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

Answer: 5

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