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Simplify: $$2(\cos^2 x - \sin^2 x)^2 \tan 2x$$ After some sketching, I arrive at: $$2 \cos 2x \sin2x$$ Now according to the answer sheet, I should simplify this further, to arrive at $\sin 4x$. But how do I derive the latter from the former? Where do I start? Hoe do I use my double-angle formulas to arrive there?

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  • $\begingroup$ Hint: $\sin(2u) = 2\cos(u)\sin(u)$. What is $u$ in this case? $\endgroup$ – MathematicsStudent1122 Dec 27 '15 at 9:46
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Notice, using double angle identity $\cos^2A-\sin^2A=\cos 2A$, one should get

$$2(\cos^2x-\sin^2x)^2\tan 2x$$ $$= 2\cos^2 2x\tan 2x$$ $$= 2\cos^2 2x\left(\frac{\sin 2x}{\cos 2x}\right)$$ $$= 2\sin 2x\cos 2x$$ using double angle identity $2\sin A\cos A=\sin 2A$, $$=\sin 2(2x)$$$$=\color{red}{\sin 4x}$$

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  • $\begingroup$ Sorry, there was a typo in the initial post. $\endgroup$ – Apeiron Dec 27 '15 at 9:45
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    $\begingroup$ Alright, so I have made correction accordingly $\endgroup$ – Harish Chandra Rajpoot Dec 27 '15 at 9:48
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$2\cos(2x)\sin(2x)=\sin(2(2x))=\sin(4x)$

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  • $\begingroup$ Yeah, this just doesn't explain why I can do this. I know $2 \cos x \sin x = \sin 2x$, but how do I get from your first step to the second. This is not obvious for me. Would $2 \cos 3x \sin 3x = \sin 6x$? $\endgroup$ – Apeiron Dec 27 '15 at 9:40
  • $\begingroup$ First step: use $2x$ instead of $x$ in the sine double angle formula Second step: $2*2x=4x$ Yes, $2 \cos 3x \sin 3x = \sin 6x$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 27 '15 at 9:42
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Use $$\cos^2x-\sin^2x=\cos2x$$ and $$\tan A=\dfrac{\sin A}{\cos A}$$

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