Question

prove that: sin8 theta - cos8 theta= (sin2 theta - cos2 theta)(1-2sin2 theta cos2theta)

Answer 1

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Answer 2

Answer:

Proved

Step-by-step explanation:

prove that: sin8 theta - cos8 theta= (sin2 theta - cos2 theta)(1-2sin2 theta cos2theta)

LHS

= Sin⁸θ - Cos⁸θ

a² - b² = (a + b) (a-b)

here a = Sin⁴θ  & b = Cos⁴θ

= (Sin⁴θ + Cos⁴θ)(Sin⁴θ - Cos⁴θ)

= (Sin⁴θ + Cos⁴θ)(Sin²θ + Cos²θ) (Sin²θ - Cos²θ)

Sin²θ + Cos²θ) = 1

= (Sin⁴θ + Cos⁴θ)(Sin²θ - Cos²θ)

a² + b² = (a + b)² - 2ab

a = Sin²θ & b = Cos²θ

= ((Sin²θ + Cos²θ)² - 2Sin²θCos²θ)(Sin²θ - Cos²θ)

= (1 - 2Sin²θCos²θ)(Sin²θ - Cos²θ)

= (Sin²θ - Cos²θ)(1 - 2Sin²θCos²θ)

= RHS

QED