Prove that the following identities :(sin8θ – cos8θ) = (sin2θ –

1.	Prove that the following identities :(sin8θ – cos8θ) = (sin2θ – cos2θ)(1 – 2sin2θ .cos2θ)
Answer» Taking LHS = sin⁸ θ – cos⁸ θ = (sin⁴ θ)² – (cos⁴ θ)² = (sin⁴ θ – cos⁴ θ)(sin⁴ θ+ cos⁴ θ ) [∵ (a² – b²) = (a + b) (a – b)] = {(sin² θ)² – (cos² θ)²}{(sin² θ)² + (cos² θ)²} = (sin² θ + cos² θ) (sin² θ – cos² θ) [(sin² θ + cos² θ) – 2 sin² θ cos² θ] [ ∵ (a² + b²) = (a +b)² – 2ab] = (1)[ sin² θ –cos² θ][(1) – 2 sin² θ cos² θ] = (sin² θ – cos² θ)(1 – 2 sin² θ cos² θ) = RHS Hence Proved

Prove that the following identities :(sin8θ – cos8θ) = (sin2θ – cos2θ)(1 – 2sin2θ .cos2θ)

Answer»

Taking LHS

= sin⁸ θ – cos⁸ θ

= (sin⁴ θ)² – (cos⁴ θ)²

= (sin⁴ θ – cos⁴ θ)(sin⁴ θ+ cos⁴ θ )

[∵ (a² – b²) = (a + b) (a – b)]

= {(sin² θ)² – (cos² θ)²}{(sin² θ)² + (cos² θ)²}

= (sin² θ + cos² θ) (sin² θ – cos² θ) [(sin² θ + cos² θ) – 2 sin² θ cos² θ]

[ ∵ (a² + b²) = (a +b)² – 2ab]

= (1)[ sin² θ –cos² θ][(1) – 2 sin² θ cos² θ]

= (sin² θ – cos² θ)(1 – 2 sin² θ cos² θ)

= RHS

Hence Proved