Question

If sin A + sin^3 A = cos^2 A, prove that cos^6 A - 4cos^4 A + 8cos^2 A = 4.

Please provide all steps. Thank You.​

Answer 1

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 2

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