I have just learned Euler's formula: $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, and I was wondering how it could be used to prove $\cos(\theta)=2\cos(\theta)^2-1$. I have begun working on the proof but I seem to be stuck once I get to the following part: $$e^{2i\theta}=(e^{i\theta})^2$$ $$e^{2i\theta}=(\cos(\theta)+i\sin(\theta))^2$$ $$e^{2i\theta}=(\cos^2(\theta)-\sin^2(\theta))+2i\sin(\theta)\cos(\theta)$$ How would I continue the rest of the proof without assuming anything about trig identities with complex numbers? Thank You!

EDIT: Using the first reply, I got the proof to: $$e^{2ix}=2\cos^2(x)+1 + 2i\sin(x)\cos(x)$$ But I am not entirely sure what else he meant and how to continue

EDIT 2: Specifically for complex angles

  • $\begingroup$ you will need $-\sin^2(x) = \cos^2(x)-1$ and equality of the real part and the imaginary part $\endgroup$ – stity Nov 24 '15 at 0:35
  • $\begingroup$ And also can't both, neither, or either part be imaginary? $\endgroup$ – sbmit2 Nov 24 '15 at 0:49
  • $\begingroup$ $\cos^2(x)-\sin^2(x)=\cos^2(x)-\left(1-\cos^2(x)\right)=2\cos^2(x)-1$ $\endgroup$ – robjohn Nov 24 '15 at 1:21

You have correctly shown that $$ e^{2i\theta}=\color{#C00000}{\cos^2(\theta)-\sin^2(\theta)}+i\,\color{#00A000}{2\sin(\theta)\cos(\theta)} $$ but using Euler's formula again, we have $$ e^{2i\theta}=\color{#C00000}{\cos(2\theta)}+i\color{#00A000}{\sin(2\theta)} $$ Look at the real parts of each and remember that $\sin^2(\theta)+\cos^2(\theta)=1$.

For complex angles, you may want to use the identity (which follows from Euler's formula) $$ \cos(z)=\frac{e^{iz}+e^{-iz}}2 $$ Then using $$ \cos^2(z)=\frac{e^{2iz}+2+e^{-2iz}}4 $$ and $$ \cos(2z)=\frac{e^{2iz}+e^{-2iz}}2 $$ will give you the desired identity.

  • $\begingroup$ How can you know that $\cos(2\theta)$ is real and not complex or imaginary? $\endgroup$ – sbmit2 Nov 24 '15 at 1:12
  • $\begingroup$ When $\theta$ is real, both $\sin(\theta)$ and $\cos(\theta)$ are also real. $\endgroup$ – robjohn Nov 24 '15 at 1:13
  • $\begingroup$ But isn't the proof attempting to prove that it works for complex angles $\theta$, or am I mistaken? $\endgroup$ – sbmit2 Nov 24 '15 at 1:13
  • $\begingroup$ I thought you were trying to use Euler's formula to show that $\cos(2\theta)=\cos^2(\theta)-1$. That is generally a formula for real $\theta$. You did not say anything about complex $\theta$. This is a very good example of why people ask for context in questions; it helps to provide proper answers. $\endgroup$ – robjohn Nov 24 '15 at 1:17
  • $\begingroup$ Oh, I'm terribly sorry for not providing that in the initial post! I have edited it to reflect the question. Sorry! $\endgroup$ – sbmit2 Nov 24 '15 at 1:18

I will use different notation if we are talking about trigonometric functions of a complex number $z$. First, Euler's identity for general complex $z$ is $$e^{iz} = \cos z + i \sin z, \quad z \in \mathbb{C}.$$ Therefore, $$\cos z = \frac{e^{iz} + e^{-iz}}{2}.$$ Consequently, $$\begin{align*} \cos 2z &= \frac{e^{2iz} + e^{-2iz}}{2} \\ &= \frac{(e^{iz})^2 + 2 + (e^{-iz})^2}{2} - 1 \\ &= \frac{(e^{iz})^2 + 2e^{iz}e^{-iz} + (e^{iz})^2}{2} - 1 \\ &= 2 \left( \frac{e^{iz} + e^{-iz}}{2} \right)^2 - 1 \\ &= 2 \cos^2 z - 1. \end{align*}$$

  • $\begingroup$ "Now comparing the real and imaginary parts gives two identities" But can't you not say that either is real or not real because the angles could be complex causing both to be complex? How can you say that there is a real and imaginary part? Thanks $\endgroup$ – sbmit2 Nov 24 '15 at 1:05
  • $\begingroup$ @sbmit2 Proof changed. $\endgroup$ – heropup Nov 24 '15 at 1:25

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