Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=3{\left(\sin x-\cos x\right)}^4+6{\left(\sin x+\cos x\right)}^2+4\left({\sin }^{6}x+{\cos }^{6}x\right) \\ =3{\left\{{\left(\sin x-\cos x\right)}^2\right\}}^2+6\left({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)+4\left\{{\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3\right\} \end{gathered}$$

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

$$\begin{gathered} =3{\left(1-\sin 2x\right)}^2+6+6s\mathrm{in}2x+4\left\{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\right\}-12{\sin }^{2}x{\cos }^{2}x \\ =3(1-\sin 2x{)}^2+6+6\sin 2x+4\left\{1-2{\sin }^{2}x{\cos }^{2}x\right\}-12{\sin }^{2}x{\cos }^{2}x \\ =3\left(\mathit{1}\mathit{-}2\sin 2x+{\sin }^{2}2x\right)+6+6\sin 2x+4-20{\sin }^{2}x{\cos }^{2}x \\ =-6\sin 2x+3{\sin }^{2}2x+13+6\sin 2x-5{\left(2\sin x\cos x\right)}^2 \\ =3{\sin }^{2}2x+13-5{\sin }^{2}2x \\ =13-2{\sin }^{2}2x=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$