Question

Prove that sin^8x-cos^8x=(1-2cos^2x)(1-2 sin^2x cos^2x)

Answer 1

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Answer 2

Question:-

$\sin^8x-\cos^8x=\left(sin^2x-\cos^2x\right)\:\left(1-2\sin^2x\cos^2x\right)$

Solution :-

Take LHS =

$=\left(\sin^8x-\cos^8x\right)= {\left(\sin^4x\right)}^2</p><p> - {\left(\cos^4x\right)}^2$

$=\left(sin^4x-\cos^4x\right) \left(sin^4x+ \cos^4x\right)$

$= \left(sin^2x-\cos^2x\right) \left(sin^2x+ \cos^2x\right)\left(sin^4x+ \cos^4x\right)$

$=\left(sin^2x-\cos^2x\right)\left(sin^4x+ \cos^4x\right)$

$=\left(sin^2x-\cos^2x\right) \left(\sin^4x +\cos^4x +2\sin^2x\cos^x-2\sin^2x\cos^2x\right)$

$= \left(sin^2x-\cos^2x\right) \left[{\left(\sin^2x + \cos^2x\right)}^2-2\sin^2x\cos^2x\right]$

$= \left(sin^2x-\cos^2x\right)\:\left(1-2\sin^2x\cos^2x\right)=RHS$

Hence proved

More to know :-

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² -2ab + b²

• a² - b² = (a + b) (a - b)