# 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 $\tan A$.

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– draks ... Mar 29 '12 at 7:55
I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name. – user21436 Mar 29 '12 at 22:28

We need to prove that: $$\frac{\sin(2A)}{1+\cos(2A)}=\tan(A)$$ Let's do LHS-RHS to prove it. I will try to make the left side equal the right side. $$\frac{\sin(2A)}{1+\cos(2A)}$$ Using double angle identites for both sine and cosine: $$\frac{2\sin(A)\cos(A)}{1+2\cos^2(A)-1}$$ How nice. The $1$ and $-1$ in the denominator cancel out. $$\frac{2\sin(A)\cos(A)}{2\cos^2(A)}$$ Cancelling out the $\cos(A)$ in the numerator and the denominator yields: $$\frac{2\sin(A)}{2\cos(A)}$$ We can also cancel out the $2$ in the numerator and the denominator. $$\frac{\sin(A)}{\cos(A)}$$ $$=\tan(A)$$ $$\text{LHS=RHS}$$ $$\displaystyle \boxed{\therefore \dfrac{\sin(2A)}{1+\cos(2A)}=\tan(A)}$$

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First, lets develop a couple of identities. Given that $\sin 2A = 2\sin A\cos A$, and $\cos 2A = \cos^2A - \sin^2 A$ we have

$$\begin{array}{lll} \tan 2A &=& \frac{\sin 2A}{\cos 2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{2\tan A}{1-\tan^2A} \end{array}$$ Similarly, we have $$\begin{array}{lll} \sec 2A &=& \frac{1}{\cos 2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{\sec^2 A}{1-\tan^2A} \end{array}$$

But sometimes it is just as easy to represent these identities as

$$\begin{array}{lll} (1-\tan^2 A)\sec 2A &=& \sec^2 A\\ (1-\tan^2 A)\tan 2A &=& 2\tan A \end{array}$$

Applying these identities to the problem at hand we have $$\begin{array}{lll} \frac{\sin 2A}{1+\cos 2A}&=& \frac{\sin 2A}{1+\cos 2A}\cdot\frac{\frac{1}{\cos 2A}}{\frac{1}{\cos 2A}}\\ &=& \frac{\tan 2A}{\sec 2A +1}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)(\sec 2A +1)}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)\sec 2A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{\sec^2 A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{(\tan^2 A+1) +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{2}\\ &=& \tan A\\ \end{array}$$ discalimer: I like JChau's answer better :)

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