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Question Number 116006 by DELETED last updated on 30/Sep/20

3(sin x−cos x)^4 +6(sin x+cos x)^2 +4(sin^6 x+cos^6 x) = _____.

$$\mathrm{3}\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} {x}+\mathrm{cos}^{\mathrm{6}} {x}\right)\:=\:\_\_\_\_\_. \\ $$

Answered by mindispower last updated on 30/Sep/20

(sin^2 (x)+cos^2 (x))^3 =1=sin^6 (x)+cos^6 (x)+3sin^2 (x)cos^2 (x) (sin(x)+cos(x))^2 =1+2sin(x)cos(x) (sin(x)−cos(x))^4 =(1−2sin(x)cos(x))^2 =1−4sin(x)cos(x)+4sin^2 (x)cos^2 (x) =3(1−4sin(x)cos(x)+4sin^2 (x)cos^2 (x)) +6(1+2sin(x)cos(x))+4(1−3sin^2 (x)cos^2 (x)) =13

$$\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{3}} \\ $$$$=\mathrm{1}={sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)+\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right) \\ $$$$\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right) \\ $$$$\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)^{\mathrm{4}} =\left(\mathrm{1}−\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\mathrm{4}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right) \\ $$$$=\mathrm{3}\left(\mathrm{1}−\mathrm{4}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$+\mathrm{6}\left(\mathrm{1}+\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\right)+\mathrm{4}\left(\mathrm{1}−\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$=\mathrm{13} \\ $$

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