I'm new to this site, english is not my mother tongue, and I'm just learning LaTeX. I'm basically a noob, so please be indulgent if I break any rule or habits.
I'm stuck at proving the following equation. I suppose I should use the formula for geometric series (n∑k=0qk=1−qn+11−q), and also use somewhere that eîθ=cos(θ)+isin(θ)
So here is the equation I have to prove :
12+cos(θ)+cos(2θ)+...+cos(nθ)=sin(n+12θ)2sin(θ2)
Thanks for your help
Answer
12+n∑r=1cos(rθ)=ℜ{12+n∑r=1eirθ}=ℜ{−12+n∑r=0eirθ}=ℜ{−12+1−e(n+1)θ1−eiθ}=ℜ{1−2e(n+1)θ+eiθ2(1−eiθ)}=ℜ{e−iθ/2−2e(n+1/2)θ+eiθ/22(e−iθ/2−eiθ/2)}=ℜ{2cos(θ2)−2cos(n+1/2)θ−2isin(n+1/2)θ−2isin(θ2)}
Multiply through by 1=ii, take the real part and there you have it. ^_^
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