I need to use that $e^{iy} = \cos y + i \sin y$ (for $y \in \mathbb{R}$) to prove that $$\cos y = \frac{e^{iy}+e^{-iy}}{2}$$ and $$\sin y = \frac{e^{iy}-e^{-iy}}{2i}$$ I'm really clueless, any suggestions would be appreciated.

  • $\begingroup$ Do you know anything about the even/odd aspects of sin and cos? Like what cos(-x) and sin(-x) are in terms of cos(X) and sin(x)? $\endgroup$ – turkeyhundt Jan 20 '15 at 20:05
  • $\begingroup$ A simple case of substitution followed by simplification will achieve what you want. $\endgroup$ – Pixel Jan 20 '15 at 20:09

$$e^{yi} = \cos y + i \sin y$$ $$e^{-yi} = \cos y - i \sin y$$


  • $\begingroup$ Thank you, I clearly hadn't put enough effort into my research. $\endgroup$ – martin Jan 20 '15 at 20:49
  • $\begingroup$ @fadaes no problem. Don't worry $\endgroup$ – rlartiga Jan 20 '15 at 20:51

You face a single equation in two unknowns, let $c$ and $s$, $$c+is=e^{iy}=z.$$ Without knowing the parity trick, you need a second relation to make a determined system. Let us try with $$c^2+s^2=1.$$

Method 1

We can factor $$c^2+s^2=(c+is)(c-is)=1,$$ so that $$c-is=\frac1{c+is}=\frac1z,$$ and, solving a linear system, $$c=\frac{z+\frac1z}2,s=\frac{z-\frac1z}{2i}.$$

Method 2

Let us make $s$ explicit from the first equation and plug it into the second $$s=\frac{z-c}i=-i(z-c),$$ $$c^2+s^2=c^2-(z-c)^2=2zc-z^2=1,$$ giving $$c=\frac{1+z^2}{2z},$$ and $$s=\frac1i\left(z-\frac{1+z^2}{2z}\right)=\frac{z^2-1}{2iz}.$$


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