Question

solve it urgent don't post wrong one I will report it​

Answer 1

To prove: sin⁸ + cos⁸ = 1 - 4sin²θ . cos²θ + 2sin⁴θ . cos⁴θ

Proof:

w.k.t., sin²θ + cos²θ = 1

On squaring both sides, we get

(sin²θ + cos²θ)² = 1²

⇒ sin⁴θ + cos⁴θ + 2sin²θ . cos²θ = 1

⇒ sin⁴θ + cos⁴θ = 1 - 2sin²θ . cos²θ

On again squaring both sides, we get

(sin⁴θ + cos⁴θ) = (1 - 2sin²θ . cos²θ)²

⇒ sin⁸θ + cos⁸θ + 2sin⁴θ . cos⁴θ = 1 + 4sin⁴θ . cos⁴θ - 4sin²θ . cos²θ

⇒ sin⁸θ + cos⁸θ = 1 + 4sin⁴θ . cos⁴θ - 4sin²θ . cos²θ - 2sin⁴θ . cos⁴θ

⇒ sin⁸θ + cos⁸θ = 1 + 2sin⁴θ . cos⁴θ - 4sin²θ . cos²θ

⇒ sin⁸θ + cos⁸θ = 1 - 4sin²θ . cos²θ + 2sin⁴θ . cos⁴θ

Hence proved

Formulas used :-

1) (a+b)² = a² + b² + 2ab

2) (a-b)² = a² + b² - 2ab