# Proving that sin(z1+z2)=sinz1cosz2+sinz2cosz1 in complex plane (Arfken)

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## Homework Statement

Prove that $\sin(z_1+z_2) = \sin z_1\cos z_2+\sin z_2\cos z_1$ such that $z_1,z_2\in\mathbb{C}$

## Homework Equations

$\sin z = \sum\limits_{n=1, \mathrm{ odd}}^\infty (-1)^{(n-1)/2}\dfrac{z^n}{n!} = \sum\limits_{s=0}^\infty (-1)^s\dfrac{z^{2s+1}}{(2s+1)!}$
$\cos z = \sum\limits_{n=0, \mathrm{ even}}^\infty (-1)^{n/2}\dfrac{z^n}{n!} = \sum\limits_{s=0}^\infty (-1)^s\dfrac{z^{2s}}{(2s)!}$

## The Attempt at a Solution

I tried to just put $\sin(z_1+z_2)$ into de sin series but I get binomial of $s$, don't think expanding it would do the trick. Maybe there's a subtle detail to solve this one.

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scottdave
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Are you allowed to use the Euler relationship between sine cosine and e ?

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Yes sir

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Solved.
I just used these series to prove that $i\sin z=\sinh iz$ and then $i\sin(z_1+z_2)=\sinh (iz_1+iz_2)\implies\sin(z_1+z_2)=\dfrac{1}{2i}(e^{iz_1+iz_2}-e^{-iz_1-iz_2})$ now it's just use the euler formula $e^{ix}=\cos x+i\sin x$.

Ray Vickson
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Solved.
I just used these series to prove that $i\sin z=\sinh iz$ and then $i\sin(z_1+z_2)=\sinh (iz_1+iz_2)\implies\sin(z_1+z_2)=\dfrac{1}{2i}(e^{iz_1+iz_2}-e^{-iz_1-iz_2})$ now it's just use the euler formula $e^{ix}=\cos x+i\sin x$.
Easier: use $e^{iz} = \cos z + i \sin z$ and $e^{i(z_1+z_2)} = e^{iz_1}\, e^{iz_2}.$

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Easier: use $e^{iz} = \cos z + i \sin z$ and $e^{i(z_1+z_2)} = e^{iz_1}\, e^{iz_2}.$
cool !! thank you

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