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Question Number 66098 by aliesam last updated on 09/Aug/19

Commented by Prithwish sen last updated on 09/Aug/19

(cosx+isinx)^5 =cos5x+isin5x = cos^5 x+5icos^4 xsinx−10cos^3 xsin^2 x−10icos^2 xsin^3 x+5cosxsin^4 x+isin^5 x compairing the real part on both sides cos5x=cos^5 x−10cos^3 xsin^2 x+5cosxsin^4 x =11cos^5 x−10cos^3 x+5cosx(1−cos^2 x)^2 =16cos^5 x−20cos^3 x+5cosx proved.

$$\left(\mathrm{cosx}+\boldsymbol{\mathrm{i}}\mathrm{sinx}\right)^{\mathrm{5}} =\mathrm{cos5x}+\boldsymbol{\mathrm{i}}\mathrm{sin5x}\:=\:\mathrm{cos}^{\mathrm{5}} \mathrm{x}+\mathrm{5}\boldsymbol{\mathrm{i}}\mathrm{cos}^{\mathrm{4}} \mathrm{xsinx}−\mathrm{10cos}^{\mathrm{3}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}−\mathrm{10}\boldsymbol{\mathrm{i}}\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{3}} \mathrm{x}+\mathrm{5cosxsin}^{\mathrm{4}} \mathrm{x}+\boldsymbol{\mathrm{i}}\mathrm{sin}^{\mathrm{5}} \mathrm{x} \\ $$$$\mathrm{compairing}\:\mathrm{the}\:\mathrm{real}\:\mathrm{part}\:\mathrm{on}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{cos5x}=\mathrm{cos}^{\mathrm{5}} \mathrm{x}−\mathrm{10cos}^{\mathrm{3}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}+\mathrm{5cosxsin}^{\mathrm{4}} \mathrm{x} \\ $$$$=\mathrm{11cos}^{\mathrm{5}} \mathrm{x}−\mathrm{10cos}^{\mathrm{3}} \mathrm{x}+\mathrm{5cosx}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} \\ $$$$=\mathrm{16cos}^{\mathrm{5}} \mathrm{x}−\mathrm{20cos}^{\mathrm{3}} \mathrm{x}+\mathrm{5cosx}\:\:\:\:\:\:\:\:\mathrm{proved}. \\ $$

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