for xto
find the phase shift. We have:
2x= 0 ⇒x= 0
So, the phase shift of f(x)is 0.
Step 4: The vertical shift of a function of the form Asin(Bx) + Cor
Acos(Bx) + Cis the value of C. In this case, the vertical shift of f(x)is
0.
Question 22
Question
Solve the equation cos(2x) = sin(x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Let’s rewrite the equation using the double angle identity: 2 cos2(x)−
1 = sin(x).
Step 3: Recall that sin(x) = 2 sin(x) cos(x).
Step 4: Substitute cos(x)for 1−cos2(x)
2in the equation: 2(1−cos2(x)
2)2−1 =
sin(x).
Step 5: Simplify the equation: 2(1−2 cos2(x)+cos4(x)
4)−1 = sin(x).
Step 6: Distribute and simplify further: 1
2−cos2(x)+ 1
2cos4(x)−1 = sin(x).
Step 7: Rearrange the equation: 1
2cos4(x)−cos2(x)−sin(x)−1
2= 0.
Step 8: Let u= cos(x)and rewrite the equation: 1
2u4−u2−2u−1
2= 0.
Step 9: This is a quadratic equation in u2. Let’s solve it: u2= 2 ±√2.
16
Step 10: Since cos(x)can only take values between -1 and 1, the solutions
for u2are u2= 2 −√2and u2= 2 + √2.
Step 11: Solve for u:u=±√2−√2and u=±√2 + √2.
Step 12: Recall that cos(x) = u, we have cos(x) = √2−√2,cos(x) =
−√2−√2,cos(x) = √2 + √2, and cos(x) = −√2 + √2.
Step 13: Find the corresponding values of xusing the unit circle and the
relationships between cos(x)and x.
Step 14: The solutions to the equation are x=π
8,x=7π
8,x=3π
8, and
x=11π
8.
Question 23
Question
Let f(x) = cos(2x)and g(x) = sin(3x). Find the exact value of f(π
6)·g(π
4).
Solution
Step 1: We first find f(π
6). Since f(x) = cos(2x), we have f(π
6)= cos (2·π
6)=
cos (π
3)=1
2.
Step 2: Next, we find g(π
4). Given that g(x) = sin(3x), we have g(π
4)=
sin (3·π
4)= sin (3π
4)=−√2
2.
Step 3: Finally, we calculate the product f(π
6)·g(π
4).
f(π
6)·g(π
4)=1
2·(−√2
2)=−√2
4
Therefore, f(π
6)·g(π
4)=−√2
4.
Question 24
Question
Evaluate the exact value of cos (5π
12 ).
Solution
Step 1: We can use the sum and difference formula for cosine to rewrite the
given angle as a sum of two common angles. Step 2: In this case, we can
rewrite 5π
12 as π
3+π
4. Step 3: Using the sum formula for cosine, cos(a+b) =
cos acos b−sin asin b, we have:
cos (5π
12 )= cos (π
3+π
4)
= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
17
Step 4: Recall that cos(π/3) = 1/2,sin(π/3) = √3/2,cos(π/4) = √2/2, and
sin(π/4) = √2/2. Step 5: Substituting these values into the formula, we get:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4.
Therefore, the exact value of cos (5π
12 )is √2−√6
4.
Question 25
Question
Prove that sin(90◦−θ) = cos(θ)for any angle θ.
Solution
To prove the given trigonometric identity, we will use the sum-to-product for-
mula for sine. The sum-to-product formula states that sin(A+B) = sin Acos B+
cos Asin Bfor any angles Aand B.
Step 1: Let A= 90◦and B=−θ. Then, we have:
sin(90◦−θ) = sin 90◦cos(−θ) + cos 90◦sin(−θ)
Step 2: Recall that sin 90◦= 1 and cos 90◦= 0. We also know that
cos(−θ) = cos θand sin(−θ) = −sin θ. Substituting these values, we get:
= 1 ·cos θ+ 0 ·(−sin θ)
= cos θ+ 0
Step 3: Therefore, sin(90◦−θ) = cos θ. Hence, the identity sin(90◦−θ) =
cos θis proven.
18
Solution
Step 1: Use the double-angle identity to express cos(2x)in terms of cos(x)and
sin(x):
cos(2x) = 2 cos2(x)−1
Step 2: Substitute this into the given equation:
2 cos2(x)−1 = sin(x)
Step 3: Rewrite sin(x)in terms of cos(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1:
2 cos2(x)−1 = √1−cos2(x)
Step 4: Square both sides to eliminate the square root:
4 cos4(x)−4 cos2(x) + 1 = 1 −cos2(x)
Step 5: Rearrange terms to form a quadratic equation in terms of cos2(x):
4 cos4(x)−5 cos2(x) = 0
Step 6: Factor out a cos2(x):
cos2(x)(4 cos2(x)−5) = 0
Step 7: Find the solutions for cos2(x):
• Setting cos2(x) = 0: This gives cos(x) = 0. So, x=π
2,3π
2.
• Setting 4 cos2(x)−5 = 0: This gives cos(x) = ±√5
2. So, x=π
3,5π
3.
Step 8: Check the solutions in the original equation to ensure they are valid:
cos(2·π
2)= cos(π) = −1= 0 = sin(π
2)
cos(2·3π
2)= cos(3π) = −1= 0 = sin(3π
2)
cos(2·π
3)= cos(2π
3)=−1
2=−√3
2= sin(π
3)
cos(2·5π
3)= cos(10π
3)=−1
2=−√3
2= sin(5π
3)
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
3,5π
3.
2
Question 3
Question
Let f(x) = sin x
1+cos x. Determine the domain of f(x).
Solution
Step 1: The function f(x) = sin x
1+cos xwill be undefined when the denominator is
equal to 0, so we need to find the values of xthat make 1 + cos x= 0.
Step 2: Solving 1 + cos x= 0 for cos x, we get
cos x=−1
Step 3: The cosine function has a range of [−1,1], so the equation cos x=−1
has a solution only when x=π.
Step 4: Therefore, the domain of f(x) = sin x
1+cos xis all real numbers xexcept
x=π. In interval notation, the domain is (−∞, π)∪(π, ∞).
Question 4
Question
Find all solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π].
Solution
Step 1: Let u= tan(x). Then the equation becomes u2−3u−1 = 0.
Step 2: Solve the quadratic equation u2−3u−1 = 0 using the quadratic
formula:
u=−(−3) ±√(−3)2−4(1)(−1)
2(1) =3±√13
2.
