Question Number 136581 by mnjuly1970 last updated on 23/Mar/21
$$\:\:\:\:\:\:\:\:…….{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:{pove}\:{that}:: \\ $$$$:::\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{\mathrm{2}{n}} }\begin{pmatrix}{\:\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}\:{cos}\left({nx}\right)\right\}=\frac{{cos}\left(\frac{{x}}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{2}}\right)}} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Mar/21
$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\left(−\frac{{e}^{{ix}} }{\mathrm{2}^{\mathrm{2}} }\right)^{{n}} +\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\left(−\frac{{e}^{−{ix}} }{\mathrm{4}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{e}^{{ix}} }}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{e}^{−{ix}} }}=\frac{{e}^{\frac{{ix}}{\mathrm{2}}} +\mathrm{1}}{\mathrm{2}\sqrt{{e}^{{ix}} +\mathrm{1}}}=\frac{\mathrm{1}+{cos}\left(\frac{{x}}{\mathrm{2}}\right)+{isin}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{1}+{cos}\left({x}\right)+{isin}\left({x}\right)}} \\ $$$$=\frac{\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{4}}\right)\left({cos}\frac{{x}}{\mathrm{4}}+{isin}\left(\frac{{x}}{\mathrm{4}}\right)\right)}{\mathrm{2}\sqrt{\mathrm{2}{cos}\frac{{x}}{\mathrm{2}}}\sqrt{{cos}\frac{{x}}{\mathrm{2}}+{isin}\frac{{x}}{\mathrm{2}}}}=\frac{{cos}\left(\frac{{x}}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}{cos}\frac{{x}}{\mathrm{2}}}}.\frac{{e}^{{ix}/\mathrm{4}} }{{e}^{{ix}/\mathrm{4}} }=\frac{{cos}\left(\frac{{x}}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}{cos}\frac{{x}}{\mathrm{2}}}} \\ $$