Question

If $\sin \mathrm{\theta }+\cos \mathrm{\theta }=m$, then prove that

${\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta }=\frac{4-3{\left(m^2-1\right)}^2}{4}$, where $m^2\le 2$

Answer 1

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