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I am trying to figure out the quick way to remember the addition formulas for $\sin$ and $\cos$ using Euler's formula:

$$\begin{align} \sin(\alpha + \beta) &= \sin \alpha\;\cos\beta + \cos\alpha\;\sin\beta \\ \cos(\alpha + \beta) &= \cos \alpha\;\cos\beta - \sin\alpha\;\sin\beta \\ \sin(\alpha-\beta) &= \sin\alpha\;\cos\beta - \cos\alpha\;\sin\beta \\ \cos(\alpha-\beta) &= \cos\alpha\;\cos\beta + \sin\alpha\;\sin\beta \end{align}$$

I'm now convinced of why it is true that $e^{ix} = \cos(x) + i\sin(x)$ but I don't know how to use this to derive these four rules.


marked as duplicate by Blue, Claude Leibovici, Ethan Bolker, Somos, Parcly Taxel Feb 26 '18 at 17:29

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  • $\begingroup$ Calculate $e^{i(\alpha+\beta)}$ and $e^{i(\alpha-\beta)}$ using the fact that $e^{i(x+y)} = e^{ix}e^{iy}$ $\endgroup$ – Dionel Jaime Feb 26 '18 at 3:29
  • $\begingroup$ If you're looking for a way to remember the identities, this answer of mine may help. $\endgroup$ – Blue Feb 26 '18 at 3:39
  • $\begingroup$ @Blue I did see that post but it almost seems like a ton more to remember than just trying to brute memorize the formulas $\endgroup$ – user525966 Feb 26 '18 at 3:52

$$e^{ix}=\cos x+i\sin x$$ $$Re\{e^{i(x+y)}\}=\cos(x+y)$$ $$=Re\{e^{ix}e^{iy}\}$$ $$=Re\{(\cos x+i\sin x)(\cos y+i\sin y)\}$$ $$=Re\{\cos x\cos y+i(\sin x\cos y+\sin y\cos x)-\sin y\sin x\}$$ $$=\cos x\cos y-\sin x\sin y$$

$$Im\{e^{i(x+y)}\}=\sin(x+y)=\sin x\cos y+\sin y\cos x$$

You can apply a similar reasoning to all the other ones, just working with the laws of exponents and comparing the real and imaginary parts.

As for why $e^{ix}=\cos x+i\sin x$ applies, this is commonly understood by considering the Taylor series expansions of $e^{ix}$ and comparing it with that of $\sin x$ and $\cos x$

  • $\begingroup$ What about subtraction? Do you have to divide these? $\endgroup$ – user525966 Feb 26 '18 at 3:38
  • $\begingroup$ Nope, just write $\cos (-x)+i\sin(-x)=\cos x-i\sin x$ as cosine is even, sine is off. $\endgroup$ – Harry Alli Feb 26 '18 at 3:39
  • $\begingroup$ But the lefthand side is still $e^{ix} / e^{iy}$? $\endgroup$ – user525966 Feb 26 '18 at 3:41
  • $\begingroup$ $e^{i(x-y)}=e^{ix}\times e^{-iy}=(\cos x + i\sin x)(\cos y-i\sin y)$ $\endgroup$ – Harry Alli Feb 26 '18 at 3:42
  • $\begingroup$ Can the tangent addition/subtraction formulas be extracted from this somehow too or is that a separate thing? $\endgroup$ – user525966 Feb 26 '18 at 4:00

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