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We are given

If $\cos(a+ib)$=$r (\cos\theta +i\sin\theta)$

then prove that $e^{2b} = \sin(a-\theta)/­\sin(a+\theta)$

I just tried and got $b = 0$ such that $\cos(a) = ra$. Will there be other solutions?

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3  
$\cos$ is defined for all complex numbers, and it is $\cos(z)=\displaystyle\frac{e^{iz}+e^{-iz}}2$. –  Berci Dec 11 '12 at 23:26
    
I also thought about that but was stuck. How do I continue from there? –  John Chang Dec 11 '12 at 23:34
    
Do you know anything about $r$? as a complex equation, it should not go away in the answer like that. I assume $a,b,r,\theta$ are real? –  Mario Carneiro Dec 11 '12 at 23:52
    
Nothing was said about r. And yes, I assume those are real too. –  John Chang Dec 11 '12 at 23:54

2 Answers 2

up vote 2 down vote accepted

$\displaystyle \cos (a+ib) = \cos a \cos ib - \sin a \sin ib = \cos a \cosh b - i \sin a \sinh b = r (\cos \theta + i \sin \theta) \\ \displaystyle r \cos \theta = \cos a \cosh b = \cos a \frac {e^{2b} + 1}{2e^b}\\ \displaystyle r \sin \theta = -\sin a \sinh b = -\sin a \frac {e^{2b} - 1}{2e^b} \\ \displaystyle \tan \theta = -\tan a \frac {e^{2b}-1}{e^{2b}+1} \\ \displaystyle \tan \theta (e^{2b}+1) = -\tan a(e^{2b}-1)\\ \displaystyle e^{2b}(\tan a + \tan \theta) = \tan a - \tan \theta \\ \displaystyle e^{2b} = \frac {\tan a - \tan \theta}{\tan a + \tan \theta} = \frac{\frac {\sin (a - \theta)}{\cos a \cos \theta}}{\frac{\sin(a + \theta)}{\cos a \cos \theta}} = \frac{\sin(a - \theta)}{\sin (a + \theta)}$

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You know that $$\frac{e^{ia-b}+e^{b-ia}}2=re^{i\theta},$$ so we can rearrange this into $$e^{-b}e^{i(a-\theta)}+e^be^{-i(a+\theta)}=2r$$ and it's conjugate, $$e^be^{i(a+\theta)}+e^{-b}e^{-i(a-\theta)}=2r.$$ If we match these equations, we get $$e^{-b}e^{i(a-\theta)}+e^be^{-i(a+\theta)}=e^be^{i(a+\theta)}+e^{-b}e^{-i(a-\theta)}$$ $$e^{-b}(e^{i(a-\theta)}-e^{-i(a-\theta)})=e^b(e^{i(a+\theta)}-e^{-i(a+\theta)})$$ $$e^{-b}\sin(a-\theta)=e^b\sin(a+\theta)$$ $$e^{2b}=\frac{\sin(a-\theta)}{\sin(a+\theta)}$$ and we're done.

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