Question

The value of 3 (sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶x + cos⁶x) is ___________.

Answer 1

3 (sin x – cos x)⁴ + 6 (sin x + cos x)² + 4 (sin⁶x + cos⁶x)
$$\begin{gathered} =3{\left[{\left(\sin x-\cos x\right)}^2\right]}^2+6\left({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)+4\left[{\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3\right] \\ =3{\left[{\sin }^{2}x+{\cos }^{2}x-2\cos x\sin x\right]}^2+6\left(1+2\sin x\cos x\right)+4\left[{\left({\sin }^{2}x+{\cos }_1^{2}x\right)}^3-3{\sin }^{2}x{\cos }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right)\right]=1 \\ =3{\left[1-2\cos x\sin x\right]}^2+6+12\sin x\cos x+4\left[1-3{\sin }^{2}x{\cos }^{2}x\right] \end{gathered}$$
$$\begin{gathered} =3\left[1+4{\cos }^{2}x{\sin }^{2}x-4\cos x\sin x\right] \\ +6+12\sin x\cos x+4-12{\sin }^{2}x{\cos }^{2}x \\ =3+12{\cos }^{2}x{\sin }^{2}x-12\cos x\sin x \\ +6+12\cos x\sin x+4-12{\sin }^{2}x{\cos }^{2}x \\ =13 \\ \mathrm{Hence},3{\left(\sin x-\cos x\right)}^4+6{\left(\sin x+\cos x\right)}^2+4\left({\sin }^{6}x+{\cos }^{6}x\right)\mathrm{has}\mathrm{value}13. \end{gathered}$$