Step 3: Now, we know that u= tan(x), so we have two cases to consider:
Case 1: tan(x) = 3+√13
2. To find the solutions in the interval [0,2π], we need
to consider all angles xsuch that tan(x) = 3+√13
2. These angles are π
6and 7π
6.
Case 2: tan(x) = 3−√13
2. The angles xthat satisfy this are 5π
6and 11π
6.
Hence, the solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π]are x=π
6,5π
6,7π
6,11π
6.
Question 5
Question
Solve the equation sin2(x) + sin(x)−2 = 0 for 0≤x≤2π.
3
Solution
To solve the equation sin2(x) + sin(x)−2=0, we can substitute sin(x)with a
variable y. The equation becomes y2+y−2 = 0, a quadratic equation. We can
then solve for yand then find the corresponding values of x.
Step 1: Solve the quadratic equation y2+y−2 = 0 for yby factoring or
using the quadratic formula.
The factors of −2that add up to 1are 2and −1, so we have:
y2+ 2y−y−2 = 0
y(y+ 2) −1(y+ 2) = 0
(y−1)(y+ 2) = 0
This gives us y= 1 or y=−2.
Step 2: Recall that sin(x) = y. So, we have:
sin(x) = 1 or sin(x) = −2
Step 3: Solve sin(x) = 1 for xwithin the interval [0,2π].
The solutions are x=π
2and x=3π
2.
Step 4: Solve sin(x) = −2for xwithin the interval [0,2π]. This equation
has no solutions as the range of sine function is [−1,1].
Therefore, the solutions to the equation sin2(x) + sin(x)−2 = 0 in the
interval [0,2π]are x=π
2and x=3π
2.
Question 6
Question
Find a general solution to the equation 2 sin(2x)−√3 cos(2x) = 1 for x∈[0,2π).
Solution
Step 1: Use the double angle identities to simplify the equation. Step 2: Solve
the resulting trigonometric equation. Step 3: Find the general solution within
the given interval.
Step 1: Using the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x)
we can rewrite the equation as:
2(2 sin(x) cos(x)) −√3(cos2(x)−sin2(x)) = 1
Step 2: Expanding the terms gives:
4 sin(x) cos(x)−√3(cos2(x)−sin2(x)) = 1
4
Using the identity cos2(x) + sin2(x) = 1 to replace cos2(x)in the equation,
we get:
4 sin(x) cos(x)−√3(1 −sin2(x)) = 1
Simplify:
4 sin(x) cos(x)−√3 + √3 sin2(x) = 1
Now, using the double angle identity for sin(2x):
2 sin(x) cos(x) = sin(2x)
we can rewrite the equation as:
2 sin(2x)−√3 + √3 sin2(x) = 1
Step 3: Now, our equation becomes:
2 sin(2x)−√3 + √3 sin2(x) = 1
Rearranging gives:
2 sin(2x) = √3−√3 sin2(x)+1
Divide by 2:
sin(2x) = √3−√3 sin2(x)+1
2
Solving for sin(x):
sin(x) = ±√1
2−1
2√7−3 sin2(x)
Thus, the general solution for x∈[0,2π)is:
x=π
12 +nπ
3or x=5π
12 +nπ
3
where nis an integer.
Question 7
Question
Compute the exact value of sin (5π
12 ).
5
Solution
Step 1: We can rewrite 5π
12 as π
12 +2π
3by common denominators.
Step 2: Using the angle addition identity sin(A+B) = sin(A) cos(B) +
cos(A) sin(B), we have:
sin (5π
12 )= sin (π
12 +2π
3).
Step 3: Recognizing that sin (π
12 )and cos (π
12 )are not familiar, we instead
use the well-known values for sin (π
6)=1
2and cos (π
6)=√3
2.
Step 4: Rewriting 2π
3as π
3+π
3and using the double angle identity sin(2A) =
2 sin(A) cos(A), we get:
sin 5π
12 = sin (π
12 +π
3+π
3).
Step 5: Apply the angle addition and double angle formulas to get:
sin 5π
12 = sin (2π
3)cos (π
12)+ cos (2π
3)sin (π
12).
Step 6: Substitute the values we know:
sin 5π
12 =(√3
2)(√3−1
2)+(−1
2)(1
2).
Step 7: Simplify the expression to get the final answer:
sin 5π
12 =3√3−√3−1
4−1
4=2√3−1
4.
Question 8
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
Solution
Step 1: We know the Pythagorean identity sin2(x) + cos2(x) = 1. We can
rewrite the given equation as sin2(x) = 1 −cos(x).
Step 2: Substitute this expression for sin2(x)into the Pythagorean identity:
(1 −cos(x))2+ cos2(x) = 1
Step 3: Expand and simplify the left-hand side of the equation:
1−2 cos(x) + cos2(x) + cos2(x) = 1
6
2 cos2(x)−2 cos(x) = 0
Step 4: Factor out a 2 cos(x)from the equation:
2 cos(x)(cos(x)−1) = 0
Step 5: Set each factor equal to zero and solve for cos(x):
2 cos(x) = 0 or cos(x)−1 = 0
Step 6: Solve the first equation 2 cos(x) = 0:
cos(x) = 0
This occurs when x=π
2or x=3π
2.
Step 7: Solve the second equation cos(x)−1 = 0:
cos(x) = 1
This occurs when x= 0.
Step 8: Thus, the solutions to the equation sin2(x) + cos(x)=1in the
interval [0,2π)are x= 0,x=π
2, and x=3π
2.
Question 9
Question
Convert the following expression to a single trigonometric function:
3 sin xcos x+ 4 sin3x
Solution
Step 1: We will first rewrite sin3xin terms of sin xand cos x. To do this,
we use the identity sin2x= 1 −cos2x, which implies sin3x= sin x·sin2x=
sin x(1 −cos2x).
Step 2: Let’s now substitute sin3xin the expression and simplify.
3 sin xcos x+ 4 sin x(1 −cos2x)
= 3 sin xcos x+ 4 sin x−4 sin xcos2x
Step 3: We can now factor out a sin xfrom the last two terms.
= sin x(3 cos x+ 4 −4 cos2x)
Step 4: To simplify further, we use the Pythagorean identity sin2x+cos2x=
1or cos2x= 1 −sin2x.
= sin x(3 cos x+ 4 −4(1 −sin2x))
7
= sin x(3 cos x+ 4 −4 + 4 sin2x)
= sin x(4 sin2x+ 3 cos x)
Step 5: We can write cos xin terms of sin xusing the Pythagorean iden-
tity sin2x+ cos2x= 1, or cos x=±√1−sin2x. Now we substitute cos x=
√1−sin2xin the expression.
