Question Number 143057 by ERA last updated on 09/Jun/21
$$\mathrm{cos}\left(\boldsymbol{\alpha}\right)×\mathrm{cos}\left(\mathrm{2}\alpha\right)×\mathrm{cos}\left(\mathrm{4}\alpha\right)×....×\mathrm{cos}\left(\mathrm{2}^{\mathrm{n}} \boldsymbol{\alpha}\right)=\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\alpha}\right)}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\alpha\right)} \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jun/21
$${cos}\left(\alpha\right).{cos}\left(\mathrm{2}\alpha\right).{cos}\left(\mathrm{4}\alpha\right)...{cos}\left(\mathrm{2}^{{n}} \alpha\right) \\ $$$$=\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{2}{sin}\left(\alpha\right)}.\frac{{sin}\left(\mathrm{4}\alpha\right)}{\mathrm{2}{sin}\left(\mathrm{2}\alpha\right)}.\frac{{sin}\left(\mathrm{8}\alpha\right)}{\mathrm{2}{sin}\left(\mathrm{4}\alpha\right)}...\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}{sin}\left(\mathrm{2}^{{n}} \alpha\right)} \\ $$$$=\frac{{sin}\left(\mathrm{2}^{{n}+\mathrm{1}} \alpha\right)}{\mathrm{2}^{{n}} {sin}\left(\alpha\right)} \\ $$