Question

sin8Q-cos8 Q
please this is urgent ​

Answer 1

Answer:

Step-by-step explanation:

Sin⁸θ - Cos⁸θ

=> (Sin⁴θ)² - (Cos⁴θ)²

=>  ( Sin⁴θ - Cos⁴θ) (Sin⁴θ + Cos⁴θ) [ ∵ a² - b² = (a - b)(a + b)]

=> [(Sin²θ)² - (Cos²θ)²][(Sin²θ)² + (Cos²θ)²]

//We know that a² - b² = (a - b)(a + b)

                         a² + b² = a² + b²+2ab - 2ab = (a + b)² - 2ab

=> [(Sin²θ - Cos²θ)(Sin²θ + Cos²θ)][(Sin²θ)² + (Cos²θ)² + 2Sin²θCos²θ -  

                                                                                             2Sin²θCos²θ]

=> [(Sin²θ - Cos²θ)(Sin²θ + Cos²θ)] [ (Sin² + Cos²θ)² - 2Sin²θCos²θ]

=> (Sin²θ - Cos²θ)( 1 - 2Sin²θCos²θ)   [∵ Sin²θ + Cos²θ = 1]