If sin A + sin^3 A = cos^2 A, prove that cos^6 A - 4cos^4 A + 8cos^2 A = 4.
2 answers:
Answer:
Step-by-step explanation:
sinA+sin
3
A=cos
2
A
⇒sinA[1+sin
2
A]=cos
2
A
⇒sinA[1+1−cos
2
A]=cos
2
A
squaring on both sides.
⇒sin
2
A[4+cos
4
A−4cos
2
A]=cos
4
A
⇒(1−cos
2
A)[4+cos
4
A−9cos
2
A]=cos
4
A
⇒4+cos
4
A−4cos
2
A−64cot
2
A−cos
6
A+9cos
4
A=cos
4
A
⇒
cos
6
A−4cos
4
A+8cos
2
A=4
Hence Prove
Answer:
Step-by-step explanation:
sinA(1+sin^2A) = cos^2A
sinA(2 -cos^2A) = cos^2A
Squaring both sides,
sin^2A(4-4cos^2A +cos^4A) = cos^4A
(1-cos^2A)(4-4cos^2A +cos^4A) = cos^4A
4-4cos^2A +cos^4A-4cos^2A+4cos^4A-cos^6A = cos^4A
4 -cos^6A +4cos^4A -8cos^2A = 0
cos^6A - 4 cos^4A + 8cos^2A = 4
hence proveproven
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