Question

The value of $3 \sin \left(\frac{1}{2} arc \,\cos \frac{1}{9}\right)+4 \cos \left(\frac{1}{2} arc\,\cos \frac{1}{8}\right)$ is equal to

Answer 1

Let $E=3 \sin \left(\frac{1}{2} \cos ^{-1} \frac{1}{9}\right)+4 \cos \left(\frac{1}{2} \cos ^{-1} \frac{1}{8}\right)$
Put $\cos ^{-1} \frac{1}{9}=\theta$ and $\cos ^{-1} \frac{1}{8}=\phi$,
where $\theta, \phi \in\left(0, \frac{\pi}{2}\right)$
$\therefore E=3 \sin \frac{\theta}{2}+4 \cos \frac{\phi}{2}$ .... (1)
Now, $\cos \theta=\frac{1}{9}=1-2 \sin ^{2} \frac{\theta}{2}$
$ \Rightarrow \sin \frac{\theta}{2}=\frac{2}{3}$ ....(2)
and $\cos \phi=\frac{1}{8}=2 \cos ^{2} \frac{\phi}{2}-1$
$ \Rightarrow \cos \frac{\phi}{2}=\frac{3}{4}$ ....(3)
$\therefore $ On using equation (2) and equation (3) in (1), we get
$E=3\left(\frac{2}{3}\right)+4\left(\frac{3}{4}\right)=2+3=5$.