I was looking for proofs for the following 2 things using Euler's identity, but was unable to find them.

I was able to derive the expressions for $sin(\theta)$, $cos(\theta)$ and double angle. Would one way of doing this be to do it geometrically, plug in for sin and cos, then equate the real and imaginary parts?

$sin(a)cos(b) = \frac{1}{2}[sin(a+b)+sin(a-b)]$


$1+cos(2a) = 2cos^{2}(a)$

Any help would be greatly appreciated.

  • $\begingroup$ Of course you can do like that: $e^{i(a+b)}=\cos(a+b)+i\sin(a+b)$ but on the other hand $e^{i(a+b)}=e^{ia}e^{ib}$. $\endgroup$ – Aretino Sep 6 '15 at 16:24
  • $\begingroup$ @Aretino, could you expand on that a little bit $\endgroup$ – Chan Hunt Sep 6 '15 at 16:47
  • 1
    $\begingroup$ $e^{i(a+b)}=e^{ia}e^{ib}=(\cos a+i\sin a)(\cos b+i\sin b)$. Multiply and then compare real and imaginary parts of the result with those of the other expression, i.e. $\cos(a+b)$ and $\sin(a+b)$. $\endgroup$ – Aretino Sep 6 '15 at 17:11

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