How do you verify the identity \[\sin \left( {\dfrac{\pi }{6} + x} \right) = \dfrac{1}{2}\left( {\cos x + \sqrt 3 \sin x} \right)\] ?

Hint: In this question, the following identity is to be used for proving the given equation:
  \[\sin (a + b) = \sin a\cos b + \cos a\sin b\]
After applying this identity, we need to use the trigonometric values of the sine and cosine of \[\dfrac{\pi }{6}\] , and further simplify the obtained expression to verify this equation.

Complete step by step solution:
To verify/prove this identity we’re going to use a common identity, which is the sin sum identity. The sin sum identity is as follows:
  \[\sin (a + b) = \sin a\cos b + \cos a\sin b\]
At LHS, we apply this identity and obtain the following result.
  \[\sin \left( {\dfrac{\pi }{6} + x} \right) = \sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x\]
 This can be further simplified by putting the values of \[\sin \dfrac{\pi }{6}\] and \[\cos \dfrac{\pi }{6}\] in the equation.
We then get,
  \[\sin \left( {\dfrac{\pi }{6} + x} \right) = \dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x\]
Now, taking \[\dfrac{1}{2}\] common, and simplifying the expression, we finally obtain
  \[\sin \left( {\dfrac{\pi }{6} + x} \right) = \dfrac{1}{2}\left( {\cos x + \sqrt 3 \sin x} \right)\] , which is the final identity to be verified.

Note: The sine expansion is a set of two properties that can be used to find the sine of sums and differences of angles.
It has been proved that the sin of sum of two angles can be expanded as the sum of the products of sin and cos of angles. This is called the angle sum identity for sin function and it is used as a trigonometric formula to expand sin function which contains sum of two angles as angle.
There is another version of this identity, which is for the sin of the difference of two angles. In that identity, the sin of the difference of two angles can be expanded as the difference of the products of sin and cos of angles.