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$\displaystyle \large \cos 2x + 1 - \sin 2x=\frac{2 \cos 2x \cos x}{\cos x + \sin x}$

I've been trying for a long time but I can't get it.

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2 Answers 2

up vote 5 down vote accepted

$$\cos 2x + 1 - \sin 2x = (\cos x + \sin x)(\cos x - \sin x) + (\cos x - \sin x)^2 $$

$$ = (\cos x - \sin x)(2 \cos x) = \frac{2 \cos 2x \cos x}{\cos x + \sin x}$$

Seems to be an identity.

So all you need is $\cos x + \sin x \neq 0$ and you are done.

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Using the identities $$ {\cos 2x=\cos^2 x -\sin^2 x},\quad \cos 2x=2\cos^2 x-1,\quad \sin 2x=2\sin x\cos x $$ We have, if $\cos x+\sin x\ne0$ $$\eqalign{ {2 {\cos 2x} \cos x\over \cos x+\sin x} &={2( {\cos^2 x-\sin^2 x}) \cos x\over \cos x+\sin x}\cr &={2(\cos x+\sin x)(\cos x-\sin x) \cos x\over \cos x+\sin x}\cr &={2 (\cos x-\sin x) \cos x }\cr &=2\cos^2 x -2\sin x\cos x\cr &=\cos 2x +1 -\sin 2x. } $$

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