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I want to know how to prove that :

$\sin(\alpha)+\sin(\alpha+\beta)+\sin(\alpha+2\beta)+...+\sin(\alpha+(n-1)\beta)=\frac{\sin(\frac{n\alpha}{2})}{\frac{n\alpha}{2}}[sin(\alpha+\frac{n-1}{2}\beta)]$

I tried by expanding $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\sin(\alpha+2\beta)=\sin(\alpha) \cos(2\beta)+\cos(\alpha)\sin(2\beta)$

And so on.

Finally I got the summation to be

$\sin(\alpha)[1+\cos\beta+\cos2\beta+\cos3\beta+...\cos(n-1)\beta]+\cos(\alpha)[\sin\beta+\sin2\beta+\sin3\beta+...+\sin(n-1)\beta]$

But I can't understand how to proceed further

Please help me out.

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    Hint, try using Euler's formula. – Jair Taylor Aug 12 '24 at 19:29
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    See also https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange's_trigonometric_identities – Vaskara_GRek_O Aug 12 '24 at 19:33
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    The formula you are trying to prove isn’t correct. Please check the original problem statement. – David Quinn Aug 12 '24 at 20:21
  • For example, $n=2$, $\alpha = 2\pi$, $\beta = \pi/2$,

    $$\begin{align} LHS&= \sin 2\pi + \sin \left(2\pi+\frac\pi2\right) = 1\ RHS &= \frac{\sin{\frac{2\cdot2\pi}2}}{\frac{2\cdot2\pi}2}\cdot\sin(\ldots) = 0 \end{align}$$

     – peterwhy Aug 12 '24 at 21:00
  • I have checked and my formula is correct. You can recheck once @DavidQuinn –  Aug 13 '24 at 02:18
  • Multiply by $2 \sin \dfrac{\beta}{2}$ and use product-to-sum formula to get a telescoping sum. – Hari Shankar Aug 13 '24 at 04:04
  • I suggest you take the hint by @HariShankar and figure out for yourself what the actual formula is meant to be, but the formula you have suggested in your question is definitely false. – David Quinn Aug 13 '24 at 08:44
  • see this - https://math.stackexchange.com/a/1633191/1157207 – Amrut Ayan Aug 14 '24 at 12:28
  • The answer should be $\frac{\sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}} \left[ \sin \left(\alpha+\frac{(n-1) \beta}{2}\right)\right]$ as shown below. – Lai Aug 14 '24 at 13:10

1 Answers1

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Using Euler’s identity $e^{ix}=\cos x+i\sin x$, we can express the sum as the imaginary part of an arithmetic series. $$ \begin{aligned} & \sin \alpha+\sin (\alpha+\beta)+\sin (\alpha+2 \beta)+\ldots+\sin (\alpha+(n-1) \beta) \\ = & \Im\left[e^{\alpha i}+e^{(\alpha+\beta) i}+e^{(\alpha+2 \beta) i}+\ldots+e^{\alpha+(n-1) \beta i}\right] \\ = & \Im\left[e^{\alpha i} \cdot \frac{e^{n \beta i}-1}{e^{\beta i}-1}\right] \\ = & \Im\left[e^{\alpha i} \cdot \frac{e^{\frac{n \beta i}{2}}\left(e^{\frac{n \beta i}{2}}-e^{-\frac{n \beta i}{2}}\right)}{e^{\frac{\beta}{2} i}\left(e^{\frac{\beta i}{2}}-e^{-\frac{\beta i}{2}}\right)}\right] \\ = & \Im\left[e^{\left(\alpha+\frac{(n-1) \beta}{2}\right) i} \cdot \frac{2 i \sin \frac{n \beta}{2}}{2 i \sin \frac{\beta}{2}}\right] \\ = & \frac{\sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}}\left[\sin \left(\alpha+\frac{(n-1) \beta}{2}\right)\right] \end{aligned} $$

Wish it helps!

Lai
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