Question

If $sin^{-1} \frac{1}{3} + sin^{-1} \frac{2}{3} = sin^{-1} x , $ then the value of 𝑥

• A. 0
• B. $\frac {(\sqrt{5} - 4 \sqrt{2})}{9}$
• C. $\frac {(\sqrt{5} + 4 \sqrt{2})}{9}$
• D. $\frac {\pi}{2}$

Answer 1

Answer: (C)

Given that, $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{1} x$
Taking sine on both sides
$\Rightarrow \left(\frac{1}{3} \sqrt{1-\frac{4}{9}}+\frac{2}{3} \sqrt{1-\frac{1}{9}}\right)=x$
$\Rightarrow \left(\frac{1}{3} \cdot \frac{\sqrt{5}}{3}+\frac{2}{3} \cdot \frac{\sqrt{8}}{3}\right)=x$
$\Rightarrow \left(\frac{\sqrt{5}+4 \sqrt{2}}{9}\right)=x$
$\therefore x=\left(\frac{\sqrt{5}+4 \sqrt{2}}{9}\right)$