Question

Prove that: sin⁸ θ – cos⁸ θ = (1 – 2cos² θ) (1 – 2sin² θ cos²θ).​

Answer 1

Answer:

LHS = sin⁸θ-cos⁸θ

= (sin⁴θ)²-(cos⁴)²

= (sin⁴θ+cos⁴θ)(sin⁴θ-cos⁴θ)

= (sin⁴θ+2sin²θ × cos²θ + cos⁴θ - 2sin²θ × cos²θ)(sin²θ + cos²θ)(sin²θ - cos²θ)

= [(sin²θ + cos²θ)² - 2sin²θ × cos²θ] ×1× (sin²θ - cos²θ)

LHS = (1 - 2 sin²θ × cos²θ) ( sin²θ+ cos²θ)

RHS = (1 - 2 sin²θ × cos²θ) ( sin² - cos²θ)

Step-by-step explanation:

Therefore, LHS = RHS