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

$$provethat\cos {}^6\theta +\sin {}^6\theta =1-\frac{3}{4}\sin {}^{2}2\theta$$

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

$$(x^6+y^6)={(x^2+y^2)}^3-3x^2{y}^2(x^2+y^2)$$$$x=cos\left(\theta \right),y=sin\left(\theta \right)weget$$$$=1-3{cos}^2\left(\theta \right){sin}^2\left(\theta \right)$$$$=1-\frac{3}{4}\left(4{sin}^2\left(\theta \right){cos}^2\left(\theta \right)\right)=1-\frac{3}{4}{\left(2sin\left(\theta \right)cos\left(\theta \right)\right)}^2$$$$=1-\frac{3}{4}{sin}^2\left(2\theta \right)$$$$$$

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

$$thanks$$

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

$$Generalyweprovethat:$$$$\mathrm{\forall }n\in {\mathbb{N}}^{\ast },{cos}^n\theta \times {sin}^n\theta =1-\frac{n}{8}{sin}^{2}2\theta$$

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