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Practice Exam 2
Conceptual Questions
1. State a Basic identity and then verify it.
(a) Identity:
Solution: One identity is csc(θ) =
1
sin(θ)
(b) Verification:
Solution: Given the point of intersection (x, y) of the terminal side of θ and the unit circle we have that sin(θ)
1
is defined to be y, while csc(θ) is defined to be y1 . Therefore
= csc(θ).
sin(θ)
2. State the definition of an odd Function and use it to determine if f (θ) = sin(θ)tan(θ) is odd, even or neither.
(a) Definition:
Solution: f (θ) is an odd function if and only if f (−θ) = −f (θ) for all θ in the domain.
(b) Odd, Even or Neither Justification of f (θ):
Solution: Since f (−θ) = sin(−θ) tan(−θ) = (− sin(θ))(− tan(θ)) = sin(θ) tan(θ) = f (θ), we have that f (θ) is
an even function.
3. State a FTI identity and then verify it.
(a) Identity:
Solution: tan2 (θ) + 1 = sec2 (θ) is one identity.
(b) Verification:
Solution: The FTI says that sin2 (θ) + cos2 (θ) = 1, therefore
( 2
)
1
1
sin (θ) + cos2 (θ) =
or that
2
cos (θ)
cos2 (θ)
sin2 (θ)
+ 1 = sec2 (θ) and hence tan2 (θ) + 1 = sec2 (θ) as desired.
cos2 (θ)
4. State a Graph identity and then verify it with a step by step graph approach.
(a) Identity:
Solution: One identity is cos( π2 − θ) = sin(θ).
(b) Verification:
Solution: Observe that cos( π2 − θ) = cos(−(θ − π2 )) = cos(θ − π2 ) since cosine is even. Now since the graph of
f (θ) = cos(θ) is
and the graph of f (θ −
π
2
is
1
which is the same as the graph of sine. Therefore cos( π2 − θ) = sin(θ).
5. State a Double Angle ID and then verify it.
(a) Identity:
Solution: One is cos(2θ) = cos2 (θ) − sin( θ).
(b) Verification:
Solution: Observe that cos(2θ) = cos(θ + θ) and by the angle sum id we have cos(2θ) = cos(θ) cos(θ) −
sin(θ) sin(θ) = cos2 (θ) − sin2 (θ) as desired.
6. Verify the ID sin(θ) cos(β) =
1
(sin(θ + β) + sin(θ − β)).
2
(a) Verification: Solution: Observe that
1
1
(sin(θ + β) + sin(θ − β)) = [(sin(θ) cos(β) + sin(β) cos(θ)) + (sin(θ) cos(β) − sin(β) cos(θ))]
2
2
and thus
1
1
(sin(θ + β) + sin(θ − β)) = [2 sin(θ) cos(β)] = sin(θ) cos(β)
2
2
2
Evaluate, Simplify and Solve Questions
π
7. If cos(θ) = x for some positive real number x < 1 with −2π ≤ θ ≤ − evaluate cot(−θ) in terms of x. Draw a
2
picture to justify your work.
Solution: Since cosine is positive in quadrants 1 and 4 but we are restricted to quadrants 1, 2 and 3 we know that
θ must terminate in quadrant 1. Since cosine is adjacent over hypotenuse we can now plot the angle and draw the
reference triangle as follows using the Pythagorean theorem to find the missing side:
x
Now since cotangent is odd we have that cot(−θ) = − cot(θ) = − √
.
1 − x2
8. Simplify the expression
v
(
u
π)
u
2
cos
θ
−
u 4 tan(−θ) sin(−θ)
2) + cos(−2θ)
(π
t√
+
2
−θ
sec
16 − 16 sin (θ)
2
Solution:
v
(
u
π)
v
u
u 4 tan(θ) sin(θ)
2 sin(θ)
u 4 tan(−θ) sin(−θ) 2 cos θ − 2
(π
) + cos(−2θ) = u
t√
+
+
+ cos(2θ)
t√
csc(θ)
sec
−θ
16 − 16 sin2 (θ)
16(1 − sin2 (θ)
2
By using the Odd and Even Function Prop and Graph IDS
√
=
4 sin(θ) sin(θ)
√
+ 2 sin(θ) sin(θ) + cos2 (θ) − sin2 (θ)
4 cos(θ) cos2 (θ)
By using the FTI, Basic IDs and double angle ID
=
√
tan2 (θ) + sin2 (θ) + cos2 (θ)
=
√
tan2 (θ) + 1
=
√
sec2 (θ) = sec(θ)
By simplifying and using Basic IDs
with FTI
3
9. Solve for all values of θ such that
(a) cos(2θ) = cos2 (θ) + 1.
Solution: Using the double angle ID for cosine we have that
cos(2θ) = cos2 (θ) + 1 ⇒ cos2 (θ) − sin2 (θ) = cos2 (θ) + 1
This results in
sin2 (θ) = −1
which has no solution. Thus θ = ∅ .
(b) Hint: Rewrite 3θ as θ + 2θ and use an angle sum identity.
sin(3θ) csc(θ) = 2
Solution: Observe that
sin(3θ) csc(θ) = 2
⇒ sin(θ + 2θ) csc(θ) = 2
1
⇒ [sin(θ) cos(2θ) + sin(2θ) cos(θ]
=2
sin(θ)
by sum angle ID
[
[
]
]
⇒ sin(θ) cos2 (θ) − sin2 (θ) + [2 sin(θ) cos(θ)] cos(θ)
1
=2
sin(θ)
By double angle ID
⇒ cos2 (θ) − sin2 (θ) + 2 cos2 (θ) = 2
by Simplifying
⇒ 4 cos2 (θ) = 3
√
√
3
3
⇒ cos(θ) = ±
=±
4
2
π
⇒ θ = ± + πk, k ∈ Z
6
4
10. Evaluate the following using ANGLE SUM IDENTITIES:
(
)
25π
(a) cos
12
(
)
(
)
25π
4π 3π
Solution: Since cos
= cos
+
we have by the angle sum ID that
12
3
4
(
cos
(
)
31π
(b) tan −
12
25π
12
)
(
4π
3π
4π
3π
= cos( ) cos( ) − sin( ) sin( ) =
3
4
3
4
31π
Solution: Since tan −
12
)
(
= tan
(
−7π π
−
3
4
31π
tan −
12
)
(
−1
2
)(
−1
√
2
)
( √ )(
)
− 3
1
√
−
2
2
)
we have by the angle sum ID that
π
−7π
√
) − tan( )
− 3−1
3
4
√
=
=
π
−7π
1− 3
) tan( )
1 + tan(
3
4
tan(
11. Evaluate the following using HALF ANGLE IDENTITIES:
(
)
13π
(a) sin
12
(
)
13π
13π
Solution: Since
is in the third quadrant we have that sin
is negative. Thus by the half angle ID
12
12
we have that
√
√
√
(
)
1 − cos( 13π
)
1 − 23
13π
6
sin
=−
= −
12
2
2
(
)
7π
(b) cot −
8
7π
Solution: Since −
is in quadrant 3 we have that cotangent is positive. Thus by half angle ID we have
8
v
u
) √
(
7π
1 + √12
1 + cos(− 4 ) u
7π
t
=
=
cot −
8
1 − cos( −7π
1 − √12
4 )
5
12. Use the bank of functions below to identify the graphs that follow



