If sinθ + cosθ = x, prove that sin^6θ + cos^6θ = 4 - 3(x^2 -1)^2

Question

If sinθ + cosθ = x, prove that sin⁶θ + cos⁶θ = \(\frac{4-3(x^2-1)^2}{4}\)

Answer 1

sinθ + cosθ = x

Squaring on both sides

(sinθ + cosθ)² = x²

⇒ sin²θ + cos²θ + 2sinθcosθ = x²

∴ sinθcosθ = \(\frac{x^2-1}{2}\) ...(1)

We know sin²θ + cos²θ = 1

combining both since

(sin²θ + cos²θ)³ = (1)³

sin⁶θ + cos⁶θ + 3sinθCos²θ (sin²θ + cos²θ) = 1

⇒ sin⁶θ + cos⁶θ = 1- 3sin²θCos²θ

= 1- \(\frac{3(x^2-1)^2}{4}\) from -(1)

sin⁶θ + cos⁶θ = \(\frac{4-3(x^2-1)^2}{4}\) 

Hence Proved.