Question

If $si{n}^{-1}\frac{1}{3}+si{n}^{-1}\frac{2}{3}=si{n}^{-1}x,$ 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
$\Rightarrow \left(\frac{1}{3}\sqrt{1-\frac{4}{9}}+\frac{2}{3}\sqrt{1-\frac{1}{9}}\right)=x$
$\Rightarrow \left(\frac{1}{3}\cdot \frac{\sqrt{5}}{3}+\frac{2}{3}\cdot \frac{\sqrt{8}}{3}\right)=x$
$\Rightarrow \left(\frac{\sqrt{5}+4\sqrt{2}}{9}\right)=x$
$\therefore x=\left(\frac{\sqrt{5}+4\sqrt{2}}{9}\right)$