Consider the expression $A+B+C=\pi $ , then prove the expression $\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\operatorname{sinC}$

Hint: Look at the condition given in the question, try to convert the whole required expression using the condition sum of all angles is \[{{180}^{\circ }}\] .
By basic knowledge of trigonometry we can say the relation between sin and its angle as:
sin (360 - x) = sin (x) and $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
and $\sin \left( 2A \right)=2\sin A\cos A$ Use the above formulae given above to solve the question in a simpler manner.

Complete step by step answer:
 Given condition:
$A+B+C=\pi ..........(i)$
It is analogous to a triangle as the sum is ${{180}^{\circ }}$ .
By basic knowledge of properties of triangles, we know that: Sum of all interior angles must be equal to ${{180}^{\circ }}$ .
Now by subtracting C on both sides of equation we get:
$A+B=\pi -C$
Now by subtracting A on both sides of equation (i) we get:
$B+C=\pi -A$
Now by subtracting B on both sides of equation (i) we get:
$A+C=\pi -B$
The value $\pi $ in degrees is equal to 180 degrees. By basic operations like subtraction we derived 3 equations. Now we can use these to solve further and then make the required expression simpler and reach our result faster.
Required expression which is to be proved in the question is:
$\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\sin C$
Take the left-hand side separately to solve and simplify it
$\sin 2A+\sin 2B-\sin 2C$
By basic knowledge of trigonometry, we know that this is true:
$\sin A=\sin \left( 2\pi -A \right)$
By using this in our equation, we convert it into the form:
$\sin \left( 2\pi -2A \right)+\sin \left( 2\pi -2B \right)-\sin \left( 2\pi -2C \right)=\sin 2\left( \pi -A \right)+\sin 2\left( \pi -B \right)-\sin 2\left( \pi -C \right)$
By basic knowledge of trigonometry, we know that these are true:
$\sin 2A=2\sin A\cos A$
We also know that $\pi -A=B+C$ by the first equations conditions:
By substituting these into equation, we convert it into the form:
\[=2\sin \left( B+C \right)\cos A+2\sin \left( A+C \right)\cos B-2\sin \left( A+B \right)\cos C\]
By basic knowledge of trigonometry, we know that this is true.
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
By substituting this into the equation above, we convert it into:
$=2\left( \sin B\cos C+\cos B\sin C \right)\cos A+2\left( \sin A\cos C+\cos A\sin C \right)\cos B-2\left( \operatorname{sinA}\cos B+\cos A\sin B \right)\cos C$
By simplifying all the terms in the equation above, we convert it into:
$\begin{align}
  & 2\sin B\cos C\cos A+2\cos A\cos B\sin C+2\sin A\cos C\cos B+2\cos A\cos B\sin C \\
 & \text{ }-2\sin A\cos B\cos C-2\cos A\sin B\cos C \\
\end{align}$
By cancelling common terms in the equation above we get:
$\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\operatorname{sinC}$
Hence, we proved the required equation asked in the question.
Note: Be careful while converting $\sin A=\sin \left( 360-A \right)$ as you do it before 2A don’t forget to write $2\pi $ instead of $4\pi $ as you are writing sin (2A) = sin (360 – 2A) you are not replacing A with 360 – A you are replacing whole angle with 360 – angle. See the difference carefully.
Be careful while cancelling terms in the last step as you see all terms must cancel except one term which is added if you miss one sign in this step then you will lead to a wrong result.