# How to prove a trigonometric identity

Show that $$\tan(A)=\frac{\sin2A}{1+\cos 2A}$$

I've tried a few methods, and it stumped my teacher.

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Failure to prove is not surprising. It is quite false. For example, take $A=\pi/4$ ($45$ degrees). –  André Nicolas Mar 15 '12 at 23:48
I think you want $\cos(2A)$ there... –  David Mitra Mar 15 '12 at 23:53
This is not a valid identity: try $A=\pi/4$ for example. –  Shane O Rourke Mar 15 '12 at 23:54

Proof without words: $\tan(A)=\dfrac{\color{red}{\sin(2A)}}{\color{blue}{1}+\color{green}{\cos(2A)}}$

$\hspace{4cm}$

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+1: Very nice.! –  Aryabhata Mar 29 '12 at 22:33
I've never seen anyone prove a trig identity like that, but they should! –  The Substitute Mar 29 '12 at 23:54

The given equality is false. Set $A = \pi/2$. (Note: this applied to an earlier version of the problem).

Perhaps what you meant was

$$\tan \frac{A}{2} = \frac{\sin A}{1 + \cos A}$$

or

$$\tan A = \frac{\sin 2A}{1 + \cos 2A}$$

which is true, by using the half/double angle formulas.

$$\frac{\sin A}{1 + \cos A} = \frac{ 2 \sin A/2 \cos A/2}{2 \cos^2 A/2} = \tan A/2$$

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This is one of my favorite identities. It is the basis for one form of $\operatorname{atan2}(x,y)=2\operatorname{atan}\left(\dfrac{y}{r+x}\right)$ which is useful if you have to compute $r=\sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1) –  robjohn Mar 29 '12 at 22:40
Nice proof, too :-) –  robjohn Mar 29 '12 at 22:48
$$\sin 2A = 2 \sin A \cos A$$
$$\cos 2A = 2 \cos^2A - 1$$
Substitute these identities and you will get $\tan A$.