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$$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$ This is what I've done: $$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$ $$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$ $$\frac{\sin x}{1+\cos x}$$ I have no idea what to do next.

edit: Solution:

$$\frac{\sin2\frac{x}{2}}{1+\cos2\frac{x}{2}}=$$ $$\frac{2\sin\frac{x}{2}cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}=$$ $$\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=$$ $$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$

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The $\frac{x}{2}$ in the answer should provide a clue. In your last step, there is still no hint of $\frac{x}{2}$. How will you change $x$ to $\frac{x}{2}$? – M. Vinay Jun 17 '14 at 13:40
First, you should use \sin and \cos, not sin and cos. As to your question, do you know the half-angle identities? One form of the half-angle identity for $\tan$ is $\tan\frac{x}{2} = \frac{\sin x}{1+\cos x}$. – rogerl Jun 17 '14 at 13:41
@M.Vinay Thanks, I'm dumb... xD – A6SE Jun 17 '14 at 13:42
@M.Vinay You can post the answer so I can accept it. – A6SE Jun 17 '14 at 13:50
@A6Tech Thanks, but there are some answers here already, you should accept one of them :) – M. Vinay Jun 17 '14 at 13:52
up vote 3 down vote accepted

$\cos 2\theta = 2\cos^2 \theta - 1 \Rightarrow 1 + \cos 2\theta = 2\cos^2 \theta$. Therefore:

$$\dfrac{\sin 2x}{1 + \cos 2x} \times \dfrac{\cos x}{1 + \cos x}\\ = \dfrac{\not 2\sin x \not\cos x}{\not 2 \not\cos^2 x} \times \dfrac{\not\cos x}{2 \cos^2 (x/2)}\\ = \dfrac{\not 2\sin(x/2)\not\cos(x/2)}{\not 2\cos^{\not 2} (x/2)}\\ = \boxed{\tan \dfrac{x}{2}} $$

Note: In the above, the $\not\square$s denote cancellations.

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By your estimate we have $$\frac{\sin x}{1+\cos x}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\cos^2\frac{x}{2}}=\tan\frac{x}{2}.$$

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

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