Question Number 116006 by DELETED last updated on 30/Sep/20
$$\mathrm{3}\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} {x}+\mathrm{cos}^{\mathrm{6}} {x}\right)\:=\:\_\_\_\_\_. \\ $$
Answered by mindispower last updated on 30/Sep/20
$$\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{3}} \\ $$$$=\mathrm{1}={sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)+\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right) \\ $$$$\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right) \\ $$$$\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)^{\mathrm{4}} =\left(\mathrm{1}−\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\mathrm{4}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right) \\ $$$$=\mathrm{3}\left(\mathrm{1}−\mathrm{4}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$+\mathrm{6}\left(\mathrm{1}+\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\right)+\mathrm{4}\left(\mathrm{1}−\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\right) \\ $$$$=\mathrm{13} \\ $$