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Question Number 134251 by Dwaipayan Shikari last updated on 01/Mar/21

$$\underset{n=1}{\sum ^n}nsin\left(n\right)$$

Commented by Dwaipayan Shikari last updated on 01/Mar/21

$$\underset{n=1}{\sum ^n}cos\left(n\theta \right)=\frac{1}{2sin\frac{\theta }{2}}(sin\frac{3\theta }{2}-sin\frac{\theta }{2}+...+sin\frac{2n+1}{2}\theta -sin\frac{2n-1}{2}\theta )$$$$=\frac{cos\left(\frac{n+1}{2}\theta \right)sin\frac{n\theta }{2}}{sin\frac{\theta }{2}}$$$$-\underset{n=1}{\sum ^n}nsin\left(n\theta \right)=\frac{d}{d\theta }\left(\frac{cos\left(\frac{n+1}{2}\theta \right)sin\frac{n\theta }{2}}{sin\frac{\theta }{2}}\right)$$$$=\frac{sin\frac{\theta }{2}(-\frac{n+1}{2}sin\frac{n\theta }{2}sin\left(\frac{n+1}{2}\theta \right))-\frac{1}{2}cos\frac{\theta }{2}cos\frac{n+1}{2}\theta sin\frac{n\theta }{2}}{{sin}^2\frac{\theta }{2}}$$$$\underset{n=1}{\sum ^n}nsin\left(n\theta \right)=\frac{(n+1)sin\frac{\theta }{2}sin\frac{n\theta }{2}sin\frac{n+1}{2}\theta +cos\frac{\theta }{2}cos\frac{n+1}{2}\theta sin\frac{n\theta }{2}}{2{sin}^2\frac{\theta }{2}}$$

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