Question

Show that $3{(\sin x-\cos x)}^4+6{(\sin x+\cos x)}^2+4({\sin }^{6}x+{\cos }^{6}x)=13$.

Answer 1

$3{(\sin x-\cos x)}^4+6{(\sin x+\cos x)}^2+4({\sin }^{6}x+{\cos }^{6}x)=3{\left({(\sin x-\cos x)}^2\right)}^2+6{(\sin x+\cos x)}^2+4({\sin }^{6}x+{\cos }^{6}x)=3{({\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x)}^2+6({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x)+4({\sin }^{6}x+{\cos }^{6}x)=3(1+4{\sin }^{2}x{\cos }^{2}x-4\sin x\cos x)+6(1+2\sin x\cos x)+4(1-3{\sin }^{2}x{\cos }^{2}x)=3+12{\sin }^2{\cos }^{2}x-12\sin x\cos x+6+12\sin x\cos x+4-12{\sin }^{2}x{\cos }^{2}x=13$