1)
Work:
sin(𝑥) = −
√3
2
𝑥 = arcsin⁡(−
√3
)
2
Principal value:
𝜋
𝑥 = −3
Exact answers restricted to 0 ≤ x < 2are: 𝑥 =
Complete exact solution: x =
Answer: x =
𝟒𝝅
+ 𝟐𝒌𝝅
𝟑
, x=
4𝜋
3
+ 2𝑘𝜋 , x =
𝟓𝝅
+
𝟑
5𝜋
3
4𝜋
3
,𝑥=
5𝜋
3
+ 2𝑘𝜋 k integer
𝟐𝒌𝝅 k integer
2)
Work:
cos(2x) – 3sin(x)cos(2x)=0
cos(2x)(1-3sin(x)) = 0
cos(2x)=0 or 1-3sin(x) = 0
2x = /2 or
sin(x) = 1/3
x = /4 or x = sin-1(1/3)
x = 0.79
Answer:
or x = 0.34 (rounded to 2 decimal places)
x = 0.79 , x = 0.34
3)
Work:
tan2(x)+tan(x)-1=0
Let u = tan (x)
u2+u-1=0
𝑢=
−1 ± √12 − 4(1)(−1) −1 ± √5
=
2(1)
2
x = tan-1(u)
x = tan-1(
−1+√5
)
2
or
x = tan-1(
−1−√5
)
2
x = 0.55 or x = -1.02
Answer: x = 0.55 or x = -1.02
4)
Proof:
𝑡𝑎𝑛2 (𝛼) − 1 + 𝑐𝑜𝑠 2 (𝛼) = 𝑡𝑎𝑛2 (𝛼) − (1 − 𝑐𝑜𝑠 2 (𝛼)) = 𝑡𝑎𝑛2 (𝛼) − 𝑠𝑖𝑛2 (𝛼)
= 𝑡𝑎𝑛2 (𝛼) (1 −
𝑠𝑖𝑛2 (𝛼)
𝑐𝑜𝑠 2 (𝛼)
2 (𝛼)
2 (𝛼)
=
𝑡𝑎𝑛
(1
−
𝑠𝑖𝑛
)
(
)) = 𝑡𝑎𝑛2 (𝛼)(1 − 𝑐𝑜𝑠 2 (𝛼))
𝑡𝑎𝑛2 (𝛼)
𝑠𝑖𝑛2 (𝛼)
= 𝑡𝑎𝑛2 (𝛼)𝑠𝑖𝑛2 (𝛼)
5)
Proof:
𝑡𝑎𝑛(𝛽) sin(𝛽) + cos(𝛽) = (
𝑠𝑖𝑛(𝛽)
𝑠𝑖𝑛2 (𝛽) + 𝑐𝑜𝑠 2 (𝛽)
1
=
= sec⁡(𝛽)
) sin(𝛽) + cos(𝛽) =
cos⁡(𝛽)
cos⁡(𝛽)
cos⁡(𝛽)
6)
Proof:
sin⁡()
2
2
𝑡𝑎𝑛()𝑐𝑜𝑠 2 () + 𝑠𝑖𝑛2 () (cos⁡()) 𝑐𝑜𝑠 () + 𝑠𝑖𝑛 () sin() 𝑐𝑜𝑠() + 𝑠𝑖𝑛2 ()
=
=
sin⁡()
sin⁡()
sin⁡()
= cos() + sin⁡()
7)
Proof:
2
1 + 𝑡𝑎𝑛(𝜃) (1 + 𝑡𝑎𝑛(𝜃)) (1 + 𝑡𝑎𝑛(𝜃)) 1 + 2 tan(𝜃) + 𝑡𝑎𝑛 (𝜃)
=
=
=
(1 − tan⁡(θ)) (1 + 𝑡𝑎𝑛(𝜃))
1 − tan⁡(θ)
1 − 𝑡𝑎𝑛2 (𝜃)
1+
𝑠𝑖𝑛2 (𝜃)
+ 2 tan(𝜃)
𝑐𝑜𝑠 2 (𝜃)
1 − 𝑡𝑎𝑛2 (𝜃)
𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃)
1
+ 2 tan(𝜃)
+ 2 tan(𝜃) 𝑠𝑒𝑐 2 (𝜃) + 2 tan(𝜃)
𝑐𝑜𝑠 2 (𝜃)
𝑐𝑜𝑠 2 (𝜃)
=
=
=
1 − 𝑡𝑎𝑛2 (𝜃)
1 − 𝑡𝑎𝑛2 (𝜃)
1 − 𝑡𝑎𝑛2 (𝜃)
8)
Proof:
𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤)
𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤)
𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤)
=
=
sin⁡(w)
sin⁡(w)
sin2 ⁡(w)
tan(𝑤) sin(𝑤) + 𝑐𝑜𝑠(𝑤) tan(𝑤)
(
) sin(𝑤) + 𝑐𝑜𝑠(𝑤) (
)
+ sin⁡(𝑤)
cos⁡(𝑤)
cos⁡(𝑤)
cos⁡(𝑤)
𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤)
cos⁡(𝑤)(𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤)) cos⁡(𝑤)(𝑠𝑖𝑛2 (𝑤) − 𝑐𝑜𝑠 2 (𝑤))
=
=
s in2 (w) + sin(w) cos⁡(w)
s in2 (w) + sin(w) cos⁡(w)
s in(w) (sin(w) + cos(w))
(
)
cos⁡(𝑤)
cos⁡(𝑤)(𝑠𝑖𝑛(𝑤) − 𝑐𝑜𝑠(𝑤))(sin(𝑤) + cos⁡(𝑤)) cos⁡(𝑤)(𝑠𝑖𝑛(𝑤) − 𝑐𝑜𝑠(𝑤))
=
s in(w) (sin(w) + cos(w))
s in(w)
=
cos(𝑤) sin⁡(𝑤) 𝑐𝑜𝑠 2 (𝑤)
cos(𝑤)
−
= cos(𝑤) − (
) cos(𝑤) = cos(𝑤) − cot(𝑤) cos⁡(𝑤)
sin⁡(𝑤)
sin(𝑤)
sin(𝑤)
9)
Work:
Let =/4 then:
sec()-cos() = sec(/4)-cos(/4)= 2 – 1/2 = 1/2
sin()sec() = (1/2 )( 2) = 1
Then for =/4 sec()-cos()  sin()sec()
10)
Work:
Using that tan(A+B) = (tan(A)+tan(B))/(1-tan(A)tan(B))
