If \[A,\,B\] and $C$ are interior angles of a triangle $ABC$ ,then show that $\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{A}{2}$

Hint:To solve this type of question we can use the property which is angle sum property of a triangle which gives sum of all sides of a triangle gives $180$ degrees. Then by doing proper rearrangement and using trigonometric identity we can prove the given question.

Complete step-by-step answer:
Given \[A,\,B\] and $C$ are interior angles of a triangle $ABC$ ,
Consider a triangle $ABC$ ,
Triangle is the smallest polygon which has three sides and three interior angles.

By using the interior angle sum property of a triangle,
Interior angle sum property of a triangle states that the sum of interior angles gives $180$ degrees.
So we get,
Sum of interior angles of a triangle $ = 180^\circ $
$
  A + B + C = 180^\circ \\
    \\
 $
By rearranging the above result we get,
$B + C = 180^\circ - A$
By multiplying on both left hand side and right hand side with $\dfrac{1}{2}$ , the above result changes to,
$\dfrac{{B + C}}{2} = \dfrac{{180^\circ - A}}{2}$
Separating the terms we get,
$\dfrac{B}{2} + \dfrac{C}{2} = \dfrac{{180^\circ }}{2} - \dfrac{A}{2}$
Perform the arithmetic operation on above result,
$\dfrac{{B + C}}{2} = 90^\circ - \dfrac{A}{2}$
Taking the left hand side from the expression,
L.H.S $ = \sin \left( {\dfrac{{B + C}}{2}} \right)$
Substitute the value of $\left( {\dfrac{{B + C}}{2}} \right)$ with $90^\circ - \dfrac{A}{2}$
Hence,
$x = \sin \left( {90^\circ - \dfrac{A}{2}} \right)$
We can perform the trigonometric identities in the above result to further step,
We know that,
$
  \cos (90 - \theta ) = \sin \theta \\
  \sin (90 - \theta ) = \cos \theta \\
 $
Applying the above result in our expansion,
$x = \sin \left( {90^\circ - \dfrac{A}{2}} \right)$
To get,
$x = \cos \dfrac{A}{2}$
$ = $ R.H.S
Which gives the right hand side,
We get the left hand side and right hand side as equal.
$\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{A}{2}$
Hence proved.

Note:We use the property of interior angle sum property of a triangle to prove the above question which gives sum of interior angles of triangle is $180$ degrees and also we have to perform some of the trigonometric identities. Students should remember the important trigonometric formulas, identities and standard trigonometric angles for solving these types of problems.