= sin x(4 sin2x+ 3√1−sin2x)
Hence, the expression 3 sin xcos x+ 4 sin3xsimplifies to sin x(4 sin2x+
3√1−sin2x).
Question 10
Question
Let f(x) = ex+e−x
2and g(x) = ex−e−x
2. Find expressions for sinh(x)and cosh(x)
in terms of f(x)and g(x).
Solution
Step 1: Recall the definitions of hyperbolic sine and hyperbolic cosine:
sinh(x) = ex−e−x
2and cosh(x) = ex+e−x
2
Step 2: We can see that g(x)is equivalent to sinh(x), and f(x)is equivalent
to cosh(x). Thus, we have:
sinh(x) = g(x)and cosh(x) = f(x)
Therefore, the expressions for sinh(x)and cosh(x)in terms of f(x)and g(x)
are:
sinh(x) = g(x)and cosh(x) = f(x)
Question 11
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
8
Solution
To solve the equation sin(2x) = cos(x), we will first use the double angle identity
for sine, sin(2x) = 2 sin(x) cos(x). Then, we can substitute this into the original
equation to get 2 sin(x) cos(x) = cos(x).
Step 1: Set sin(x)equal to 0or 1.
If cos(x)=0, then x=π
2,3π
2. However, x=3π
2is outside the interval
0≤x≤2π, so we reject this solution.
If sin(x) = 1, then x=π
2.
Step 2: Check the value of x=π
2in the equation sin(2x) = cos(x).
For x=π
2, we have sin(π) = cos(π
2), which is true. Therefore, the solution
to the equation is x=π
2.
Hence, the solution to the equation sin(2x) = cos(x)for 0≤x≤2πis
x=π
2.
Question 12
Question
Given that sin(θ) = −3
5and θis in Quadrant IV, find the exact values of cos(θ),
tan(θ),csc(θ),sec(θ), and cot(θ).
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant IV, we can use the Pythagorean
identity to find cos(θ).
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
Therefore, cos(θ) = ±4
5. Since θis in Quadrant IV, we have cos(θ) = 4
5.
Step 2: Next, we can find tan(θ)using the definitions of tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=−3
5
4
5
=−3
4
Step 3: To find csc(θ), we use the definition csc(θ) = 1
sin(θ).
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
Step 4: Next, we find sec(θ)using the definition sec(θ) = 1
cos(θ).
sec(θ) = 1
cos(θ)=1
4
5
=5
4
9
Step 5: Lastly, we find cot(θ)using the definition cot(θ) = 1
tan(θ).
cot(θ) = 1
tan(θ)=1
−3
4
=−4
3
Therefore, the exact values of cos(θ),tan(θ),csc(θ),sec(θ), and cot(θ)are
4
5,−3
4,−5
3,5
4, and −4
3respectively.
Question 13
Question
Suppose sin(θ) = 3
5and cos(θ)<0. Find the values of the other five trigono-
metric functions of θ.
Solution
Step 1: Since sin(θ) = 3
5and cos(θ)<0, we know that θlies in the second
quadrant. By drawing a triangle in the second quadrant, we can determine the
value of the third side using the Pythagorean theorem. Step 2: Let’s assume
the hypotenuse is 5(since the sine of θis 3
5). Using the Pythagorean theorem,
we find that the opposite side is 3. Step 3: Now, we can determine the adjacent
side using the cosine function. Since cosine is negative in the second quadrant,
the adjacent side will be negative. Therefore, the adjacent side is −4. Step 4:
Now we can find the values of the other five trigonometric functions of θ. Step
5: cos(θ) = −4
5,tan(θ) = 3
−4,cot(θ) = −4
3,sec(θ) = −5
4,csc(θ) = 5
3.
Question 14
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ)
10
Solution
Start with the left-hand side (LHS):
LHS =sin(θ)
1−cos(θ)
=sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)
=sin(θ)(1 + cos(θ))
1−cos2(θ)
=sin(θ)(1 + cos(θ))
sin2(θ)(using Pythagorean identity)
=sin(θ) + sin(θ) cos(θ)
sin2(θ)
=sin(θ)
sin(θ)+sin(θ) cos(θ)
sin(θ)
= 1 + cos(θ)
=1 + sin(θ)
sin(θ)(using Pythagorean identity)
Therefore, we have shown that sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ), which completes
the proof.
Question 15
Question
Evaluate the following trigonometric expression:
tan (5π
12 )·cot (π
12)
Solution
Step 1: First, let’s express 5π
12 in terms of π
12 :
5π
12 =2π
12 +3π
12 =π
6+π
4=3π
12 +4π
12 =3π
12 +π
3
Step 2: Since tan (π
4)= 1 and tan (π
6)=√3
3, we have:
tan (5π
12 )= tan (3π
12 +π
3)=tan (3π
12 )+ tan (π
3)
1−tan (3π
12 )·tan (π
3)
11
=tan (π
4)+ tan (π
6)
1−tan (π
4)·tan (π
6)=1 + √3
3
1−1·√3
3
=3 + √3
3−√3= 2 + √3
Step 3: Similarly, we know that cot (π
6)=√3
3, so:
cot (π
12)= cot (π
6−π
4)=cot (π
6)cot (π
4)+ 1
cot (π
6)+ cot (π
4)
=
√3
3·1+1
√3
3+ 1 =√3+3
√3+3 = 1
Step 4: Therefore, we can now evaluate the expression:
tan (5π
12 )·cot (π
12)= (2 + √3) ·1 = 2 + √3
So, tan (5π
12 )·cot (π
12 )= 2 + √3.
Question 16
Question
Solve the equation for 0≤x≤2π:2 sin(2x) + √3 = 0.
Solution
Step 1: Rewrite the equation to isolate sin(2x):
2 sin(2x) = −√3
Step 2: Divide by 2 to solve for sin(2x):
sin(2x) = −√3
2
Step 3: Recall the angle where sin(π
3)=√3
2. We can now find the solutions
for 2x:
2x=11π
6+ 2πk or 2x=7π
6+ 2πk
Step 4: Solve for x:
x=11π
12 +πk or x=7π
12 +πk
Therefore, the solutions to the equation 2 sin(2x) + √3 = 0 for 0≤x≤2π
are x=11π
12 and x=7π
12 .
12
Question 17
Question
Find the exact value of tan (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles: π
3and π
4. Step
2: Thus, we have 5π
12 =π
3+π
4. Step 3: Using the angle addition identity for
tangent, we have
tan (5π
12 )=tan (π
3+π
4)
1−tan (π
3
tan(π
4).Step 4: Since tan(π/3) = √3and tan(π/4) = 1, we can substitute in
these values:
tan (5π
12 )=tan (π
3+π
4)
1−√3.
