I was told that I could prove the cosine addition formula using complex numbers, but I must be missing something, and I was wondering if anyone could offer some insight into where I'm going wrong. Basically right now I have the following:

$$\cos (b) - \cos (a) = 2\cdot \sin (\frac{a+b}{2})\cdot \sin(\frac{a-b}{2})$$ $$\cos (b) - \cos (a) = 2e^{i\frac{a+b}{2} + i\frac{a-b}{2}}$$ $$\cos (b) - \cos (a) = 2e^{ia}$$

Unfortunately that doesn't seem to make any sense if I take either the real or imaginary parts. So I was wondering if there were some extra steps that I've missed ... I'm figuring there must be. All of the internet references I can find for this identity use other trig-identities, not complex numbers, so they unfortunately haven't been much help in this situation (or at least don't seem to offer much help).

Update: Apparently I was confused ... I thought I was being told I could complexify the right-hand side for the proof, when in-fact I needed to use substitution. Thanks @anon for pointing that out :-)


You replaced $\sin\theta$ with $\exp(i\theta)$ for $\theta=(a+b)/2$ and $\theta=(a-b)/2$, but these things are not equal.

From $e^{i\theta}=\cos\theta+i\sin\theta$ we may conclude the actual formulas:

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

Now give it a try with the correct substitutions.

  • $\begingroup$ Okay, I see I mis-understood the problem ... I thought I was being told that I could complexify the right-hand side for the proof, rather than substituting the equivalent complex values using Euler's formulas as you've done. Thanks for the pointer :-) $\endgroup$
    – Jason
    Mar 11 '12 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.