 f1 (θ) = sin(θ) + cos(3θ), g1 (θ) = sin(θ) + cos(5θ), h1 (θ) = sin(2θ) + cos(7θ) 

f2 (θ) = sin(3θ) + cos(θ), g2 (θ) = sin(5θ) + cos(θ), h2 (θ) = sin(7θ) + cos(2θ)


 f3 (θ) = 2θ + tan(θ), g3 (θ) = 1 θ + tan(2θ), h3 (θ) = 1 θ + tan( 1 θ)

2
2
2
,
(a)
1
1
θ + tan( θ)
2
2
6
(b)
sin(θ) + cos(5θ)
(c)
sin(2θ) + cos(7θ)
7
13. For the function f (θ) = cos(2θ) + cos(4θ):
(a) Find the period of f (θ)
Solution: The period of cos(2θ) is
f (θ) = LCM (π, π2 ) = π.
2π
2
= π, while the period of cos(4θ) is
2π
4
=
π
2.
Therefore the period of
(b) Solve for all θ such that f (θ) = 0. Hint: (Use a sum to product ID)
Solution: Using the sum to product ID we have that
f (θ) = cos(2θ) + cos(4θ)
6θ
−2θ
)
= 2 cos( ) cos(
2
2
= 2 cos(3θ) cos(−θ)
= 2 cos(3θ) cos(θ)
Thus f (θ) = 0 implies 2 cos(3θ) cos(θ) = 0 or that
cos(3θ) = 0
This results in 3θ =
π
2
+ πk or θ =
π
2
or
cos(θ) = 0
+ πk for k ∈ Z. This results in θ =
π
6
+ π3 k and
π
2
14. Simplify. Hint: Be sure to use an angle sum ID and a product to sum ID.
(sin(θ) + sin(3θ))
csc(θ) sec(θ
− 2 cos(3θ) cos(θ) + cos(4θ)
2
Solution: Observe that
csc(θ) sec(θ
− 2 cos(3θ) cos(θ) + cos(4θ)
2
[
]
1
1
= [2 sin(2θ) cos(−θ)]
−2
cos(4θ) + cos(2θ) + cos(4θ)
2 sin(θ) cos(θ)
2
(sin(θ) + sin(3θ))
Using the sum to product ID and the prod to sum ID
=
2 sin(θ) cos2 (θ)
− cos(4θ) − cos(2θ) + cos(4θ)
sin(θ) cos(θ)
by using the double angle ID, the Even function ID and simplyfying
= 2 cos(θ) − cos(2θ)
8
+ πk.