How do you simplify \[\sin \left( {\arccos \left( x \right)} \right)\]?

Hint: In the given question, we have been given a trigonometric function. The argument of the given function is \[\arccos \]. For solving this question, we need to know what this argument is. It is the inverse of the cosine function. It means that the \[\arccos \] function returns the angle whose cosine value is given. Then, we are going to use this definition to define the range of the argument of the \[\arccos \] function. After defining the range, we are going to combine all the observations to find the value and to simplify the given expression.

Complete step-by-step answer:
From the square sum formula,
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
So, if
\[x \in \left[ { - 1,1} \right]\] (the range of sine and cosine functions), then
\[\arccos \left( x \right) \in \left[ {0,\pi } \right]\]
Thus, \[\sin \left( {\arccos \left( x \right)} \right) \ge 0\]
Hence, \[\sin \left( {\arccos \left( x \right)} \right) = \sin \left( \theta \right) = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - {x^2}} \]

Additional Information:
In the given question, we used the \[\sqrt {1 - {{\cos }^2}\theta } \] for solving the question in the last step after we also substituted the value of \[\cos \theta = x\] because we had shown that
\[\sin \left( {\arccos \left( x \right)} \right) > 0\]

Note: In the given question we had been asked to simplify \[\sin \left( {\arccos \left( x \right)} \right)\]. To solve that, we need to know the meaning of the \[\arccos \] function – it is the inverse of cosine function. Then, we defined the range of the arguments, combined them using the identity involving sine and cosine and then found their value. After doing that, we just put in the values and simplified them.