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Question Number 75983 by mathocean1 last updated on 21/Dec/19

show that cos^6 x+sin^6 x=(1/8)(5+3cos4x)

$$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{5}+\mathrm{3cos4x}\right) \\ $$

Commented by MJS last updated on 21/Dec/19

answer is part of the solution of the other question

$$\mathrm{answer}\:\mathrm{is}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{question} \\ $$

Commented by mathmax by abdo last updated on 22/Dec/19

cos^6 x +sin^6 x =(cos^2 x)^3 +(sin^2 x)^3 =(cos^2 x+sin^2 x)(cos^4 x −cos^2 x sin^2 x +sin^4 x) =cos^4 x +sin^4 x −cos^2 x sin^2 x =(cos^2 x +sin^2 x)^2 −3cos^2 x sin^2 x =1−(3/4)sin^2 (2x) =1−(3/4)(((1−cos(4x))/2)) =1−(3/8) +(3/8)cos(4x) =(5/8) +(3/8)cos(4x) so the result is proved.

$${cos}^{\mathrm{6}} {x}\:+{sin}^{\mathrm{6}} {x}\:=\left({cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \:+\left({sin}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \\ $$$$=\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)\left({cos}^{\mathrm{4}} {x}\:−{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{4}} {x}\right) \\ $$$$={cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} {x}\:−{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\:=\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{3}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\:=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{8}}\:+\frac{\mathrm{3}}{\mathrm{8}}{cos}\left(\mathrm{4}{x}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{8}}\:+\frac{\mathrm{3}}{\mathrm{8}}{cos}\left(\mathrm{4}{x}\right)\:\:{so}\:{the}\:{result}\:{is}\:{proved}. \\ $$

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