# 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.

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

• Orodruin
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Yes sir

Gold Member
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
Homework Helper
Dearly Missed
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}.##

• Felipe Lincoln
Gold Member
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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