Question

Prove that 3(sin theta -cos theta)^4+6(sin theta+cos theta)^2 +4(sin^6theta+cos^6theta) is independent of theta.

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ 3{\left(\mathrm{sinx}-\mathrm{cosx}\right)}^4+6{\left(\mathrm{sinx}+\mathrm{cosx}\right)}^2+4\left({\left({\sin }^2\mathrm{x}\right)}^3+{\left({\cos }^2\mathrm{x}\right)}^3\right) \\ =3{\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}-2\mathrm{sinxcosx}\right)}^2+6\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}+2\mathrm{sinxcosx}\right)+4\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)\left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}-{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\right) \\ =3{\left(1-\sin 2\mathrm{x}\right)}^2+6+6\sin 2\mathrm{x}+4\left[{\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)}^2-3{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\right] \\ =3+3{\sin }^{2}2\mathrm{x}-6\sin 2\mathrm{x}+6\sin 2\mathrm{x}+6+4-12{\sin }^2{\mathrm{xcos}}^2\mathrm{x} \\ =13+3{\sin }^{2}2\mathrm{x}-3\left(4{\sin }^2{\mathrm{xcos}}^2\mathrm{x}\right) \\ =13+3{\sin }^{2}2\mathrm{x}-3\left({\sin }^{2}2\mathrm{x}\right) \\ =13 \\ \mathrm{So}\mathrm{it}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{angle}\mathrm{x} \\ \mathrm{Regards} \end{gathered}$$