Math 1205 Trigonometry Review Answer Sheet
1. a. sin(−
b. sin(
€
5π
) = −1
2
5π
)=
2
cos(−
1
Name: ANSWER KEY
5π
)= 0
2
cos(
5π
)=
2
0
c. sin(360°) = 0
cos(360°) = 1
d. sin(−π ) = 0
cos(−π ) = −1
€
€
5π
9π
13π
−3π
−7π
−11π
b. θ2 =
or
or
... or
or
or
...
2
2
2
2
2
2
π
2. a. θ1 =
2
€
3. a.
€
tan
1
=
3
π
=
6
3
3€
b. cot
π
= 1
4
4. Quadrant(s) I & II
6. Quadrant(s)
II
€
€
5. Quadrant(s)
II & III
7. Quadrant(s) I, II, III, & IV (all 4 quadrants)
8. The sign of the following:
 7π 
 is negative,
 6 
sin
 7π 
 is negative
 6 
 7π 
 is positive.
 6 
cos
and tan
9. For each of the following, give the quadrant that the angle is in, φ and the answer.
€
a. sin(−
3π
)
4
π
4
Q III
−1
2
=
− 2
2
€
€
b. cos(
29 π
)
6
c. tan(420°)
Q II
d. sec(−
9π
)
4
e. csc(510°)
3
3
€
€
π
4
€
π
Q II
QI
2
€
€
€
19π
f. cot(
)
3
6
− 3
π
Q IV
€
€
€
QI
€
€
π
6
π
€
3
€
€
2
2
1
3
=
3
3
10. Quadrant IV
Answer: sec θ =
2
2 3
=
3
3
11. For each of the following, find the quadrants (there will be 2), φ and θ1 and θ 2 .
€
€
€
1
a. cos θ = −
2
QII & QIII
b. tan θ = − 3
QII & QIV
c. sin θ =
1
2
QI & QII
φ=
€
€
π
4
θ1 =
3π
5π
and θ 2 =
4
4
€
φ=
π
3 €
θ1 =
2π
5π
and θ 2 =
3
3
φ=
π
6 €
θ1 =
π
5π
and θ 2 =
6
6
12. Find the quadrants (there will be 2), φ and θ1 and θ 2 and then generalize to find all the
€
values for θ if cosθ =
€
π
6
φ =€
QI & QIV
€
3
2
€
Generally: θ1 =
€
€
θ1 =
π
11π
and θ 2 =
6
6
π
11π
+ (n)2π and θ 2 =
+ (n)2π where n is an integer .
6
6
€
13. Show all work for a through c.
a.
€
sin2π x = 0 for x in (0,2)
sin φ = 0
φ1 = 0 ± 2nπ and φ 2 = π ± 2nπ where n is an integer
OR
φ = 0 ± nπ where n is an integer
2π x = ±nπ where n is an integer
n
x =±
where n is an integer
2
1
3
x = , 1,
for x in (0,2)
2
2
c.
π
cos x = 1 for x in (−7,7)
3
cos φ = 1
φ = 0 ± 2nπ where n is an integer
π
x = ±2nπ where n is an integer
3
x = ±6n where n is an integer
x = −6,0,6 for x in (−7,7)
€
b.
π
1
tan x = −
for x in (1,4)
2
3
1
tan φ = −
3
5π
11π
φ1 =
± 2nπ and φ 2 =
± 2nπ where n is an integer
6
6
OR
5π
± nπ where n is an integer
6
π
5π
x=
± nπ where n is an integer
2
6
5
x = ± 2n where n is an integer
3
5 11
x= ,
for x in (1,4)
3 3
φ=