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Question Number 153758 by yeti123 last updated on 10/Sep/21

$$S=\frac{\underset{k=1}{\sum ^n}\sin \left({\theta }_k\right)}{\underset{k=1}{\sum ^n}\cos \left({\theta }_k\right)};\mathrm{where}{\left({\theta }_k\right)}_{k=1}^n\mathrm{is}\mathrm{an}\mathrm{arithmetic}\mathrm{progression}.$$$$\mathrm{show}\mathrm{that}S=\tan \left(\overline{\theta }\right)$$$$\mathrm{where}\overline{\theta }=\frac{1}{n}\underset{k=1}{\sum ^n}{\theta }_k\mathrm{is}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}\mathrm{of}\left({\theta }_k\right)$$

Answered by mindispower last updated on 10/Sep/21

$${\theta }_k=ak+b$$$$\underset{k=1}{\sum ^n}{e}^{i{\theta }_k}=e^{ib}.e^{ia}\frac{1-{\left(e^{ia}\right)}^n}{1-e^{ia}}=S$$$$=e^{i(a+b)}.\frac{e^{i(\frac{na}{2}-\frac{a}{2})}cos\left(\frac{na}{2}\right)}{cos\left(\frac{a}{2}\right)}$$$$=e^{i(\frac{a}{2}(n+1)+b)}\frac{cos\left(n\frac{a}{2}\right)}{cos\left(\frac{a}{2}\right)}n$$$$\mathrm{\Sigma }sin\left({\theta }_k\right)=\frac{cos\left(\frac{na}{2}\right)}{cos\left(\frac{a}{2}\right)}sin\left(\frac{1}{n}(\frac{n(n+1)a}{2}+bn)\right)$$$$\mathrm{\Sigma }cos\left({\theta }_k\right)=\frac{cos\left(\frac{na}{2}\right)}{cos\left(\frac{a}{2}\right)}cos(\left(\frac{1}{n}(\frac{n(n+1)a}{2}+bn)\right)$$$$\mathrm{\Sigma }{\theta }_k=\underset{k=1}{\sum ^n}(ak+b)=nb+\frac{a}{2}(n(n+1)$$$$\frac{\mathrm{\Sigma }sin\left({\theta }_k\right)}{\mathrm{\Sigma }cos\left({\theta }_k\right)}=\frac{\frac{cos\left(\frac{na}{2)}\right)}{cos\left(\frac{a}{2}\right)}sin\left(\frac{1}{n}(n(n+1)\frac{a}{2}+nb)\right)}{\frac{cos\left(\frac{na}{2}\right)}{cos\left(\frac{a}{2}\right)}cos(\frac{1}{n}(\left(\frac{n+1}{2}\right)na+nb)}$$$$=tg\left(\frac{1}{n}\left(\mathrm{\Sigma }{\theta }_k\right)\right)=tg\left({\stackrel{-}{\theta }}_k\right)$$$$$$

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