Question

3(sin^4theta + cos ^4 theta)=2(sin^6 theta+cos^6theta)+1

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ \mathrm{Prove}:3({\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta })=2({\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta })+1 \\ \mathrm{Now}, \\ \mathbf{LHS}\mathbf{:}\mathbf{} \\ \mathbf{3}\mathbf{(}{\mathbf{sin}}^{\mathbf{4}}\mathbf{\theta }\mathbf{+}{\mathbf{cos}}^{\mathbf{4}}\mathbf{\theta }\mathbf{)} \\ =3\left[\right({\sin }^2\mathrm{\theta }{)}^2+\left({\cos }^2\mathrm{\theta }{)}^2\right] \\ =3\left[\right({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }{)}^2-2{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }]\{\mathrm{Because},{\mathrm{a}}^2+{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b}{)}^2-2\mathrm{ab}\} \\ =3[1-2{\sin }^2\mathrm{\theta }(1-{\sin }^2\mathrm{\theta }\left)\right] \\ \mathbf{=}\mathbf{3}\mathbf{-}\mathbf{6}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }\mathbf{+}\mathbf{6}{\mathbf{sin}}^{\mathbf{4}}\mathbf{\theta }\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{i}\mathbf{\right)} \\ \mathrm{Now}, \\ \mathbf{RHS}\mathbf{:} \\ \mathbf{2}\mathbf{(}{\mathbf{sin}}^{\mathbf{6}}\mathbf{\theta }\mathbf{+}{\mathbf{cos}}^{\mathbf{6}}\mathbf{\theta }\mathbf{)}\mathbf{+}\mathbf{1} \\ =2\left[\right({\sin }^2\mathrm{\theta }{)}^3+\left({\cos }^2\mathrm{\theta }{)}^3\right]+1 \\ =2\left[\right({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }{)}^3-3{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta })]+1\{\mathrm{Because},{\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b}{)}^3-3\mathrm{ab}(\mathrm{a}+\mathrm{b}\left)\right\} \\ =2[1-3{\sin }^2\mathrm{\theta }(1-{\sin }^2\mathrm{\theta }\left)\right]+1 \\ =2-6{\sin }^2\mathrm{\theta }+6{\sin }^4\mathrm{\theta }+1 \\ \mathbf{=}\mathbf{3}\mathbf{-}\mathbf{6}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }\mathbf{+}\mathbf{6}{\mathbf{sin}}^{\mathbf{4}}\mathbf{\theta }\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{ii}\mathbf{\right)} \\ \mathbf{We}\mathbf{}\mathbf{see}\mathbf{}\mathbf{that}\mathbf{;}\mathbf{}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{;} \\ \mathbf{\Rightarrow }\mathbf{LHS}\mathbf{=}\mathbf{RHS}\mathbf{;}\mathbf{}\mathbf{Hence}\mathbf{}\mathbf{Proved} \\ \mathrm{Regards}. \end{gathered}$$