Question Number 151559 by peter frank last updated on 21/Aug/21
$$\mathrm{prove}\:\mathrm{that} \\ $$$$\left(\mathrm{1}+\mathrm{cos}\:\theta+\mathrm{isin}\:\theta\right)^{\mathrm{n}} \\ $$$$+\:\left(\mathrm{1}+\mathrm{cos}\:\theta−\mathrm{isin}\:\theta\right)^{\mathrm{n}} =\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{cos}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{n}\theta}{\mathrm{2}} \\ $$
Answered by Olaf_Thorendsen last updated on 22/Aug/21
$$\mathrm{S}_{{n}} \:=\:\left(\mathrm{1}+\mathrm{cos}\theta+{i}\mathrm{sin}\theta\right)^{{n}} +\left(\mathrm{1}+\mathrm{cos}\theta−{i}\mathrm{sin}\theta\right)^{{n}} \\ $$$$\mathrm{S}_{{n}} \:=\:\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}} +\left(\mathrm{1}+{e}^{−{i}\theta} \right)^{{n}} \\ $$$$\mathrm{S}_{{n}} \:=\:\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}} +\mathrm{conj}\left(\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}} \right) \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{2Re}\left(\mathrm{1}+{e}^{{i}\theta} \right)^{{n}} \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{2Re}\left({e}^{{i}\frac{{n}\theta}{\mathrm{2}}} \left({e}^{{i}\frac{\theta}{\mathrm{2}}} +{e}^{−{i}\frac{\theta}{\mathrm{2}}} \right)^{{n}} \right) \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{2Re}\left({e}^{{i}\frac{{n}\theta}{\mathrm{2}}} \mathrm{2}^{{n}} \mathrm{cos}^{{n}} \left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{S}_{{n}} \:=\:\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\left(\frac{{n}\theta}{\mathrm{2}}\right)\mathrm{cos}^{{n}} \left(\frac{\theta}{\mathrm{2}}\right) \\ $$
Commented by peter frank last updated on 22/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by peter frank last updated on 22/Aug/21
