<s>[INST] How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$. 
Can anybody help me?!
Thank you!
 [/INST] L.H.S=$3\sin^4x-2\sin^6x$
=$3(\sin^2x)^2-2(\sin^2x)^3$
=$3(1-\cos^2x)^2 -2 (1-\cos^2x)^3$
=$3(1+\cos^4x -2\cos^2x) -2 (1-\cos^6x -3 \cos^2x+3\cos^4x)$
=$ 3+3\cos^4x -6\cos^2x -2+2\cos^6x +6\cos^2x-6\cos^4x)$
=$1-3\cos^4x+2\cos^6x$
=R.H.S
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