Question

(sin^8x-cos^8x)=(sin^2x-cos^2x)(1-2sin^2x.cos^2x)​

Answer 1

Answer:

Sin ⁸ x –Cos ⁸ x = ( 1 – 2 Cos²x) ( 1 – 2 Sin²x Cos²x)

LHS

Sin ⁸ x –Cos ⁸ x

= ( Sin⁴ x ) ² – (cos⁴ x) ²

= ( Sin⁴ x - Cos⁴ x ) ( Sin⁴ x + Cos⁴ x )

= ( Sin² x - Cos² x ) ( Sin² x + Cos² x ) ( Sin⁴ x + Cos⁴ x )

= ( Sin² x - Cos² x )(1) ( Sin⁴ x + Cos⁴ x ) as Sin² x + Cos² x=1

writing Sin² x= 1-Cos² x & adding and subtracting 2 Sin² xCos² x in second factor

=(1 - Cos² x -Cos² x )( Sin⁴ x + Cos⁴ x+ 2 Sin² xCos² x - 2 Sin² xCos² x)

= (1 -2 Cos² x) [( Sin² x + Cos² x )² - 2 Sin² xCos² x]

= (1 -2 Cos² x) [( 1)² - 2 Sin² xCos² x] as Sin² x + Cos² x=1

= ( 1 – 2 Cos²x) ( 1 – 2 Sin²x Cos²x)

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