ScommaMaruj

2022-07-06

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

iskakanjulc

Expert

$\phantom{\rule{0ex}{0ex}}3{\left(\mathrm{sin}\left(x\right)-\mathrm{cos}\right)}^{4}+4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)+6{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}=13\phantom{\rule{0ex}{0ex}}3{\left({\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}\right)}^{2}+4\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x\right)+6\left({\mathrm{sin}}^{2}x+2\mathrm{sin}x\mathrm{cos}x+{\mathrm{cos}}^{2}x\right)=13\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3{\left({\mathrm{sin}}^{2}x-2\mathrm{sin}x\mathrm{cos}x+{\mathrm{cos}}^{2}x\right)}^{2}+4\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x\right)+6\left(1+\mathrm{sin}2x\right)=13\phantom{\rule{0ex}{0ex}}3{\left(1-\mathrm{sin}2x\right)}^{2}+4\left({\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)}^{2}-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)+6\left(1+\mathrm{sin}2x\right)=13\phantom{\rule{0ex}{0ex}}3{\left(1-\mathrm{sin}2x\right)}^{2}+4\left(1-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)+6\left(1+\mathrm{sin}2x\right)=13\phantom{\rule{0ex}{0ex}}3{\left(1-\mathrm{sin}2x\right)}^{2}+4\left(1-\frac{3}{4}{\mathrm{sin}}^{2}2x\right)+6\left(1+\mathrm{sin}2x\right)=13\phantom{\rule{0ex}{0ex}}3-6\mathrm{sin}2x+3{\mathrm{sin}}^{2}2x+4-3{\mathrm{sin}}^{2}x+6+6\mathrm{sin}2x=13$
So :
13=13

Do you have a similar question?