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

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