8-6 Double and Half Angle Formulas
MATH 162 TRIG
Example 1:
45
1−cos⁡(45)
2
2
sin( ) = ±√
√
√2
1− 2
2
=√
2−√2
2
2
at 45̊ the cosine is positive
=√
2−√2 1
2
Example 2:
2(1-cos2x) – cosx – 1 = 0
2-2cos2x – cosx – 1 = 0
0 = 2cos2x + cosx – 1
cos2x + cosx – 2 = 0
(cosx + 2)(cosx – 1)=0 divide by 2
cosx = -1 cosx = ½
𝜋
x = 3 , 𝜋,
5𝜋
3
Example 3:
15- sin =(1-sin2)
sin2 - sin + 14 = 0
No solution
because it doesn’t factor
∙ =√
2
2−√2
4
=
√2−√2
2
TEAGUE
Example 4:
Quadrant III
-
9

√145
-8
=2 (
a) sin(2) = 2sincos
b) cos(2) = cos -sin  = (
2

c) sin(2) = √

1−
d) cos(2) = √
Example 5:
=√
2
1+
−9
√145
=√
2
√
)(
145
√145
√
2
√145−9
√145
1
√2
√2
𝜋 3𝜋 5𝜋 7𝜋
= 4,
4
,
4
,
4
−8
√145
2
) =⁡
17
145
=√
145+9√145 1
=√
145−9√145 1
145
cos(2) + 6sin2 = 3
1 - sin2 + 6sin2 = 3
4sin2 = 2
sin2 = ½
sin = ±√2 ⁡⁡ ∙
144
) = ⁡ 145
145
) −(
√145+9
√145
2
−9
2
−9
2
−9
√145
−8
=±
√2
2
145
⁡⁡⁡⁡⁡⁡
∙2=√
∙2=√
145+9√145
290
145−9√145
290
Example 6: -tan(2) + 2cos = 0
−
sin2⁡
+ 2 cos 
cos2⁡
multiply all by cos2
-sin2 + 2coscos2 convert all (2)
-2sincos + 2cos(1-2 sin2) factor out -2cos
-2cos(sin - 1 + 2sin2) = 0 rearrange to factor
-2cos(2sin2 + sin - 1) = 0 factor the trinomial
-2cos(2sin - 1)(sin + 1) = 0
set each equal to 0
-2cos = 0
2sin - 1 = 0 sin + 1 = 0
cos = 0
sin = ½
sin = -1
𝜋 𝜋 5𝜋 3𝜋
= 6,2,
6
,
2
Example 7: cos[2sin-1(-½)]= 0
using cos(2α) where α = sin-1(-½)
convert cos(2α) to 1-2 sin2 α
(1-2 sin2 α) = 0
then replace α with sin-1(-½)
(1-2 sin2 (sin-1(-½)) = 0
(1-2(-½)2) = ½
Example 8: sin2(½cos-1(12⁄13))= 0
sin2α where α = ½cos-1(12⁄13)
using formula cos(2α) = 1-2sin2 α
2sin2 α = 1 - cos(2α)
sin2 α =⁡
1−⁡cos(2α)⁡⁡
2
1−cos(2(½cos−1 (12⁄13)))⁡
2
1−⁡(12⁄13)⁡⁡
2
1
= 26
replace α with ½cos-1(12⁄13)
=
1−⁡cos(cos−1 (12⁄13))⁡⁡
2
Example 9: tan(2cos-1(4⁄5))= 0
tan2α where α = cos-1(4⁄5) then tanα = 3⁄4 using a triangle
using formula tan(2α) =
2tanα⁡⁡
1−tan2 α⁡
2⁡(3⁄4)⁡⁡
1−(3⁄4)
2
=
Example 10: √3cosx + 1 = -cos(2x)
24
7
convert cos(2x) to 2cos2x - 1
√3cosx + 1 = -(2cos2x – 1) distribute the negative
get in factor form
√3cosx + 1 = -2cos2x + 1
2cos2x + √3cosx = 0
factor out cosx
cosx(2cosx + √3) = 0 set each equal to 0
cosx = 0
2cosx - √3 = 0
cosx = 0
cosx =
𝜋 3𝜋 5𝜋 7𝜋
= 2,
2
,
6
,
6
√3
2