1. Tardigrade
  2. Question
  3. Mathematics
  4. If cos 2 θ=(1/3)( a 2-1) and tan 2=(θ/2)= tan 2 / 3 α, then cos 2 / 3 α+ sin 2 / 3 α=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Trigonometric Functions

Solution:

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