Question

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

Answer 1

Consider $4({\sin }^{6}x+{\cos }^{6}x)$

$=4[{\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3]$

$=4\left[({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x)\right]$

$=4[{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x{\cos }^{2}x]$

$=4[1-3{\sin }^{2}x{\cos }^{2}x]$

$=4-12{\sin }^{2}x{\cos }^{2}x$ ........$\left(1\right)$

$6{[\sin x+\cos x]}^2$

$=6[{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x]$

$=6[1+2\sin x\cos x]$

$=6+12\sin x\cos x$ ......$\left(2\right)$

$3{(\sin x-\cos x)}^4$

$=3{\left[{(\sin x-\cos x)}^2\right]}^2$

$=3{[{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x]}^2$

$=3{[1-2\sin x\cos x]}^2$

$=3[1-4\sin x\cos x+4{\sin }^{2}x{\cos }^{2}x]$

$=3-12\sin x\cos x+12{\sin }^{2}x{\cos }^{2}x$ .....$\left(3\right)$

Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$3{(\sin x-\cos x)}^4+4({\sin }^{6}x+{\cos }^{6}x)+6{[\sin x+\cos x]}^2$

$=3-12\sin x\cos x+12{\sin }^{2}x{\cos }^{2}x+4-12{\sin }^{2}x{\cos }^{2}x+6+12\sin x\cos x$

$=13$

Hence proved