MHB Proving Identity: $\sin^8A-\cos^8A$

[Silver Bolt]

$\sin^8\left({A}\right)-\cos^8\left({A}\right)=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$

$L.H.S=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$

$ =(\sin\left({A}\right)+\cos\left({A}\right)) (\sin\left({A}\right)-\cos\left({A}\right)(\cos^2\left({A}\right)-\sin^2\left({A}\right))^2$

Any ideas?

 

[kaliprasad]

$\sin^8\left({A}\right)-\cos^8\left({A}\right)=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$

$L.H.S=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$

$ =(\sin\left({A}\right)+\cos\left({A}\right)) (\sin\left({A}\right)-\cos\left({A}\right)(\cos^2\left({A}\right)-\sin^2\left({A}\right))^2$

Any ideas?

$\sin^8\left({A}\right)-\cos^8\left({A}\right)$
= $(\sin^4\left({A}\right)+\cos^4\left({A}\right))(\sin^4\left({A}\right)-\cos^4\left({A}\right)$ using $a^2-b^2= (a+b)(a-b)$
= $(\sin^4\left({A}\right)+\cos^4\left({A}\right))(\sin^2\left({A}\right)+\cos^2\left({A}\right)(\sin^2\left({A}\right)-
\cos^2\left({A}\right))$ using $a^2-b^2= (a+b)(a-b)$
= $(\sin^4\left({A}\right)+\cos^4\left({A}\right))(\sin^2\left({A}\right)-
\cos^2\left({A}\right))$
= $((\sin^2\left({A}\right)+\cos^2\left({A}\right))^2-2 \sin^2\left({A}\right)\cos^2\left({A}\right)(\sin^2\left({A}\right)-
\cos^2\left({A}\right))$ using $a^2+b^2 = + (a+b)^2 - 2ab$
= $((\sin^2\left({A}\right)+\cos^2\left({A}\right))^2-2 \sin^2\left({A}\right)\cos^2\left({A}\right)(\sin^2\left({A}\right)-
\cos^2\left({A}\right))$
= $(1-2 \sin^2\left({A}\right)\cos^2\left({A}\right)(\sin^2\left({A}\right)-
\cos^2\left({A}\right))$

The rational is that we need to reduce the power to squares of sin and cos