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Question Number 79751 by Rio Michael last updated on 27/Jan/20

prove that cos^6 θ + sin^6 θ = 1 − (3/4) sin^2 2θ

$${prove}\:{that}\:\mathrm{cos}\:^{\mathrm{6}} \theta\:+\:\mathrm{sin}\:^{\mathrm{6}} \theta\:=\:\mathrm{1}\:−\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta \\ $$

Answered by mind is power last updated on 27/Jan/20

(x^6 +y^6 )=(x^2 +y^2 )^3 −3x^2 y^2 (x^2 +y^2 )  x=cos(θ),y=sin(θ) we get  =1−3cos^2 (θ)sin^2 (θ)  =1−(3/4)(4sin^2 (θ)cos^2 (θ))=1−(3/4)(2sin(θ)cos(θ))^2   =1−(3/4)sin^2 (2θ)

$$\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$${x}={cos}\left(\theta\right),{y}={sin}\left(\theta\right)\:{we}\:{get} \\ $$$$=\mathrm{1}−\mathrm{3}{cos}^{\mathrm{2}} \left(\theta\right){sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{4}{sin}^{\mathrm{2}} \left(\theta\right){cos}^{\mathrm{2}} \left(\theta\right)\right)=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{sin}\left(\theta\right){cos}\left(\theta\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}{sin}^{\mathrm{2}} \left(\mathrm{2}\theta\right) \\ $$$$ \\ $$

Commented by Rio Michael last updated on 27/Jan/20

thanks

$${thanks} \\ $$

Answered by Henri Boucatchou last updated on 28/Jan/20

 Generaly   we  prove  that :    ∀n∈N^∗ ,  cos^n θ×sin^n θ=1−(n/8)sin^2 2θ

$$\:{Generaly}\:\:\:{we}\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\forall{n}\in\mathbb{N}^{\ast} ,\:\:{cos}^{{n}} \theta×{sin}^{{n}} \theta=\mathrm{1}−\frac{{n}}{\mathrm{8}}{sin}^{\mathrm{2}} \mathrm{2}\theta \\ $$

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