Step 5: Using the sum-to-product identity, we have
tan (5π
12 )=
tan(π
3)+tan(π
4)
1−tan(π
3)tan(π
4)
1−√3.
Step 6: Simplifying further gives us
tan (5π
12 )=
√3+1
1−√3
1−√3.
Step 7: Rationalizing the denominator, we get
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3).
Step 8: Simplifying the numerator gives us
tan (5π
12 )=4+2√3
−2.
Step 9: Therefore, the exact value of tan (5π
12 )is −2−√3.
Question 18
Question
Let f(x) = sin(3x)and g(x) = cos(2x), find the values of xin the interval [0,2π]
that satisfy the equation f(x) = g(x).
13
Solution
Step 1: Set up the equation f(x) = sin(3x) = g(x) = cos(2x).
Step 2: Since sin(3x) = cos (π
2−3x), rewrite the equation as cos (π
2−3x)=
cos(2x).
Step 3: In order for cos (π
2−3x)to equal cos(2x), the angles inside the cosine
function must be equal or their difference must be a multiple of 2π. Thus, we
have two cases:
• Case 1: π
2−3x= 2x
• Case 2: π
2−3x=−2x
Step 4: Solve Case 1: π
2−3x= 2x
π
2= 5x
x=π
10
Step 5: Solve Case 2: π
2−3x=−2x
π
2=x
Step 6: Check if the solutions fall within the interval [0,2π]:
0≤π
10 ≤2π
0≤π≤20π
However, πfalls outside the interval, so the only solution that satisfies x∈
[0,2π]is x=π
10 .
Question 19
Question
Solve the equation sin2(x)−cos(x) = 0 for xin the interval [0,2π].
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin2(x)−cos(x) = 0
sin2(x) = cos(x)
14
sin2(x) = √1−sin2(x)
Step 2: Square both sides of the equation to eliminate the square root.
sin2(x) = 1 −sin2(x)
2 sin2(x) = 1
sin2(x) = 1
2
Step 3: Solve for sin(x)by taking the square root of both sides, considering
the interval [0,2π].
sin(x) = ±√1
2=±√2
2
Step 4: Determine the possible values of xin the interval [0,2π]. Since sin(x)
is positive in the first and second quadrants and negative in the third and fourth
quadrants, the solutions are:
x=π
4,3π
4,5π
4,7π
4
Question 20
Question
Let θbe an angle in standard position. If sin θ=3
5and cos θ < 0, determine
the exact value of tan θ.
Solution
Step 1: Since sin θ=3
5, we can use the Pythagorean identity sin2θ+ cos2θ= 1
to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=−4
5(since cos θ < 0)
Step 2: Now, we can find tan θusing the definition tan θ=sin θ
cos θ.
tan θ=sin θ
cos θ=
3
5
−4
5
=3
5·(−5
4)=−3
4
Therefore, the exact value of tan θis −3
4.
15
Question 21
Question
Let f(x) = 3 sin(2x) + 4 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), recall that the amplitude of a function of
the form Asin(Bx)or Acos(Bx)is |A|. The amplitude of f(x)is the absolute
value of the coefficient of sin(2x)or cos(2x). In this case, the amplitude is
|3|= 3.
Step 2: The period of a function of the form sin(Bx)or cos(Bx)is 2π
|B|.
Therefore, the period of f(x)is 2π
2=π.
Step 3: To find the phase shift of f(x), set 2xequal to 0 and solve for xto
find the phase shift. We have:
2x= 0 ⇒x= 0
So, the phase shift of f(x)is 0.
Step 4: The vertical shift of a function of the form Asin(Bx) + Cor
Acos(Bx) + Cis the value of C. In this case, the vertical shift of f(x)is
0.
Question 22
Question
Solve the equation cos(2x) = sin(x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Let’s rewrite the equation using the double angle identity: 2 cos2(x)−
1 = sin(x).
Step 3: Recall that sin(x) = 2 sin(x) cos(x).
Step 4: Substitute cos(x)for 1−cos2(x)
2in the equation: 2(1−cos2(x)
2)2−1 =
sin(x).
Step 5: Simplify the equation: 2(1−2 cos2(x)+cos4(x)
4)−1 = sin(x).
Step 6: Distribute and simplify further: 1
2−cos2(x)+ 1
2cos4(x)−1 = sin(x).
Step 7: Rearrange the equation: 1
2cos4(x)−cos2(x)−sin(x)−1
2= 0.
Step 8: Let u= cos(x)and rewrite the equation: 1
2u4−u2−2u−1
2= 0.
Step 9: This is a quadratic equation in u2. Let’s solve it: u2= 2 ±√2.
16
Step 10: Since cos(x)can only take values between -1 and 1, the solutions
for u2are u2= 2 −√2and u2= 2 + √2.
Step 11: Solve for u:u=±√2−√2and u=±√2 + √2.
Step 12: Recall that cos(x) = u, we have cos(x) = √2−√2,cos(x) =
−√2−√2,cos(x) = √2 + √2, and cos(x) = −√2 + √2.
Step 13: Find the corresponding values of xusing the unit circle and the
relationships between cos(x)and x.
Step 14: The solutions to the equation are x=π
8,x=7π
8,x=3π
8, and
x=11π
8.
Question 23
Question
Let f(x) = cos(2x)and g(x) = sin(3x). Find the exact value of f(π
6)·g(π
4).
Solution
Step 1: We first find f(π
6). Since f(x) = cos(2x), we have f(π
6)= cos (2·π
6)=
cos (π
3)=1
2.
Step 2: Next, we find g(π
4). Given that g(x) = sin(3x), we have g(π
4)=
sin (3·π
4)= sin (3π
4)=−√2
2.
Step 3: Finally, we calculate the product f(π
6)·g(π
4).
f(π
6)·g(π
4)=1
2·(−√2
2)=−√2
4
Therefore, f(π
6)·g(π
4)=−√2
4.
Question 24
Question
Evaluate the exact value of cos (5π
12 ).
Solution
Step 1: We can use the sum and difference formula for cosine to rewrite the
given angle as a sum of two common angles. Step 2: In this case, we can
rewrite 5π
12 as π
3+π
4. Step 3: Using the sum formula for cosine, cos(a+b) =
cos acos b−sin asin b, we have:
cos (5π
12 )= cos (π
3+π
4)
= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
17
Step 4: Recall that cos(π/3) = 1/2,sin(π/3) = √3/2,cos(π/4) = √2/2, and
sin(π/4) = √2/2. Step 5: Substituting these values into the formula, we get:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4.
