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I have searched a lot and I couldn't find answer for the below sum.

$$S_n = \sin x \cos x +\sin 2x \cos 2x + \sin 3x \cos 3x + \ldots + \sin nx \cos nx$$

How can I solve the above problem?

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    $\begingroup$ HINT: $$\sin2A=2\sin A\cos A$$ and use math.stackexchange.com/questions/17966/… $\endgroup$ – lab bhattacharjee Mar 20 '17 at 17:56
  • $\begingroup$ @labbhattacharjee: can you please post it as an answer. so that I can combine your hint and answer posted in question you mentioned. $\endgroup$ – confusedDeveloper Mar 20 '17 at 18:02
  • $\begingroup$ If u've understood the method, please feel free to supply the answer & optionally accept it $\endgroup$ – lab bhattacharjee Mar 20 '17 at 18:04
  • $\begingroup$ BTW I am not good in maths. That was for my son :) He was able to solve the problem by your hint. Thanks a lot. $\endgroup$ – confusedDeveloper Mar 20 '17 at 18:13
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$$ S_n = \frac{1}{2}\left(\sin(2x)+\sin(4x)+\ldots+\sin(2nx)\right) \tag{1}$$

$$ S_n \sin(x) = \frac{1}{4}\left[\left(\cos(x)-\cos(3x)\right)+\ldots\left(\cos((2n-1)x)-\cos((2n+1)x)\right)\right]\tag{2} $$

$$ S_n \sin(x) = \frac{\cos(x)-\cos((2n+1)x)}{4}\tag{3} $$

$$ S_n = \color{red}{\frac{\cos(x)-\cos((2n+1)x)}{4\sin x}}=\frac{\sin(nx)\sin((n+1)x)}{2\sin x}\tag{4}$$

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