<s>[INST] Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct. 
 [/INST] $\sin^4x-\sin^6x=\sin^4x(1-\sin^2x)=\sin^4x \cos^2x=\sin^2x(\sin^2x \cos^2x)$ $$=\frac{\sin^2 x\,\sin^2 2x}{4}=\left(\frac{\sin x \sin 2x}{2}\right)^2=\left(\frac{\cos x-\cos3x}{4}\right)^2$$
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