Therefore, the exact value of cos (5π
12 )is √2−√6
4.
Question 25
Question
Prove that sin(90◦−θ) = cos(θ)for any angle θ.
Solution
To prove the given trigonometric identity, we will use the sum-to-product for-
mula for sine. The sum-to-product formula states that sin(A+B) = sin Acos B+
cos Asin Bfor any angles Aand B.
Step 1: Let A= 90◦and B=−θ. Then, we have:
sin(90◦−θ) = sin 90◦cos(−θ) + cos 90◦sin(−θ)
Step 2: Recall that sin 90◦= 1 and cos 90◦= 0. We also know that
cos(−θ) = cos θand sin(−θ) = −sin θ. Substituting these values, we get:
= 1 ·cos θ+ 0 ·(−sin θ)
= cos θ+ 0
Step 3: Therefore, sin(90◦−θ) = cos θ. Hence, the identity sin(90◦−θ) =
cos θis proven.
18
Solution
Step 1: Use the double-angle identity to express cos(2x)in terms of cos(x)and
sin(x):
cos(2x) = 2 cos2(x)−1
Step 2: Substitute this into the given equation:
2 cos2(x)−1 = sin(x)
Step 3: Rewrite sin(x)in terms of cos(x)using the Pythagorean identity
sin2(x) + cos2(x) = 1:
2 cos2(x)−1 = √1−cos2(x)
Step 4: Square both sides to eliminate the square root:
4 cos4(x)−4 cos2(x) + 1 = 1 −cos2(x)
Step 5: Rearrange terms to form a quadratic equation in terms of cos2(x):
4 cos4(x)−5 cos2(x) = 0
Step 6: Factor out a cos2(x):
cos2(x)(4 cos2(x)−5) = 0
Step 7: Find the solutions for cos2(x):
• Setting cos2(x) = 0: This gives cos(x) = 0. So, x=π
2,3π
2.
• Setting 4 cos2(x)−5 = 0: This gives cos(x) = ±√5
2. So, x=π
3,5π
3.
Step 8: Check the solutions in the original equation to ensure they are valid:
cos(2·π
2)= cos(π) = −1= 0 = sin(π
2)
cos(2·3π
2)= cos(3π) = −1= 0 = sin(3π
2)
cos(2·π
3)= cos(2π
3)=−1
2=−√3
2= sin(π
3)
cos(2·5π
3)= cos(10π
3)=−1
2=−√3
2= sin(5π
3)
Therefore, the solutions to the equation cos(2x) = sin(x)for 0≤x≤2πare
x=π
3,5π
3.
2
Question 3
Question
Let f(x) = sin x
1+cos x. Determine the domain of f(x).
Solution
Step 1: The function f(x) = sin x
1+cos xwill be undefined when the denominator is
equal to 0, so we need to find the values of xthat make 1 + cos x= 0.
Step 2: Solving 1 + cos x= 0 for cos x, we get
cos x=−1
Step 3: The cosine function has a range of [−1,1], so the equation cos x=−1
has a solution only when x=π.
Step 4: Therefore, the domain of f(x) = sin x
1+cos xis all real numbers xexcept
x=π. In interval notation, the domain is (−∞, π)∪(π, ∞).
Question 4
Question
Find all solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π].
Solution
Step 1: Let u= tan(x). Then the equation becomes u2−3u−1 = 0.
Step 2: Solve the quadratic equation u2−3u−1 = 0 using the quadratic
formula:
u=−(−3) ±√(−3)2−4(1)(−1)
2(1) =3±√13
2.
Step 3: Now, we know that u= tan(x), so we have two cases to consider:
Case 1: tan(x) = 3+√13
2. To find the solutions in the interval [0,2π], we need
to consider all angles xsuch that tan(x) = 3+√13
2. These angles are π
6and 7π
6.
Case 2: tan(x) = 3−√13
2. The angles xthat satisfy this are 5π
6and 11π
6.
Hence, the solutions to the equation tan2(x)−3 tan(x)−1 = 0 on the interval
[0,2π]are x=π
6,5π
6,7π
6,11π
6.
Question 5
Question
Solve the equation sin2(x) + sin(x)−2 = 0 for 0≤x≤2π.
3
Solution
To solve the equation sin2(x) + sin(x)−2=0, we can substitute sin(x)with a
variable y. The equation becomes y2+y−2 = 0, a quadratic equation. We can
then solve for yand then find the corresponding values of x.
Step 1: Solve the quadratic equation y2+y−2 = 0 for yby factoring or
using the quadratic formula.
The factors of −2that add up to 1are 2and −1, so we have:
y2+ 2y−y−2 = 0
y(y+ 2) −1(y+ 2) = 0
(y−1)(y+ 2) = 0
This gives us y= 1 or y=−2.
Step 2: Recall that sin(x) = y. So, we have:
sin(x) = 1 or sin(x) = −2
Step 3: Solve sin(x) = 1 for xwithin the interval [0,2π].
The solutions are x=π
2and x=3π
2.
Step 4: Solve sin(x) = −2for xwithin the interval [0,2π]. This equation
has no solutions as the range of sine function is [−1,1].
Therefore, the solutions to the equation sin2(x) + sin(x)−2 = 0 in the
interval [0,2π]are x=π
2and x=3π
2.
Question 6
Question
Find a general solution to the equation 2 sin(2x)−√3 cos(2x) = 1 for x∈[0,2π).
Solution
Step 1: Use the double angle identities to simplify the equation. Step 2: Solve
the resulting trigonometric equation. Step 3: Find the general solution within
the given interval.
Step 1: Using the double angle identities:
sin(2x) = 2 sin(x) cos(x)and cos(2x) = cos2(x)−sin2(x)
we can rewrite the equation as:
2(2 sin(x) cos(x)) −√3(cos2(x)−sin2(x)) = 1
Step 2: Expanding the terms gives:
4 sin(x) cos(x)−√3(cos2(x)−sin2(x)) = 1
4
Using the identity cos2(x) + sin2(x) = 1 to replace cos2(x)in the equation,
we get:
4 sin(x) cos(x)−√3(1 −sin2(x)) = 1
Simplify:
4 sin(x) cos(x)−√3 + √3 sin2(x) = 1
Now, using the double angle identity for sin(2x):
2 sin(x) cos(x) = sin(2x)
we can rewrite the equation as:
2 sin(2x)−√3 + √3 sin2(x) = 1
Step 3: Now, our equation becomes:
2 sin(2x)−√3 + √3 sin2(x) = 1
Rearranging gives:
2 sin(2x) = √3−√3 sin2(x)+1
Divide by 2:
sin(2x) = √3−√3 sin2(x)+1
2
Solving for sin(x):
sin(x) = ±√1
2−1
2√7−3 sin2(x)
Thus, the general solution for x∈[0,2π)is:
x=π
12 +nπ
3or x=5π
12 +nπ
3
where nis an integer.
