Name: Solutions
Due Date: Monday, June 1st.
Written Homework 2
Directions: Treat these assignments as if they were take home test questions as they will be graded by me as such.
1. Simplify and show all work at each step
sec (θ) cot(θ)
(a) √
9 + 9 tan2 (θ)
Solution: Using the Trig ID 1 + tan2 (θ) = sec2 (θ) we have
sec (θ) cot(θ)
sec (θ) cot(θ)
√
= √
2
9 + 9 tan (θ)
3 1 + tan2 (θ)
=
sec (θ) cot(θ)
√
3 sec2 (θ)
=
sec (θ) cot(θ)
3 sec(θ)
=
(b) cos
(π
2
)
− θ sin
(π
2
)
−θ −
cot(θ)
3
cos(−θ)
− sin(2θ)
csc(−θ)
(π
)
(π
)
1
− θ = sin(θ) and sin
− θ = cos(θ), the basic ID csc(θ) =
,
2
2
sin(θ)
the double angle formula sin(2θ) = 2 sin(θ) cos(θ), and finally the fact that cosine is an even function, while
sine and co-secant are odd we get that
Solution: Using the Graph ID’s cos
cos
(π
)
(π
) cos(−θ)
cos(−θ)
− θ sin
−θ −
− sin(2θ) = sin(θ) cos(θ) −
− sin(2θ)
2
2
csc(−θ)
csc(−θ)
cos(θ)
= sin(θ) cos(θ) −
− 2 sin(θ) cos(θ)
1
sin(θ)
= 2 sin(θ) cos(θ) − 2 sin(θ) cos(θ)
= 0
1
2. Find all solutions to the following. Be sure to show all work to receive full credit.
(a) cos(2θ) = − sin2 (θ)
Solution: Rewriting then using the double angle identity for cosine we have that
cos(2θ) = − sin2 (θ) ⇒ cos(2θ) + sin2 (θ) = 0 ⇒ cos2 (θ) − sin2 (θ) + sin2 (θ) = 0.
Or that
cos(2θ) = − sin2 (θ) ⇒ cos2 (θ) = 0
Square rooting both sides of the last equality gives us
cos(θ) = 0
Solving for theta results in
θ=
π
+ πk, k ∈ Z
2
(b) sin(2θ) = 2 cos(θ)
Solution: First we rewrite the equation, then use the angle sum ID for sine, then factor, and finally solve as
follows:
sin(2θ) = 2 cos(θ) ⇒ sin(2θ) − 2 cos(θ) = 0
⇒ 2 sin(θ) cos(θ) − 2 cos(θ) = 0
⇒ 2 cos(θ) (sin(θ) − 1) = 0
⇒ 2 cos(θ) = 0 and
(sin(θ) − 1) = 0
⇒ cos(θ) = 0 and
sin(θ) = 1
⇒ θ=
2
π
+ πk, k ∈ Z
2
3. Evaluate the following. Be sure to show all work to receive full credit.
(
)
37π
(a) sin
12
37π
is bigger then 2π and that a coterminal angle is
Solution: One way to do this is to notice that
12
(
)
(
)
37π
13π
37π
13π
− 2π =
. Thus sin
= sin
, which may be easier to deal with.
12
12
12
12
**Remember you don’t have to do it this way you could have gone straight to finding multiples of
that combine to get you 37π
12 .**
In our case we find multiples that add to
13π
12 .
One set of multiples is
4π
3
and
π
4
since
16π − 3π
13π
4π π
− =
=
3
4
12
23
Thus we have that
(
( )
)
(
)
(
)
( )
(π)
(π )
37π
13π
4π π
4π
4π
sin
= sin
= sin
−
= sin
cos
− sin
cos
12
12
3
4
3
4
4
3
and since
(
it follows that sin
37π
12
)
=
√
− 3 1
1 1
·√ +√
2
2
22
3
π
4
and
π
3
(
)
−17π
(b) tan
12
Solution: I could use the same trick to work with a positive coterminal angle or even use the idea that tangent
is an odd function and pull out the negative sign. However, lets just do it the standard way and work directly
−17π
π
5π
−17π
with
. Notice that 3 − 4 · 5 = −17 so I know that −
=
. Using this pair of angles we have
12
4
3
12
that
( )
(π)
5π
(
)
(
)
− tan
tan
−17π
π 5π
4
3
( )
tan
= tan
−
=
(π )
5π
12
4
3
1 + tan
· tan
4
3
Then since
(
it follows that tan
−17π
12
)
=
√
1+ 3
√
1− 3
4
4. Verify the following identity. Be sure to show all work to receive full credit.
(a) cos(θ − β) sec(θ) csc(β) = cot(β) + tan(θ)
Solution: We will start with the left side of the equality and use the angle difference ID for cosine of cos(θ−β) =
1
1
cos(θ) cos(β) + sin(θ) sin(β) and the basic ID’s sec(θ) =
and csc(β) =
as follows.
cos(θ)
sin(β)
cos(θ − β) sec(θ) csc(β) = (cos(θ) cos(β) + sin(θ) sin(β)) sec(θ) csc(β)
(
= (cos(θ) cos(β) + sin(θ) sin(β)) ·
(
=
cos(θ) cos(β)
cos(θ) sin(β)
)
(
+
1
cos(θ) sin(β)
sin(θ) sin(β)
cos(θ) sin(β)
)
)
= cot(β) + tan(θ)
⇒ cos(θ − β) sec(θ) csc(β) = cot(β) + tan(θ)
(b) cot(2θ) =
1
2
as desired.
(cot(θ) − tan(θ))
cos(θ)
, then we will use the double
sin(θ)
angle ID’s for sine and cosine, split the fraction, simplify and use the Basic ID’s for cotangent and tangent as
follows:
Solution: Starting with the left side we will first use the Basic ID cot(θ) =
cot(2θ) =
=
cos(2θ)
sin(2θ)
cos2 (θ) − sin2 (θ)
2 sin(θ) cos(θ)
1
=
2
1
=
2
(
(
cos2 (θ)
sin2 (θ)
−
sin(θ) cos(θ) sin(θ) cos(θ)
cos(θ) sin(θ)
−
sin(θ)
cos(θ
=
1
(cot(θ) − tan(θ))
2
⇒ cot(2θ) =
1
(cot(θ) − tan(θ))
2
5
)
)
as desired.