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Show that $8\cos(x)(\sin(2x)^2-\cos(2x)^{2})(\cos(x)^{2}- \sin(x)^2) = 1$

I am a bit confusing how to show this equation. I consider $$8\cos(x)(\sin(2x)^2-\cos(2x)^{2})(\cos(x)^{2}- \sin(x)^2) = 1$$ $$8\cos(x)(4\sin^2(x)\cos^{2}(x)-(2\cos^2(x)-1)(\cos(x)^{2}+ \cos(x)-1) = 1$$ $$8\cos(x)(4((1-\cos^2)\cos^{2}(x)-(2\cos^2(x)-1)(2\cos(x)^{2}-1) = 1$$ But it's look very complicated. Maybe you have simply method?

EDIT:

In general I tried show that: $$8\cos(x)(\sin^{4}(2x)-\cos^{4}(2x))= \frac{1+\tan(x)^{2}}{1-\tan(x)^{2}}$$ which implies that $$8\cos(x)(\sin(2x)^2-\cos(2x)^{2})(\cos(x)^{2}- \sin(x)^2) =1 $$ it's correct?

Blabla
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    "Show that ..."? It is clearly not true for all $x$, eg take $x=0$. Do you want to know the $x$ for which it is true? – almagest Jan 21 '20 at 11:53

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We may write the LHS as $$-8\cos(x)\cos(2x)\cos(4x)$$ and by Proving: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $ we find that $$-8\cos(x)\cos(2x)\cos(4x)=-\frac {\sin (8 x)}{\sin(x) }$$ which is NOT identically $1$.

Robert Z
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  • I tried compute general formula where is reduce to your answer, $8\cos(x)(\sin^{4}(2x)-\cos^{4}(2x))= \frac{1+\tan(x)^{2}}{1-\tan(x)^{2}}$ maybe it's wrong? – Blabla Jan 21 '20 at 12:11
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    @Blabla It's wrong. Take $x=\pi/4$: the LHS is bounded whereas the RHS is not bounded because $\tan(\pi/4)=1$ – Robert Z Jan 21 '20 at 12:27
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This identity cannot be true, as you can check with $x=\frac\pi2$.