Question 7
Question
Compute the exact value of sin (5π
12 ).
5
Solution
Step 1: We can rewrite 5π
12 as π
12 +2π
3by common denominators.
Step 2: Using the angle addition identity sin(A+B) = sin(A) cos(B) +
cos(A) sin(B), we have:
sin (5π
12 )= sin (π
12 +2π
3).
Step 3: Recognizing that sin (π
12 )and cos (π
12 )are not familiar, we instead
use the well-known values for sin (π
6)=1
2and cos (π
6)=√3
2.
Step 4: Rewriting 2π
3as π
3+π
3and using the double angle identity sin(2A) =
2 sin(A) cos(A), we get:
sin 5π
12 = sin (π
12 +π
3+π
3).
Step 5: Apply the angle addition and double angle formulas to get:
sin 5π
12 = sin (2π
3)cos (π
12)+ cos (2π
3)sin (π
12).
Step 6: Substitute the values we know:
sin 5π
12 =(√3
2)(√3−1
2)+(−1
2)(1
2).
Step 7: Simplify the expression to get the final answer:
sin 5π
12 =3√3−√3−1
4−1
4=2√3−1
4.
Question 8
Question
Solve the equation sin2(x) + cos(x) = 1 for xin the interval [0,2π).
Solution
Step 1: We know the Pythagorean identity sin2(x) + cos2(x) = 1. We can
rewrite the given equation as sin2(x) = 1 −cos(x).
Step 2: Substitute this expression for sin2(x)into the Pythagorean identity:
(1 −cos(x))2+ cos2(x) = 1
Step 3: Expand and simplify the left-hand side of the equation:
1−2 cos(x) + cos2(x) + cos2(x) = 1
6
2 cos2(x)−2 cos(x) = 0
Step 4: Factor out a 2 cos(x)from the equation:
2 cos(x)(cos(x)−1) = 0
Step 5: Set each factor equal to zero and solve for cos(x):
2 cos(x) = 0 or cos(x)−1 = 0
Step 6: Solve the first equation 2 cos(x) = 0:
cos(x) = 0
This occurs when x=π
2or x=3π
2.
Step 7: Solve the second equation cos(x)−1 = 0:
cos(x) = 1
This occurs when x= 0.
Step 8: Thus, the solutions to the equation sin2(x) + cos(x)=1in the
interval [0,2π)are x= 0,x=π
2, and x=3π
2.
Question 9
Question
Convert the following expression to a single trigonometric function:
3 sin xcos x+ 4 sin3x
Solution
Step 1: We will first rewrite sin3xin terms of sin xand cos x. To do this,
we use the identity sin2x= 1 −cos2x, which implies sin3x= sin x·sin2x=
sin x(1 −cos2x).
Step 2: Let’s now substitute sin3xin the expression and simplify.
3 sin xcos x+ 4 sin x(1 −cos2x)
= 3 sin xcos x+ 4 sin x−4 sin xcos2x
Step 3: We can now factor out a sin xfrom the last two terms.
= sin x(3 cos x+ 4 −4 cos2x)
Step 4: To simplify further, we use the Pythagorean identity sin2x+cos2x=
1or cos2x= 1 −sin2x.
= sin x(3 cos x+ 4 −4(1 −sin2x))
7
= sin x(3 cos x+ 4 −4 + 4 sin2x)
= sin x(4 sin2x+ 3 cos x)
Step 5: We can write cos xin terms of sin xusing the Pythagorean iden-
tity sin2x+ cos2x= 1, or cos x=±√1−sin2x. Now we substitute cos x=
√1−sin2xin the expression.
= sin x(4 sin2x+ 3√1−sin2x)
Hence, the expression 3 sin xcos x+ 4 sin3xsimplifies to sin x(4 sin2x+
3√1−sin2x).
Question 10
Question
Let f(x) = ex+e−x
2and g(x) = ex−e−x
2. Find expressions for sinh(x)and cosh(x)
in terms of f(x)and g(x).
Solution
Step 1: Recall the definitions of hyperbolic sine and hyperbolic cosine:
sinh(x) = ex−e−x
2and cosh(x) = ex+e−x
2
Step 2: We can see that g(x)is equivalent to sinh(x), and f(x)is equivalent
to cosh(x). Thus, we have:
sinh(x) = g(x)and cosh(x) = f(x)
Therefore, the expressions for sinh(x)and cosh(x)in terms of f(x)and g(x)
are:
sinh(x) = g(x)and cosh(x) = f(x)
Question 11
Question
Solve the equation sin(2x) = cos(x)for 0≤x≤2π.
8
Solution
To solve the equation sin(2x) = cos(x), we will first use the double angle identity
for sine, sin(2x) = 2 sin(x) cos(x). Then, we can substitute this into the original
equation to get 2 sin(x) cos(x) = cos(x).
Step 1: Set sin(x)equal to 0or 1.
If cos(x)=0, then x=π
2,3π
2. However, x=3π
2is outside the interval
0≤x≤2π, so we reject this solution.
If sin(x) = 1, then x=π
2.
Step 2: Check the value of x=π
2in the equation sin(2x) = cos(x).
For x=π
2, we have sin(π) = cos(π
2), which is true. Therefore, the solution
to the equation is x=π
2.
Hence, the solution to the equation sin(2x) = cos(x)for 0≤x≤2πis
x=π
2.
Question 12
Question
Given that sin(θ) = −3
5and θis in Quadrant IV, find the exact values of cos(θ),
tan(θ),csc(θ),sec(θ), and cot(θ).
Solution
Step 1: Since sin(θ) = −3
5and θis in Quadrant IV, we can use the Pythagorean
identity to find cos(θ).
cos2(θ) = 1 −sin2(θ) = 1 −(−3
5)2
= 1 −9
25 =16
25
Therefore, cos(θ) = ±4
5. Since θis in Quadrant IV, we have cos(θ) = 4
5.
