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

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

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

Answer»

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.