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Question Number 111426 by weltr last updated on 03/Sep/20

Answered by Dwaipayan Shikari last updated on 03/Sep/20

$$4({sin}^6\alpha +{cos}^6\alpha )-3{cos}^{2}2\alpha$$$$=4{({sin}^2\alpha +{cos}^2\alpha )}^3-12{sin}^2\alpha {cos}^2\alpha -3{cos}^{2}2\alpha$$$$=4-3{sin}^{2}2\alpha -3{cos}^{2}2\alpha$$$$=4-3=1$$

Commented by weltr last updated on 03/Sep/20

$$thankyou$$

Answered by bobhans last updated on 03/Sep/20

$${\left(\sin {}^2\mathrm{x}\right)}^3+{\left(\cos {}^2\mathrm{x}\right)}^3=1.(\sin {}^4\mathrm{x}-{\left(\sin \mathrm{xcos}\mathrm{x}\right)}^2+\cos {}^4\mathrm{x})$$$$=1-2{\left(\sin \mathrm{xcos}\mathrm{x}\right)}^2-{\left(\sin \mathrm{xcos}\mathrm{x}\right)}^2$$$$=1-\frac{3}{4}{\left(2\sin \mathrm{xcos}\mathrm{x}\right)}^2=1-\frac{3}{4}{\sin }^{2}2\mathrm{x}$$$$(\ast )4\{1-\frac{3}{4}\sin {}^{2}2\mathrm{x}\}-3\cos {}^{2}3\mathrm{x}$$$$=4-3\sin {}^{2}2\mathrm{x}-3\cos {}^{2}2\mathrm{x}=4-3\left(1\right)=1$$

Commented by weltr last updated on 03/Sep/20

$$thankyou$$

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