Question

If sin x + sin^2x + sin 3x = 1. then find value of cos^6x-cos^4x +8 cos^2x?​

Answer 1

Correct Question: If sinx + sin²x + sin³x = 1 then find value of cos⁶x - 4cos⁴x + 8 cos²x.

Solution: sinx + sin²x + sin³x = 1

→ sin²x + sinx + sin³x = 1

→ sin²x + sinx(1 + sin²x) = 1

→ sinx(1 + sin²x) = 1 - sin²x

→ sinx(1 + sin²x) = cos²x

Square both sides because we want one term cos⁴x.

→ [sinx(1 + sin²x)]² = cos⁴x

→ sin²x(1 + sin²x)² = cos⁴x

→ (1 - cos²x)(2 - cos²x)² = cos⁴x

→ (1 - cos²x)(4 + cos⁴x - 4cos²x) = cos⁴x

→ 4 + cos⁴x - 4cos²x - 4cos²x - cos⁶x + 4cos⁴x = cos⁴x

→ 4 - 8 cos²x - cos⁶x + 4cos⁴x = 0

→ 4 = cos⁶x + 8cos²x - 4cos⁴x

Answer is 4.