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Question Number 73665 by aliesam last updated on 14/Nov/19

$$if$$$$$$$${cos}^2\left(\theta \right)=\frac{m^2-1}{3},{tan}^3\left(\frac{\theta }{2}\right)=tan\left(a\right)$$$$$$$$provethat$$$$$$$$\sqrt[3]{{cos}^2\left(a\right)}+\sqrt[3]{{sin}^2\left(a\right)}=\sqrt[3]{{\left(\frac{2}{m}\right)}^2}$$

Answered by Tanmay chaudhury last updated on 14/Nov/19

$$cos\theta =\sqrt{\frac{m^2-1}{3}}\to \frac{1-{tan}^2\frac{\theta }{2}}{1+{tan}^2\frac{\theta }{2}}=\sqrt{\frac{m^2-1}{3}}=\frac{a}{1}$$$$\frac{1+{\mathit{t}\mathit{a}\mathit{n}}^2\frac{\theta }{2}}{1-{tan}^2\frac{\theta }{2}}=\frac{1}{a}\to \frac{2}{2{tan}^2\frac{\theta }{2}}=\frac{1+a}{1-a}\to {tan}^2\frac{\theta }{2}=\frac{1-a}{1+a}$$$$tan\frac{\theta }{2}=\sqrt{\frac{1-a}{1+a}[so}tan\alpha ={tan}^3\left(\frac{\theta }{2}\right)={\left(\frac{1-a}{1+a}\right)}^{\frac{3}{2}}$$$${cos}^2\alpha =\frac{1}{1+{tan}^2\alpha }=\frac{1}{1+{\left(\frac{1-a}{1+a}\right)}^3}=\frac{{(1+a)}^3}{1+3a+3a^2+a^3+1-3a+3a^2-a^3}$$$${cos}^2\alpha =\frac{{(1+a)}^3}{2+6a^2}\to {\mathit{s}\mathit{i}\mathit{n}}^2\alpha =\frac{2+6a^2-1-3a-3a^2-a^3}{2+6a^2}$$$${sin}^2\alpha =\frac{1-3a+3a^2-a^3}{2+6a^2}=\frac{{(1-a)}^3}{2+6a^2}$$$${\left({cos}^2\alpha \right)}^{\frac{1}{3}}+{\left({sin}^2\alpha \right)}^{\frac{1}{3}}$$$$=\frac{1+a+1-a}{{(2+6a^2)}^{\frac{1}{3}}}=\frac{2}{{\{2+6\times \frac{m^2-1}{3})}^{\frac{1}{3}}}=\frac{2}{{(2+2m^2-2)}^{\frac{1}{3}}}$$$$=\frac{2^{1-\frac{1}{3}}}{{\mathit{m}}^{\frac{2}{3}}}={\left(\frac{2}{\mathit{m}}\right)}^{\frac{2}{3}}\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}$$

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