Question

Prove the following identities (1-17)

${\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta }=1-3{\sin }^2\mathrm{\theta }{\cos }^2\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^6\theta +{\cos }^6\theta \\ ={\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3 \\ =\left({\sin }^2\theta +{\cos }^2\theta \right)\left[{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2-{\sin }^2\theta {\cos }^2\theta \right]\left[\because {\mathrm{a}}^3+{\mathrm{b}}^3=\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}\right)\right] \\ =1\times \left[{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}^2-2{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \right]\left[\because {\sin }^{\mathit{2}}\theta \mathit{+}{\cos }^{\mathit{2}}\theta \mathit{}=\mathit{}1\mathrm{and}{\mathrm{a}}^2+{\mathrm{b}}^2={\left(\mathrm{a}+\mathrm{b}\right)}^2-2\mathrm{ab}\right] \\ =1^2-3{\sin }^2\theta {\cos }^2\theta \\ =1-3{\sin }^2\theta {\cos }^2\theta \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$