How do you find the exact value of \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{7\pi }}{6}} \right)} \right)\]?

Hint: Here the question is related to inverse trigonometry. Since the function contains both trigonometric and inverse trigonometric functions it doesn't cancel, because the value of an angle doesn’t lie in the range of the sine trigonometric ratio. First we write the given angle in the form of sum of allied angles and then by ASTC(all, sin, tan, cos or all, silver, tea, cups) rule we simplify the function and hence we can obtain the solution.

Complete step by step answer:
In trigonometry we have six trigonometry ratios namely, sine, cosine, tangent, cosecant, secant and cotangent. For this trigonometric ratio we have inverse trigonometric ratios also.
To solve these kinds of problems we have to know the principal values, range and domain of the inverse trigonometric ratios.
In inverse trigonometry there is a rule for the sine trigonometric ratio, \[{\sin ^{ - 1}}(\sin (\theta )) = \theta \] where \[\dfrac{{ - \pi }}{2} \leqslant \theta \leqslant \dfrac{\pi }{2}\]
Here the value of \[\theta = \dfrac{{7\pi }}{6}\], this value doesn’t lies between \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\]. So we write the value of \[\theta \] as \[\theta = \pi + \dfrac{\pi }{6}\]
Therefore the given term is written as \[{\sin ^{ - 1}}\left( {\sin \left( {\pi + \dfrac{\pi }{6}} \right)} \right)\]
The \[\pi \] value is \[{180^ \circ }\], the \[\pi + \theta \] lies in the third quadrant. The sine trigonometry ratio is negative in the third quadrant.
\[ \Rightarrow {\sin ^{ - 1}}\left( { - \sin \left( {\dfrac{\pi }{6}} \right)} \right)\]
Take out the negative sign out of the bracket, so we have
\[ \Rightarrow - {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{6}} \right)} \right)\]
Here the value of \[\theta = \dfrac{\pi }{6}\], this value lies between \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\].
\[ \Rightarrow - \dfrac{\pi }{6}\]
Therefore, the exact value of \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{7\pi }}{6}} \right)} \right)\] is $- \dfrac{\pi }{6}$.

Note:
Inverse Trigonometric Functions in Maths is simply defined as the inverse of some basic trigonometric functions such as sine, cosine, tan, sec, cosec and cot. The other names of Inverse trigonometric functions are arcus functions, anti-trigonometric functions or the cyclometric functions. The ASTC rule will say the sign of the trigonometric ratios in each quadrant.