Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on a problem but it's been a while since I last saw trig identities so I'd love some help or being pointed in the right direction.

Basically, I'd like to understand where this identity comes from;

$$\tan(2t) = \dfrac{2\tan(t)}{1 - \tan^2(t)}$$

Thanks for any help you can give - if it's useful to know the context of the problem, I'm writing a bit of code that converges on $\pi$ faster than Leibniz's series. (Please don't give too much away about the rest of the problem though, I'd like to get there myself :) )

share|cite|improve this question
up vote 5 down vote accepted

If we agree on the two identities:

$$\sin \left( 2 \theta \right) = 2 \sin \left(\theta \right) \cos \left(\theta \right)\\ \cos \left(2 \theta \right)=\cos^2\left(\theta \right)-\sin^2\left(\theta \right) $$ then the rest of it is straight-forward.

$$\tan \left(2\theta \right)=\frac{\sin\left(2\theta \right)}{\cos\left(2\theta \right)}=\frac{ 2 \sin \left(\theta \right) \cos \left(\theta \right)}{\cos^2\left(\theta \right)-\sin^2\left(\theta \right)}=\frac{ 2 \sin \left(\theta \right) \cos \left(\theta \right)}{\cos^2\left(\theta \right)}\frac{1}{1-\tan^2\left(\theta \right)}=\frac{2 \tan{\left(\theta \right)}}{1-\tan^2\left(\theta \right)}$$

Those two identities can be proved in one step using Complex-Numbers. It is a well known identity that $e^{i \theta}=\cos{\left(\theta \right)}+i\sin{\left(\theta \right)}$.

Now consider this:


On the other hand:

$$e^{2i \theta}=\left(e^{i\theta} \right)^2=\left(\cos{\left(\theta \right)}+i\sin{\left(\theta \right)} \right)^2=\left(\cos^2\left(\theta \right)-\sin^2\left(\theta \right)\right)+i\left(2 \sin \left(\theta \right) \cos \left(\theta \right)\right)$$


share|cite|improve this answer
Thanks for the great answer! – Jacobadtr May 27 '13 at 20:04
You are welcome. – Ali May 28 '13 at 5:19

If you know that $\sin(a+b)=\sin a\cos b+\cos a\sin b$ and $\cos(a+b)=\cos a\cos b-\sin a\sin b$, you can apply that to $\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}$ as a function of $\tan a$ and $\tan b$. Once you've got that, consider the case where $a$ and $b$ are both the same number, and you've got it.

share|cite|improve this answer
For variants of the code that you are writing, you may find the formula $\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$ handy. This more general formula also comes out of Michael Hardy's answer. – André Nicolas May 27 '13 at 18:29
@AndréNicolas : That formula is exactly what I had in mind when I wrote "as a function of $\tan a$ and $\tan b$. – Michael Hardy May 27 '13 at 18:32
I was trying to hint at expressions for $\arctan(1)$ other than the one OP has in mind, while not giving away too much, as per request. – André Nicolas May 27 '13 at 18:39

This follows from the fact that $$ \tan(2t) = \frac{\sin(2t)}{\cos(2t)}, $$ and $$ \sin(2t) = 2\sin(t)\cos(t), $$ $$ \cos(2t) = \cos^2(t) - \sin^2(t). $$ These two identities can be found using the formula $$ e^{it} = \cos(t) + i\sin(t), $$ so that $$ \begin{align} e^{i 2t} &= \cos (2t) + i\sin (2t) \\ &= \left(e^{it}\right)^2 = [\cos(t) + i\sin(t)]^2 = [\cos^2(t) - \sin^2(t)] + i[2\sin(t)\cos(t)]. \end{align} $$

share|cite|improve this answer

I think the one of the easiest ways is starting from:

$$\tan2t = \dfrac{\sin2t}{\cos2t} = \dfrac{2\sin t\cos t}{\cos^2t-\sin^2t}$$

From there, you can divide both the numerator and denominators by something and you get the result.

$$\dfrac{2\sin t\cos t}{\cos^2t-\sin^2t}\times \dfrac{\ \frac{1}{A}\ }{\frac{1}{A}}$$

share|cite|improve this answer
Indeed. Also, (for user78514) look up double angle identities. See Pulsar's solution – Frudrururu May 27 '13 at 18:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.