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This was a part of a larger question. I have the solution with me, but I have a problem with the way they wrote it.

The expression was written as $$\frac{\sin (\frac{x+29x}{2})\sin(\frac{2x.15}{2})}{\sin\frac{2x}{2}}$$

Is this a formula that I am not aware of, or was it somehow derived from the specific information

Aditya
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  • Hint: multiply by $\cos x$ and express $2\sin nx\cos x$ as the difference of two sines. The resulting series telescopes. – almagest Jan 11 '20 at 14:38
  • See https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Jan 11 '20 at 14:44

3 Answers3

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Note

\begin{align} &2\sin x\sin x=1 -\cos2x\\ &2\sin x\sin 3x= \cos 2x - \cos4x\\ &2\sin x\sin 5x= \cos 4x - \cos6x\\ &\cdots\\ & 2\sin x \sin 29x= \cos 28x -\cos30x \end{align}

Sum up both sides

$$2\sin x\ ( \sin x+\sin 3x+\sin 5x+....+\sin 29x) = 1-\cos30x$$ to arrive at $$\sin x+\sin 3x+\sin 5x +\cdots+\sin 29x=\frac{1-\cos30x}{2\sin x}=\frac{\sin^2 15x}{\sin x}$$

Quanto
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  • So can it be treated as a formula, or does if need to be derived every time – Aditya Jan 11 '20 at 14:49
  • @Aditya - It may be unfamiliar to many, which should not be taken as a well-known formula. – Quanto Jan 11 '20 at 14:54
  • But can it be treated as one, because I don’t really need to show my work to any teacher, so would it apply to every problem? – Aditya Jan 11 '20 at 15:09
  • @Aditya - It should be okay to assume that your teacher knows about it, unless she/he requires you to derive – Quanto Jan 11 '20 at 15:15
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$$\sum_{k=1}^{29}\sin\left(kx\right)=\Im\left[\sum_{k=1}^{29}\left(e^{ikx}\right)\right]$$$$=\Im\left[e^{ix}\frac{e^{i\left(29\right)k}-1}{e^{ix}-1}\right]=\Im\left[e^{ix}\cdot\frac{\left(e^{i\left(\frac{29}{2}\right)x}\right)e^{i\left(\frac{29}{2}\right)x}-e^{-i\left(\frac{29}{2}\right)x}}{e^{i\left(\frac{1}{2}\right)x}\left(e^{i\left(\frac{1}{2}\right)x}-e^{-i\left(\frac{1}{2}\right)x}\right)}\right]$$$$=\Im\left[e^{i\left(15\right)x}\cdot\frac{2i\sin\left(\frac{29}{2}x\right)}{2i\sin\left(\frac{1}{2}x\right)}\right]$$$$=\Im\left[(\cos\left(\left(15\right)x\right)+i\sin\left(\left(15\right)x\right))\frac{2i\sin\left(\frac{29}{2}x\right)}{2i\sin\left(\frac{1}{2}x\right)}\right]$$$$=\sin\left(\left(15\right)x\right)\frac{\sin\left(\frac{29}{2}x\right)}{\sin\left(\frac{1}{2}x\right)}$$

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Note that $$\sum_{r=1}^n\sin(a+(r-1)b)=\dfrac{\sin\left(\dfrac{2a+(n-1)b}{2}\right)\sin\left(\dfrac{nb}{2}\right)}{\sin\left(\dfrac{b}{2}\right)}$$ To prove, multiply and divide by $2\sin\left(\dfrac{b}{2}\right)$ in LHS, apply the identity $$2\sin x\sin y=\cos(x-y)-\cos(x+y)$$ and do the telescopic sum.

Martund
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