sin 3 A
6.
1 + 2 cos 2 A
Answers
Answered by
2
Answer:
sinA+sin
2
A+sin
3
A=1, then find the value of cos
6
A−4cos
4
A+8cos
2
A.
ANSWER
sinA+sin
2
A+sin
3
A=1
sinA+sin
3
A=1−sin
2
A=cos
2
A
sinA(1+sin
2
A)=cos
2
A
sinA(2−cos
2
A)=cos
2
A (since, sin
2
A=1−cos
2
A)
Squaring both sides,
sin
2
A(4−4cos
2
A+cos
4
A)=cos
4
A
(1−cos
2
A)(4−4cos
2
A+cos
4
A)=cos
4
A
4−4cos
2
A+cos
4
A−4cos
2
A+4cos
4
A−cos
6
A=cos
4
A
4−cos
6
A+4cos
4
A−8cos
2
A=0
cos
6
A−4cos
4
A+8cos
2
A=4
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