Question

Sin8A-cos8A =(sin2A-cos2A)(1-2sin2Acos2A)

Answer 1

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Answer 2

Answer:

We have to prove the relation:

$\sin^8\theta-\cos^8\theta=(\sin^2\theta-\cos^2\theta)(1-2\sin^2\theta\cos^2\theta)$

We will evaluate the left hand side term to obtain the right hand side term.

On taking Left hand side:

$\sin^8\theta-\cos^8\theta=(\sin^4\theta)^2-(\cos^4\theta)^2\\\\=(\sin^4\theta-\cos^4\theta)(\sin^4\theta+\cos^4\theta)$

Since, we know that:

$a^2-b^2=(a+b)(a-b)$

$=[(\sin^2\theta)^2-(\cos^2\theta)^2][(\sin^2\theta)^2+(\cos^2\theta)^2]\\\\=(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta]$

Also we know that:

$\sin^2\theta+\cos^2\theta=1$

and

$a^2+b^2=(a+b)^2-2ab$ )

$=(\sin^2\theta-\cos^2\theta)(1-2\sin^2\theta\cos^2\theta)$

Hence, the expression is proved.

Hence,  we get

$\sin^8\theta-\cos^8\theta=(\sin^2\theta-\cos^2\theta)(1-2\sin^2\theta\cos^2\theta)$