Question

If ${\cos }^2\theta =\frac{1}{3}\left(a^2-1\right)$ and ${\tan }^2=\frac{\theta }{2}={\tan }^{2/3}\alpha$, then ${\cos }^{2/3}\alpha +{\sin }^{2/3}\alpha =$

• A. $2a^{2/3}$
• B. ${\left(\frac{2}{a}\right)}^{2/3}$
• C. ${\left(\frac{2}{a}\right)}^{1/3}$
• D. $2a^{1/3}$

Answer 1

Answer: (B)

${\cos }^2\theta =\frac{a^2-1}{3},{\tan }^2\frac{\theta }{2}={\tan }^{2/3}\alpha$
${\tan }^3\frac{\theta }{2}=\tan \alpha \Rightarrow \frac{{\sin }^3\frac{\theta }{2}}{{\cos }^3\frac{\theta }{2}}=\frac{\sin \alpha }{\cos \alpha }$
$=\frac{{\sin }^3\frac{\theta }{2}}{\sin \alpha }=\frac{{\cos }^2\frac{\theta }{2}}{\cos \alpha }=k$
${\sin }^3\frac{\theta }{2}=k\sin \alpha ...$(i)
${\cos }^3\frac{\theta }{2}=k\cos \alpha ...$(ii)
$k^{2/3}{\sin }^{2/3}\alpha +k^{2/3}\cos \alpha =1$
${\sin }^{2/3}\alpha +{\cos }^{2/3}\alpha =k^{2/3}$
Squaring and adding (i) and (ii)
$k^2\left({\sin }^2\alpha +{\cos }^2\alpha \right)={\sin }^6\frac{\theta }{2}+{\cos }^6\frac{\theta }{2}$
$k^2=1-\frac{3}{4}{\sin }^2\theta =1-\frac{3}{4}\left(1-{\cos }^2\theta \right)$
$k^2=\frac{1}{4}+\frac{3}{4}\frac{a^2-1}{3}\Rightarrow k^2=\frac{a^2}{4}$
${\sin }^{2/3}\alpha +{\cos }^{2/3}\alpha ={\left(\frac{2}{a}\right)}^{2/3}$
and $\cos \sqrt{2}-\cos \sqrt{3}$
$=2\sin \left(\frac{\sqrt{3}+\sqrt{2}}{2}\right)\times \sin \left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)>0$