Step 2: Next, we can find tan(θ)using the definitions of tan(θ) = sin(θ)
cos(θ).
tan(θ) = sin(θ)
cos(θ)=−3
5
4
5
=−3
4
Step 3: To find csc(θ), we use the definition csc(θ) = 1
sin(θ).
csc(θ) = 1
sin(θ)=1
−3
5
=−5
3
Step 4: Next, we find sec(θ)using the definition sec(θ) = 1
cos(θ).
sec(θ) = 1
cos(θ)=1
4
5
=5
4
9
Step 5: Lastly, we find cot(θ)using the definition cot(θ) = 1
tan(θ).
cot(θ) = 1
tan(θ)=1
−3
4
=−4
3
Therefore, the exact values of cos(θ),tan(θ),csc(θ),sec(θ), and cot(θ)are
4
5,−3
4,−5
3,5
4, and −4
3respectively.
Question 13
Question
Suppose sin(θ) = 3
5and cos(θ)<0. Find the values of the other five trigono-
metric functions of θ.
Solution
Step 1: Since sin(θ) = 3
5and cos(θ)<0, we know that θlies in the second
quadrant. By drawing a triangle in the second quadrant, we can determine the
value of the third side using the Pythagorean theorem. Step 2: Let’s assume
the hypotenuse is 5(since the sine of θis 3
5). Using the Pythagorean theorem,
we find that the opposite side is 3. Step 3: Now, we can determine the adjacent
side using the cosine function. Since cosine is negative in the second quadrant,
the adjacent side will be negative. Therefore, the adjacent side is −4. Step 4:
Now we can find the values of the other five trigonometric functions of θ. Step
5: cos(θ) = −4
5,tan(θ) = 3
−4,cot(θ) = −4
3,sec(θ) = −5
4,csc(θ) = 5
3.
Question 14
Question
Prove the following trigonometric identity:
sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ)
10
Solution
Start with the left-hand side (LHS):
LHS =sin(θ)
1−cos(θ)
=sin(θ)
1−cos(θ)·1 + cos(θ)
1 + cos(θ)
=sin(θ)(1 + cos(θ))
1−cos2(θ)
=sin(θ)(1 + cos(θ))
sin2(θ)(using Pythagorean identity)
=sin(θ) + sin(θ) cos(θ)
sin2(θ)
=sin(θ)
sin(θ)+sin(θ) cos(θ)
sin(θ)
= 1 + cos(θ)
=1 + sin(θ)
sin(θ)(using Pythagorean identity)
Therefore, we have shown that sin(θ)
1−cos(θ)=1 + sin(θ)
sin(θ), which completes
the proof.
Question 15
Question
Evaluate the following trigonometric expression:
tan (5π
12 )·cot (π
12)
Solution
Step 1: First, let’s express 5π
12 in terms of π
12 :
5π
12 =2π
12 +3π
12 =π
6+π
4=3π
12 +4π
12 =3π
12 +π
3
Step 2: Since tan (π
4)= 1 and tan (π
6)=√3
3, we have:
tan (5π
12 )= tan (3π
12 +π
3)=tan (3π
12 )+ tan (π
3)
1−tan (3π
12 )·tan (π
3)
11
=tan (π
4)+ tan (π
6)
1−tan (π
4)·tan (π
6)=1 + √3
3
1−1·√3
3
=3 + √3
3−√3= 2 + √3
Step 3: Similarly, we know that cot (π
6)=√3
3, so:
cot (π
12)= cot (π
6−π
4)=cot (π
6)cot (π
4)+ 1
cot (π
6)+ cot (π
4)
=
√3
3·1+1
√3
3+ 1 =√3+3
√3+3 = 1
Step 4: Therefore, we can now evaluate the expression:
tan (5π
12 )·cot (π
12)= (2 + √3) ·1 = 2 + √3
So, tan (5π
12 )·cot (π
12 )= 2 + √3.
Question 16
Question
Solve the equation for 0≤x≤2π:2 sin(2x) + √3 = 0.
Solution
Step 1: Rewrite the equation to isolate sin(2x):
2 sin(2x) = −√3
Step 2: Divide by 2 to solve for sin(2x):
sin(2x) = −√3
2
Step 3: Recall the angle where sin(π
3)=√3
2. We can now find the solutions
for 2x:
2x=11π
6+ 2πk or 2x=7π
6+ 2πk
Step 4: Solve for x:
x=11π
12 +πk or x=7π
12 +πk
Therefore, the solutions to the equation 2 sin(2x) + √3 = 0 for 0≤x≤2π
are x=11π
12 and x=7π
12 .
12
Question 17
Question
Find the exact value of tan (5π
12 ).
Solution
Step 1: We can express 5π
12 as the sum of two special angles: π
3and π
4. Step
2: Thus, we have 5π
12 =π
3+π
4. Step 3: Using the angle addition identity for
tangent, we have
tan (5π
12 )=tan (π
3+π
4)
1−tan (π
3
tan(π
4).Step 4: Since tan(π/3) = √3and tan(π/4) = 1, we can substitute in
these values:
tan (5π
12 )=tan (π
3+π
4)
1−√3.
Step 5: Using the sum-to-product identity, we have
tan (5π
12 )=
tan(π
3)+tan(π
4)
1−tan(π
3)tan(π
4)
1−√3.
Step 6: Simplifying further gives us
tan (5π
12 )=
√3+1
1−√3
1−√3.
Step 7: Rationalizing the denominator, we get
tan (5π
12 )=(√3 + 1)(1 + √3)
(1 −√3)(1 + √3).
Step 8: Simplifying the numerator gives us
tan (5π
12 )=4+2√3
−2.
Step 9: Therefore, the exact value of tan (5π
12 )is −2−√3.
Question 18
Question
Let f(x) = sin(3x)and g(x) = cos(2x), find the values of xin the interval [0,2π]
that satisfy the equation f(x) = g(x).
13
Solution
Step 1: Set up the equation f(x) = sin(3x) = g(x) = cos(2x).
Step 2: Since sin(3x) = cos (π
2−3x), rewrite the equation as cos (π
2−3x)=
cos(2x).
Step 3: In order for cos (π
2−3x)to equal cos(2x), the angles inside the cosine
function must be equal or their difference must be a multiple of 2π. Thus, we
have two cases:
• Case 1: π
2−3x= 2x
• Case 2: π
2−3x=−2x
Step 4: Solve Case 1: π
2−3x= 2x
π
2= 5x
x=π
10
Step 5: Solve Case 2: π
2−3x=−2x
π
2=x
Step 6: Check if the solutions fall within the interval [0,2π]:
0≤π
10 ≤2π
0≤π≤20π
However, πfalls outside the interval, so the only solution that satisfies x∈
[0,2π]is x=π
10 .
Question 19
Question
Solve the equation sin2(x)−cos(x) = 0 for xin the interval [0,2π].
