TRIG Final Review
1)
180 – (20+50) = 110̊
sin20
8
sin20
8
=
sin110
=
sin50
a
c
= 21.98
= 17.92
2)
𝜋
𝑐𝑜𝑠 =
22
√3
6
√3
2
𝜋
2
6
√3
 𝑠𝑒𝑐 =
−5= −
11
3
3)
(𝑠𝑖𝑛
√3
1+
) = ±√ 2 = √
150
2
2+√3
2
2
2
=√
2+√3 1
2
∙ =√
2+√3
2
4
=
√2−√3
2
Sin is positive at 150̊
4)
𝑐𝑜𝑠
5)
3𝜋
4
=−
√2
2
𝑠𝑖𝑛−1 (−
√2
)
2
1

√10
-3
sin =
−3
√10
𝑜𝑟
−3√10
10
= −
𝜋
4
6)
8

1
a) sin =
1
8
b) cot 2  = 63
c) sec(90-) =8 b) sec 2  =
same as csc
√63
64
63
7)
tan(85.833)=
ℎ
80
h = 1098.05 feet
8)
s = ½(9+2+10) = 10.5
K = √10.5(10.5 − 9)(10.5 − 2)(10.5 − 10)
K = 8.18
9)
S = r
1
5 = r( )
3
r = 15 feet
10)
30̊
12 ft
12 ft
180 – 30 = 150 then 150/2 = 75
sin30
x
11)
=
sin75
12
75̊
= 6.21 𝑓𝑒𝑒𝑡
Amplitude is 8 Period is
2𝜋
2
=𝜋
12)
tan=
15
8
= 61.9°
tan=
8
15
= 28.1°
c2 = 152+ 82 c = 17
13)
10(1-cos) = sin2
10-10cos = 1-cos2
cos2 -10cos + 9 = 0
(cos - 9)(cos - 1) = 0
cos = 1
=0
14)
1.47
15)
=
1
2
16)
cosine is + in Quad I & IV 2- 0.90
8cos - 5 = 0
cos =
5
 = 0.90 and 5.39
8
17)
sin( + ) = sincos  + cossin
2𝜋
sin(
3
𝜋
2𝜋
4
3
+ ) = sin
𝜋
2𝜋
4
3
𝑐𝑜𝑠 + cos
√3 √2
∙
2
2
𝑠𝑖𝑛
−1 √2
∙ )
2
2
+(
𝜋
4
1
= (√6 − √2)
4
18)
4sec + 1 = 9
sec =2
5
cos =
8
𝜋 5𝜋
= ,
3
3
1
2
19)
sinθ−cosθ
sinθ
sinθ+cosθ
−
cosθ
sinθcosθ−cos2 θ
sinθcosθ
−
−1(cos2 θ+sin2 θ)
sinθcosθ
=
= (cosθ)
sin2 θ+sinθcosθ
sinθcosθ
sinθ−cosθ
sinθ
=
− (sinθ)
sinθ+cosθ
cosθ
=
sinθcosθ−cos2 θ−sin2 θ−sinθcosθ
sinθcosθ
−1
sinθcosθ
20)
=-
𝜋
4
21)
= √5
22)
𝜋
𝑐𝑜𝑠 =
6
√3
2
23)
24)
7 =
2𝜋
𝜔
=
2
7

2⁄
7
=
−1
8
=
−1
28
2
1
7
28
y = 5sin( 𝑥 +
)
sin2 θ − 16sinθ + 63 = 0
25)
(sinθ − 7)(sinθ − 9) = 0
sinθ = 1 sinθ =
θ=
𝜋
9
7
2
26)
sin( − ) = sincos  − cossin
𝜋
5𝜋
3
6
sin( −
27)
28)
𝜋
) = sin(- ) = -1
2
29)
30)
v = (10-9)i + (3-9)j
v = i - 6j