tan(/4-) = (tan(/4)+tan(-))/(1-tan(/4)tan(-)) = (1-tan())/(1+tan())
Answer: tan(/4-) =(1-tan())/(1+tan())
11)
Work:
Using that cos(A+B) =cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
cos(+/3) = cos()cos(/3) - sin()sin(/3) = cos()/2 - 3sin()/2
Answer: cos(+/3) = cos()/2 - 3sin()/2
12)
Work:
Using that: cos(-)=cos()
cos(-83) = cos(83)
Answer: cos(83)
13)
Work:
cos and sin are cofunctions
sin(125)=cos(90-125) = cos(-35) = cos(35)
Answer: cos(35)
14)
Work:
Since sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
sin(195) = sin(180+15) = sin(180)cos(15)+cos(180)sin(15)
= 0cos(15)-sin(15) = -sin(15)
sin(30)=1/2
sin(2A) =2sin(A)cos(A)
Taking A= 15
sin(30) = 2sin(15)cos(15) = ½
sin(15)cos(15)= ¼
sin(15) √1 − 𝑠𝑖𝑛2 (15) = 1/4
sin2(15) (1 − 𝑠𝑖𝑛2 (15)) = 1/16
Taking x = sin2(15)
x(1-x)=1/16
-x2+x-1/16=0
(-1  (1-1/4))/(-2) =(-1  (3/4))/(-2) = ½ (3)/4
1 √3
sin(15) = √ −
2
4
1
Then sin(195) = −√2 −
𝟏
𝟐
Answer: −√ −
15)
√𝟑
𝟒
√3
4
16)
Work:
cot(2) = 5/12  cos(2)/sin(2) = 5/12  sin(2)/cos(2) = 12/5
 tan(2)=12/5
Using that: tan(2A) = 2tan(A)/(1-tan2(A))
tan(2)=2tan()/(1-tan2()) = 12/5
5tan() = 6(1-tan2())
-6 tan2() -5 tan()+6 =0
tan⁡(𝜃) =
5 ± √(−5)2 − 4(−6)(6) 5 ± 13
=
2(−6)
−12
3
2
tan⁡(𝜃) = − 2 𝑜𝑟⁡⁡tan⁡(𝜃) = 3 ,
Then tan⁡(𝜃) =
since 0 ≤ 2≤0 ≤ ≤tan() >0
2
3
Since tan2(𝜃)+1 = sec2(𝜃)  (2/3)2 +1 = sec2(𝜃)
sec(𝜃) = (13/9) 1/cos(𝜃) =13 / 3  cos() =  3 / 13
since 0 ≤ 2≤0 ≤ ≤cos() >0
Then: cos() = 3 / 13
sin() = tan()cos() = (2/3)(3/ 13) = 2/ 13
Answer: sin() = 2/ 13 ,
cos() = 3 / 13 ,
tan() = 2/3
17)
Work:
4 2
9
3
cos() = 4/5  sin(𝛼) = √1 − 𝑐𝑜𝑠 2 (𝛼) = √1 − (5) = √25 = ± 5
Since a is in the 1st quadrant sin() = 3/5
Using sin(2A)= 2sin(A)cos(A)
sin(2)= 2sin()cos() = = 2(3/5)(4/5) = 24/25
Answer: 24/25
18)
Work:
5 2
144
12
sin() = 5/13  cos(𝛽) = √1 − 𝑠𝑖𝑛2 (𝛽) = √1 − (13) = √169 = ± 13
Since  is in the 2st quadrant cos() = -12/13
Then tan() = sin()/cos() = (5/13)/(-12/13) = -5/12
tan(2)= 2tan()/(1-tan2()) = 2(-5/12)/(1-(-5/12)2) = (-10/12)/(1-25/144)) = (-5/6)/(119/144) =
-5(144)/6(119) = -5(24)/119 = -120/119
Answer: -120/119
19)
Work:
sin(2x)+sin(x)=0
0 ≤ x ≤ 2
Since sin(2A) = 2sin(A)cos(A)  sin(2x) = 2sin(x)cos(x)
We have to solve:
2sin(x)cos(x)+sin(x) = 0
sin(x)(2cos(x)+1)=0
sin(x)=0 or 2cos(x)+1=0
sin(x)=0 or cos(x)=-1/2
x = 0,,2 , or x = 2/3, 4/3
Answer: 0, 2/3, , 4/3 ,2
20)
Work:
Since: 2sin(a)sin(b) = cos(a-b) - cos(a+b)
2sin(37)sin(26) = cos(37-26) – cos(37+26) = cos(11)-cos(63)
Answer: cos(11)-cos(63)