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I have seen many questions talking about closed-form formulas for, say, $\sin(a)+\sin(a+d)+\sin(a+2d)+\dots+\sin(a+[n-1]d)$, however my question is dealing with each of those terms being squared, which seems to make it harder for me. I have tried going the complex number route, in other words turning the sum into taking the imaginary part of a sum with e, but the algebra just explodes. Would anyone be able to "lead me down the right path"?

Any help is greatly appreciated!

Gregory
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nak17
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    Since $\sin^2(x) = \frac{1 - \cos{2x}}{2}$, sum over square of sine of an arithmetic progression can be reduced to sum of cosines of another progression. – achille hui Oct 19 '20 at 15:53
  • $$\sum _{k=0}^{n-1} \sin ^2(a+k d)=\frac{\sin (2 a-d)+2 n \sin d-\sin (2 a+2 d n-d)}{4 \sin d}$$ Found using Wolfram Mathematica – Raffaele Oct 19 '20 at 17:12

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