Prove the given result involving the trigonometric functions: $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\dfrac{\sqrt{3}}{2}$.

Hint: We start solving the problem by considering the LHS (Left Hand Side) of the given result. We can then compare the given LHS (Left Hand Side) with $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$ and find the values of A and B. We then substitute these values in $\sin \left( A+B \right)$ and make necessary calculations to complete the required result.

Complete step by step answer:
According to the problem, we are asked to prove the given result $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Let us consider LHS (Left Hand Side) and prove that its value is equal to the given value of RHS (Right Hand Side).
So, let us consider $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}$.
We can see that $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}$ resembles the form $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$.
By comparing $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}$ with $\sin A\cos B+\cos A\sin B$, we get $A={{38}^{\circ }}$ and $B={{22}^{\circ }}$.
So, we get $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\sin \left( {{38}^{\circ }}+{{22}^{\circ }} \right)$.
$\Rightarrow \sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\sin \left( {{60}^{\circ }} \right)$.
We know that the value of $\sin \left( {{60}^{\circ }} \right)$ is equal to $\dfrac{\sqrt{3}}{2}$.
$\Rightarrow \sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
We can see that the value of LHS (left Hand side) is equal to the given value of RHS (Right Hand side).
∴ We have proved the given result $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\dfrac{\sqrt{3}}{2}$.

Note: Whenever we get this type of problems, we check whether the given equation fits with any of the trigonometric identities to complete the given proof. We can also prove the given result as shown below:
We have given $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}$.
Let us multiply and divide with 2.
So, we get $\sin {{38}^{\circ }}\cos {{22}^{\circ }}+\cos {{38}^{\circ }}\sin {{22}^{\circ }}=\dfrac{1}{2}\times \left( 2\sin {{38}^{\circ }}\cos {{22}^{\circ }}+2\cos {{38}^{\circ }}\sin {{22}^{\circ }} \right)$.