Solution
Step 1: Rewrite the equation using the Pythagorean identity sin2(x)+cos2(x) =
1.
sin2(x)−cos(x) = 0
sin2(x) = cos(x)
14
sin2(x) = √1−sin2(x)
Step 2: Square both sides of the equation to eliminate the square root.
sin2(x) = 1 −sin2(x)
2 sin2(x) = 1
sin2(x) = 1
2
Step 3: Solve for sin(x)by taking the square root of both sides, considering
the interval [0,2π].
sin(x) = ±√1
2=±√2
2
Step 4: Determine the possible values of xin the interval [0,2π]. Since sin(x)
is positive in the first and second quadrants and negative in the third and fourth
quadrants, the solutions are:
x=π
4,3π
4,5π
4,7π
4
Question 20
Question
Let θbe an angle in standard position. If sin θ=3
5and cos θ < 0, determine
the exact value of tan θ.
Solution
Step 1: Since sin θ=3
5, we can use the Pythagorean identity sin2θ+ cos2θ= 1
to find cos θ.
cos2θ= 1 −sin2θ= 1 −(3
5)2
= 1 −9
25 =16
25
cos θ=−4
5(since cos θ < 0)
Step 2: Now, we can find tan θusing the definition tan θ=sin θ
cos θ.
tan θ=sin θ
cos θ=
3
5
−4
5
=3
5·(−5
4)=−3
4
Therefore, the exact value of tan θis −3
4.
15
Question 21
Question
Let f(x) = 3 sin(2x) + 4 cos(2x). Find the amplitude, period, phase shift, and
vertical shift of the function f(x).
Solution
Step 1: To find the amplitude of f(x), recall that the amplitude of a function of
the form Asin(Bx)or Acos(Bx)is |A|. The amplitude of f(x)is the absolute
value of the coefficient of sin(2x)or cos(2x). In this case, the amplitude is
|3|= 3.
Step 2: The period of a function of the form sin(Bx)or cos(Bx)is 2π
|B|.
Therefore, the period of f(x)is 2π
2=π.
Step 3: To find the phase shift of f(x), set 2xequal to 0 and solve for xto
find the phase shift. We have:
2x= 0 ⇒x= 0
So, the phase shift of f(x)is 0.
Step 4: The vertical shift of a function of the form Asin(Bx) + Cor
Acos(Bx) + Cis the value of C. In this case, the vertical shift of f(x)is
0.
Question 22
Question
Solve the equation cos(2x) = sin(x)for xin the interval [0,2π].
Solution
Step 1: Recall the double angle identity for cosine: cos(2θ) = 2 cos2(θ)−1.
Step 2: Let’s rewrite the equation using the double angle identity: 2 cos2(x)−
1 = sin(x).
Step 3: Recall that sin(x) = 2 sin(x) cos(x).
Step 4: Substitute cos(x)for 1−cos2(x)
2in the equation: 2(1−cos2(x)
2)2−1 =
sin(x).
Step 5: Simplify the equation: 2(1−2 cos2(x)+cos4(x)
4)−1 = sin(x).
Step 6: Distribute and simplify further: 1
2−cos2(x)+ 1
2cos4(x)−1 = sin(x).
Step 7: Rearrange the equation: 1
2cos4(x)−cos2(x)−sin(x)−1
2= 0.
Step 8: Let u= cos(x)and rewrite the equation: 1
2u4−u2−2u−1
2= 0.
Step 9: This is a quadratic equation in u2. Let’s solve it: u2= 2 ±√2.
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Step 10: Since cos(x)can only take values between -1 and 1, the solutions
for u2are u2= 2 −√2and u2= 2 + √2.
Step 11: Solve for u:u=±√2−√2and u=±√2 + √2.
Step 12: Recall that cos(x) = u, we have cos(x) = √2−√2,cos(x) =
−√2−√2,cos(x) = √2 + √2, and cos(x) = −√2 + √2.
Step 13: Find the corresponding values of xusing the unit circle and the
relationships between cos(x)and x.
Step 14: The solutions to the equation are x=π
8,x=7π
8,x=3π
8, and
x=11π
8.
Question 23
Question
Let f(x) = cos(2x)and g(x) = sin(3x). Find the exact value of f(π
6)·g(π
4).
Solution
Step 1: We first find f(π
6). Since f(x) = cos(2x), we have f(π
6)= cos (2·π
6)=
cos (π
3)=1
2.
Step 2: Next, we find g(π
4). Given that g(x) = sin(3x), we have g(π
4)=
sin (3·π
4)= sin (3π
4)=−√2
2.
Step 3: Finally, we calculate the product f(π
6)·g(π
4).
f(π
6)·g(π
4)=1
2·(−√2
2)=−√2
4
Therefore, f(π
6)·g(π
4)=−√2
4.
Question 24
Question
Evaluate the exact value of cos (5π
12 ).
Solution
Step 1: We can use the sum and difference formula for cosine to rewrite the
given angle as a sum of two common angles. Step 2: In this case, we can
rewrite 5π
12 as π
3+π
4. Step 3: Using the sum formula for cosine, cos(a+b) =
cos acos b−sin asin b, we have:
cos (5π
12 )= cos (π
3+π
4)
= cos (π
3)cos (π
4)−sin (π
3)sin (π
4)
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Step 4: Recall that cos(π/3) = 1/2,sin(π/3) = √3/2,cos(π/4) = √2/2, and
sin(π/4) = √2/2. Step 5: Substituting these values into the formula, we get:
cos (5π
12 )=1
2·√2
2−√3
2·√2
2
=√2
4−√6
4
=√2−√6
4.
Therefore, the exact value of cos (5π
12 )is √2−√6
4.
Question 25
Question
Prove that sin(90◦−θ) = cos(θ)for any angle θ.
Solution
To prove the given trigonometric identity, we will use the sum-to-product for-
mula for sine. The sum-to-product formula states that sin(A+B) = sin Acos B+
cos Asin Bfor any angles Aand B.
Step 1: Let A= 90◦and B=−θ. Then, we have:
sin(90◦−θ) = sin 90◦cos(−θ) + cos 90◦sin(−θ)
Step 2: Recall that sin 90◦= 1 and cos 90◦= 0. We also know that
cos(−θ) = cos θand sin(−θ) = −sin θ. Substituting these values, we get:
= 1 ·cos θ+ 0 ·(−sin θ)
= cos θ+ 0
Step 3: Therefore, sin(90◦−θ) = cos θ. Hence, the identity sin(90◦−θ) =
cos θis proven.
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