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Question Number 97135 by  M±th+et+s last updated on 06/Jun/20

$$provethat:$$$$sin\left(16x\right)cot\left(x\right)=1+2cos\left(2x\right)+2cos\left(4x\right)+2cos\left(6x\right)+...+2cos\left(16x\right)$$

Commented by prakash jain last updated on 07/Jun/20

$$\cos x=\frac{1}{2}(e^{ix}+e^{-ix})$$$$\mathrm{RHS}=1+\underset{k=1}{\sum ^8}(e^{2kix}+e^{-2kix})$$$$=e^{-16ix}+e^{-14kx}+...+e^{16ix}$$$$=\frac{e^{-16ix}(e^{34ix}-1)}{e^{2ix}-1}=\frac{e^{18ix}-e^{-16ix}}{e^{2ix}-1}$$$$$$$$\sin 16x\cot x=\frac{e^{16ix}-e^{-16ix}}{2}\times \frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}$$$$=\frac{e^{16x}-e^{-16x}}{2}\times \frac{e^{2ix}+1}{e^{2ix}-1}$$$$=\frac{e^{18ix}-e^{-14ix}+e^{16ix}-e^{-16ix}}{2(e^{2ix}-1)}$$$$\mathrm{Equality}\mathrm{does}\mathrm{not}\mathrm{seem}\mathrm{be}\mathrm{valid}.$$$$\mathrm{I}\mathrm{will}\mathrm{recheck}\mathrm{my}\mathrm{calculation}\mathrm{again}.$$

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