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Question Number 113408 by ZiYangLee last updated on 13/Sep/20

Commented by ZiYangLee last updated on 13/Sep/20

$$\mathrm{help}..$$

Answered by Dwaipayan Shikari last updated on 13/Sep/20

$$sin3\theta +sin5\theta +...$$$$=\frac{1}{2sin\theta }(2sin3\theta sin\theta +2sin5\theta sin\theta +....)$$$$=\frac{1}{2sin\theta }(cos2\theta -cos4\theta +cos4\theta -cos6\theta +.....+cos2n\theta -cos(2n+2)\theta )$$$$=\frac{1}{2sin\theta }(cos2\theta -cos(2n+2)\theta )$$$$=\frac{1}{sin\theta }\left(sin(n+2)\theta sin\left(n\theta \right)\right)$$

Commented by ZiYangLee last updated on 14/Sep/20

$$\mathrm{Tks}!$$

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