Question

Sin inverse 1 by 3 + Sin inverse 2 by 3 is equal to Sin inverse X find the value of x

Answer 1

Dear Student,

$$\begin{gathered} {\sin }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{2}{3}\right)={\sin }^{-1}\mathrm{x} \\ \Rightarrow {\sin }^{-1}\left[\frac{1}{3}\sqrt{1-{\left(\frac{2}{3}\right)}^2}+\frac{2}{3}\sqrt{1-{\left(\frac{1}{3}\right)}^2}\right]={\sin }^{-1}\mathrm{x} \\ \Rightarrow {\sin }^{-1}\left[\frac{1}{3}\sqrt{1-\frac{4}{9}}+\frac{2}{3}\sqrt{1-\frac{1}{9}}\right]={\sin }^{-1}\mathrm{x} \\ \Rightarrow {\sin }^{-1}\left[\frac{1}{3}\times \frac{\sqrt{5}}{3}+\frac{2}{3}\times \frac{2\sqrt{2}}{3}\right]={\sin }^{-1}\mathrm{x} \\ \Rightarrow {\sin }^{-1}\left[\frac{\sqrt{5}+4\sqrt{2}}{9}\right]={\sin }^{-1}\mathrm{x} \\ \Rightarrow \mathrm{x}=\frac{\sqrt{5}+4\sqrt{2}}{9} \\ \boxed{ \\ \mathbf{Using}\mathbf{}\mathbf{:}\mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]} \end{gathered}$$
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