Prove the following : ^8 – ^8 = ( ^2 – ^2 ) (1 – 2sin^2 ^2 )

L.H.S. = ^8 – ^8 = ( ^4 )^2 – ( ^4 )^2 = ( ^4 – ^4 ) ( ^4 + ^4 ) = [( ^2 )^2 – ( ^2 )^2 ] . [( ^2 )^2 + ( ^2 )^2 ] = ( ^2 + ^2 ) ( ^2 – ^2 ). [( ^2 + ^2 )^2 – 2sin^2 . ^2 ] …[Y a^2 + b^2 = (a + b)^2 – 2ab] = (1) ( ^2 – ^2 ) (1^2 – 2sin^2 ^2 ) = ( ^2 – ^2 ) (1 – 2sin^2 ^2 ) = R.H.S.

Prove the following : ^8 – ^8 = ( ^2 – ^2 ) (1

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Question

Prove the following : $\sin^8\theta – \cos^8\theta = (\sin^2 \theta – \cos^2 \theta ) (1 – 2sin^2 \theta \cos^2 \theta )$

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Answer

L.H.S. $= \sin^8\theta – \cos^8\theta $
$= (\sin^4\theta )^2 – (\cos^4\theta )^2$
$= (\sin^4\theta – \cos^4\theta ) (\sin^4\theta + \cos^4\theta )$
$= [(\sin^2 \theta )^2 – (\cos^2 \theta )^2 ]$
$. [(\sin^2 \theta )^2 + (\cos^2 \theta )^2 ]$
$= (\sin^2 \theta + \cos^2 \theta ) (\sin^2 \theta – \cos^2 \theta ). [(\sin^2 \theta + \cos^2 \theta )^2 – 2sin^2 \theta .\cos^2 \theta ] …[Y a^2 + b^2 = (a + b)^2 – 2ab]$
$= (1) (\sin^2 \theta – \cos^2 \theta ) (1^2 – 2sin^2 \theta \cos^2 \theta )$
$= (\sin^2 \theta – \cos^2 \theta ) (1 – 2sin^2 \theta \cos^2 \theta )$
= R.H.S.

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