Browse Questions

What is the value of the function: $\large \frac{1}{4} $$\sin^{-1}\large\frac{2\sqrt 2}{3}$$+\sin^{-1}\large\frac{1}{3}$

(A) $\large \frac{\pi}{16}$ (B) $\large \frac{-\pi}{8}$ (C) $\large \frac{\pi}{8}$ (D) $\large \frac{\pi}{2}$

Toolbox:
• $sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$
• $sin^{-1}1=\frac{\pi}{2}$
Given $\frac{1}{4} \bigg[ sin^{-1}\frac{2\sqrt 2}{3}+sin^{-1}\frac{1}{3} \bigg]$
We know that $sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$
By taking $x=\frac{2\sqrt 2}{3}\:and\:y=\frac{1}{3}$
$\sqrt{1-y^2}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$
$\sqrt{1-x^2}=\sqrt{1-\frac{8}{9}}=\frac{1}{3}$
$\Rightarrow\:$ $\frac{1}{4} \bigg[ sin^{-1} \bigg( \frac{2\sqrt 2}{3} \sqrt{1-\frac{1}{9}} +\frac{1}{3} \sqrt{1-\frac{8}{9}} \bigg) \bigg]$
$=\frac{1}{4}sin^{-1}\bigg(\frac{2\sqrt{2}}{3}.\frac{2\sqrt{2}}{3}+\frac{1}{3}.\frac{1}{3}\bigg)$
$= \frac{1}{4}sin^{-1} \bigg(\frac{8}{9}+\frac{1}{9}\bigg)$
$\frac{1}{4}sin^{-1}1=\frac{1}{4}.\frac{\pi}{2}=\frac{\pi}{8}$