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Trig Identities Prove: \[\sin^4x+\cos^4x=\sin^2x(\csc^2x-2\cos^2x)\]
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write \(\sin^4x+\cos^4x\) as \((\sin^2x)^2 +(\cos^2x)^2\) and recall the identity \(a^2+b^2=(a+b)^2-2ab\)
\[\begin{align}\sin^4x+\cos^4x &=(\sin^2x)^2+(\cos^2x)^2\\~\\ &=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\~\\ &=(1)^2-2\sin^2x\cos^2x\\~\\ &=1 - 2\sin^2x\cos^2x\\~\\ \end{align}\]
factor out \(\sin^2x\)
I got it now! Thank you so much!
yw
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