Question

If a cos³θ + 3a sin²θ cos θ = m and a sin³θ + 3a sin θ cos²θ = n, prove that
(m + n)^2/3 + (m − n)^2/3 = 2a^2/3.

Answer 1

$$\begin{gathered} \text{We have}a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta =m \\ \text{and}a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta =n \\ \mathrm{Now},(m+n)=a{\cos }^3\theta +3a{\sin }^2\theta \cos \theta +a{\sin }^3\theta +3a\sin \theta {\cos }^2\theta \\ =a[({\cos }^3\theta +{\sin }^3\theta )+3\sin \theta \cos \theta (\sin \theta +\cos \theta )] \\ \Rightarrow \left(\frac{m+n}{a}\right)=[({\cos }^3\theta +{\sin }^3\theta )+3\sin \theta \cos \theta (\sin \theta +\cos \theta )] \\ \Rightarrow m+n={a(\cos \theta +\sin \theta )}^3 \\ \therefore {(m+n)}^{\frac{2}{3}}={(\cos \theta +\sin \theta )}^2{a}^{\frac{2}{3}}...\left(\text{i}\right) \\ \text{Similarly,}{(m-n)}^{\frac{2}{3}}={(\cos \theta -\sin \theta )}^2{a}^{\frac{2}{3}}...\left(\text{ii}\right) \\ \mathrm{Adding}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ {(m+n)}^{\frac{2}{3}}+{(m-n)}^{\frac{2}{3}}=a^{\frac{2}{3}}[{\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta ] \\ =a^{\frac{2}{3}}\left[2\right] \\ =2a^{\frac{2}{3}} \end{gathered}$$