Question

If $\sin \mathit{}x+\cos \mathit{}x=m$, then prove that ${\sin }^{6}x+{\cos }^{6}x=\frac{4-3{\left(m^2-1\right)}^2}{4}$, where $m^2\le 2$

Answer 1

$$\begin{gathered} \sin x+\cos x=m\left(\mathrm{Given}\right) \\ \mathrm{To}\mathrm{prove}:{\sin }^{6}x+{\cos }^{6}x=\frac{4-3(m^2-1{)}^2}{4},\mathrm{where}{m}^2\le 2 \\ \mathrm{Proof}: \\ \mathrm{LHS}: \\ {\sin }^{6}x+{\cos }^{6}x \\ ={\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3 \\ ={\left({\sin }^{2}x+{\cos }^{2}x\right)}^3-3{\sin }^{2}x{\cos }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right) \\ =1-3{\sin }^{2}x{\cos }^{2}x \\ \mathrm{RHS}: \\ \frac{4-3(m^2-1{)}^2}{4} \\ =\frac{4-3{\left[{\left(\sin x+\cos x\right)}^2-1\right]}^2}{4} \\ =\frac{4-3{\left[{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x-1\right]}^2}{4} \\ =\frac{4-3{\left[{\sin }^{2}x-\left(1-{\cos }^{2}x\right)+2\sin x\cos x\right]}^2}{4} \\ =\frac{4-3\times 4{\sin }^{2}x{\cos }^{2}x}{4} \\ =1-3{\sin }^{2}x{\cos }^{2}x \\ \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved} \end{gathered}$$