How do you verify the identity $\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y$ ?

Hint: We are asked to verify that $\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y$.
To solve this we will first learn how sin behaves when applied on the sum of two numbers and on the difference of two numbers.
We will use $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ and
We will use $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$
We start out by considering the left hand side part of the problem, we use the above formula and simplify and will record to the right side of the given equation.

Complete step by step answer:
We are given that we have to verify that $\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y$.
To do that solution we will need to know how one will be able to use the function sin on the sum of two numbers and also on the difference of two numbers.
We have to learn that what does $\sin \left( x+y \right)$ expand into and we will also know the knowledge that how does the $\sin \left( x-y \right)$ will expand itself into .
We know that –
$\sin \left( A+B \right)$ is given as $\sin A\cos B+\cos A+\sin B$
While $\sin \left( A-B \right)$ is given as $\sin A\cos B-\cos A\sin B$
We consider the left hand side.
Now we are asked to find the value of $\sin \left( x+y \right)+\sin \left( x-y \right)$
So,
Using above formula
Considering A as x and B as y, we get –
$\sin \left( x+y \right)+\sin \left( x-y \right)$ will become
$\sin \left( x+y \right)+\sin \left( x-y \right)=\sin x\cos y+\cos x\sin y+\sin x\cos y-\cos x\sin y$
We can see that –
$+\cos x\sin y-\cos x\sin y=0$ , so we get –
$\begin{align}
  & \sin \left( x+y \right)+\sin \left( x-y \right)=\sin x\cos y+\sin x\cos y+0 \\
 & =\sin x\cos y+\sin x\cos y \\
\end{align}$
Now, as $\sin x\cos y$ adding two times it become $2\sin x\cos y$ ,
So, we get –
$\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y$
We got the right hand side so we got our answer.
L.H.S=R.H.S
Hence proved.

Note: We can cross check this identity
$\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y$is valid
For any x and y
Let $x={{60}^{\circ }}$ and $y={{30}^{\circ }}$
So, $x+y={{60}^{\circ }}+{{30}^{\circ }}={{90}^{\circ }}$
And $x-y={{60}^{\circ }}-{{30}^{\circ }}={{30}^{\circ }}$
So,
First left hand side value
$\sin \left( x+y \right)+\sin \left( x-y \right)=\sin \left( {{90}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)$
As $\sin {{90}^{\circ }}=1$ and $\sin {{30}^{\circ }}=\dfrac{1}{2}$ So we get –
$\sin \left( x+y \right)+\sin \left( x-y \right)=1+\dfrac{1}{2}=\dfrac{3}{2}$
Now considering right hand side
Putting $x={{60}^{\circ }}$ and $y={{30}^{\circ }}$ , we get –
$2\sin x\cos y=2\times \sin {{60}^{\circ }}\times \cos {{30}^{\circ }}$
As $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$
So,
$2\sin x\cos y=2\times \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}$
By simplifying, we get –
$=\dfrac{2\times \sqrt{3}\times \sqrt{3}}{4}$
As $\sqrt{3}\times \sqrt{3}=3$ so we get –
$2\sin x\cos y=\dfrac{2\times 3}{4}=\dfrac{3}{2}$ ,
So, we get –
L.H.S=R.H.S
Hence, this identity stands true.