Answer
The required value is $\frac{4\sqrt{3}+3}{10}$
Work Step by Step
Step 1: As $\sin \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Hypotenuse}\left( \text{H} \right)}\,\,$ $\cos \theta =\frac{\text{Base}\left( \text{B} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\tan \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Base}\left( \text{B} \right)}$
Let us suppose, ${{\cos }^{-1}}\frac{1}{2}=\alpha $
$\begin{align}
& \text{This}\,\text{implies,}\,\frac{1}{2}=\cos \alpha \\
& \cos \alpha =\frac{1}{2}=\frac{\text{B}}{\text{H}} \\
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{2}^{2}}={{\text{P}}^{2}}+{{1}^{2}} \\
& {{\text{P}}^{2}}=4-1=3 \\
& \text{P}=\sqrt{3} \\
\end{align}$
$\text{So,}\,\sin \alpha =\frac{\text{P}}{\text{H}}=\frac{\sqrt{3}}{2}$
Step 2: Similarly, ${{\sin }^{-1}}\frac{3}{5}=\beta $
$\begin{align}
& \frac{3}{5}=\sin \beta \\
& \sin \beta =\frac{3}{5}=\frac{\text{P}}{\text{H}} \\
\end{align}$
By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$
$\begin{align}
& {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\
& {{5}^{2}}={{\text{3}}^{2}}+{{\text{B}}^{2}} \\
& {{\text{B}}^{2}}=25-9=16 \\
& \text{B}=\sqrt{16}=4 \\
\end{align}$
$\text{So,}\,\cos \beta =\frac{\text{B}}{\text{H}}=\frac{4}{5}$
Step 3: Using steps 1 and 2, $\sin \left( {{\cos }^{-1}}\frac{1}{2}+{{\sin }^{-1}}\frac{3}{5} \right)$ can be written as:
$\sin \left( {{\cos }^{-1}}\frac{1}{2}+{{\sin }^{-1}}\frac{3}{5} \right)=\sin \left( \alpha +\beta \right)$
By using the trigonometric identity,
$\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
From above steps, $\sin \alpha =\frac{\sqrt{3}}{2}$ $\cos \alpha =\frac{1}{2}$ $\cos \beta =\frac{4}{5}$ $\sin \beta =\frac{3}{5}$
$\begin{align}
& \,\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\
& =\left( \frac{\sqrt{3}}{2}\times \frac{4}{5} \right)+\left( \frac{1}{2}\times \frac{3}{5} \right) \\
& =\frac{4\sqrt{3}}{10}+\frac{3}{10} \\
& =\frac{4\sqrt{3}+3}{10}
\end{align}$
