Problem 1. (25 pts)
a) Let Θ be an angle in the 2nd quadrant, such that sin Θ = 3/4. Find
cos Θ and tan Θ.
Since
sin2 Θ + cos2 Θ = 1,
we have
7
9
= .
16
16
Since Θ is in the 2nd quadrant, cos Θ < 0 and hence
r
√
7
7
cos Θ = −
=−
.
16
4
Finally,
sin Θ
3
tan Θ =
= −√ .
cos Θ
7
cos2 Θ = 1 − sin2 Θ = 1 −
b) Let Θ be an angle in the 3rd quadrant, such that tan Θ = 2. Find
cos Θ and sin Θ.
Since
1 + tan2 Θ = sec2 Θ,
we have
sec2 Θ = 1 + 4 = 5.
Since, on the other hand,
sec Θ =
we have
1
,
cos Θ
1
1
= .
2
sec Θ
5
Furthermore, cos Θ < 0 because Θ is in the 3rd quadrant. Thus
1
cos Θ = − √ .
5
Finally,
2
sin Θ = tan(Θ) · cos(Θ) = − √ .
5
cos2 Θ =
1
2
Problem 2. (25 pts)
a) Find the exact value of sin(arctan(−3)).
Denote
arctan(−3) = Θ.
Then Θ is in the interval (−π/2, π/2) and
tan Θ = −3.
In particular,
tan Θ < 0,
hence Θ is in the 4th quadrant. To find sin Θ, observe first that
1
1
1
1
cos2 Θ =
=
=
= .
2
2
sec Θ
1 + tan Θ
1+9
10
Since Θ is in the 4th quadrant, cos Θ > 0 and therefore
1
cos Θ = √ .
10
Finally,
3
sin Θ = tan(Θ) · cos(Θ) = − √ .
10
b) Find the exact value of sin(arccos(−1/3)).
Denote
arccos(−1/3) = Θ.
Then Θ is in the interval [0, π] and
cos Θ = −1/3.
In particular,
cos Θ < 0,
hence Θ is in the 2nd quadrant. Since
sin2 Θ + cos2 Θ = 1,
we have
1
8
= .
9
9
Since Θ is in the 2nd quadrant, sin Θ > 0 and therefore
r
√
8
2 2
sin Θ =
=
.
9
3
sin2 Θ = 1 − cos2 Θ = 1 −
3
Problem 3. (25 pts)
√
a) Find all the solutions of the equation sin x = −1/ 2 in the interval
[−π, 3π].
The general solution of the equation is
x = x0 + 2nπ
or x = −x0 + (2n + 1)π,
where
√
√
x0 = arcsin(−1/ 2) = − arcsin(1/ 2) = −π/4,
and n = 0, ±1, ±2, . . . Thus in the interval [−π, 3π] we have the following solutions:
π π
π
3π π 5π 7π
π
x = − π, − , + π, − + 2π = − , − , , .
4
4 4
4
4
4 4 4
√
b) Find all the solutions of the equation cos x = −1/ 2 in the interval
[−π, 3π].
The general solution of the equation is
x = ±x0 + 2nπ,
where
√
√
x0 = arccos(−1/ 2) = π − arccos(1/ 2) = 3π/4,
and n = 0, ±1, ±2, . . . Thus in the interval [−π, 3π] we have the following solutions:
3π 3π 3π
3π
3π 3π 5π 11π
x = − , ,−
+ 2π,
+ 2π = − , , ,
.
4 4
4
4
4 4 4 4
4
Problem 4. (25 pts) Verify the identity
(1 + tan x)2 cos x = sec x + 2 sin x.
(1 + tan x)2 cos x = (1 + 2 tan x + tan2 x) cos x
= (sec2 x + 2 tan x) cos x = sec2 (x) cos(x) + 2 tan(x) cos(x)
= sec2 (x)
1
sin(x)
+2
cos(x) = sec(x) + 2 sin(x).
sec(x)
cos(x)