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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 :) )

Jerry
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jm22b
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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:

$$e^{2i\theta}=\cos{2\theta}+i\sin{2\theta}$$

On the other hand:

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

Bingo!

Ali
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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.

Michael Hardy
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  • 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
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    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
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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 $$ \begin{align} e^{i 2t} &= \cos (2t) + i\sin (2t) \\ &= \left(e^{it}\right)^2 = [\cos(t) + i\sin(t)]^2 = [\cos^2(t) - \sin^2(t)] + i[2\sin(t)\cos(t)]. \end{align} $$

Pulsar
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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.

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

Jerry
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  • Indeed. Also, (for user78514) look up double angle identities. See Pulsar's solution –  May 27 '13 at 18:26