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**Hint:**We know that ${\sin ^{ - 1}}x = \arcsin \left( x \right)$, such that here we have to evaluate in short $\cos \left( {{{\sin }^{ - 1}}\left( { - \dfrac{2}{3}} \right)} \right)$.

We also know the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ which can be used to evaluate the given question. So by using the above identities and information can evaluate the given question.

**Complete step by step solution:**

Given

$\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right)......................................\left( i \right)$

Now we know to evaluate the value of$\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right)$. For that purpose we can use various basic trigonometric identities such as ${\sin ^2}\theta + {\cos^2}\theta = 1$.

Such that let:

$

p = \arcsin \left( { - \dfrac{2}{3}} \right) \\

\Rightarrow \sin p = \left( { - \dfrac{2}{3}} \right)........................\left( {ii} \right) \\

$

Since${\sin ^{ - 1}}x = \arcsin \left( x \right)$.

Now we have the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$, from which we can find $\cos p$.

So on substituting the values in the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we can write:

$

{\cos ^2}p = 1 - {\sin ^2}p \\

= 1 - {\left( { - \dfrac{2}{3}} \right)^2} \\

= 1 - \left( {\dfrac{4}{9}} \right) \\

= \dfrac{{9 - 4}}{9} \\

= \dfrac{5}{9}..................................\left( {iii} \right) \\

$

So we get: ${\cos ^2}p = \dfrac{5}{9}$

Now substituting back the value of $p$we can write:

$

{\cos ^2}\left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \dfrac{5}{9} \\

\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \pm \dfrac{{\sqrt 5

}}{3}.........................\left( {iv} \right) \\

$

Such that there are two possibilities for the value of $\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)}

\right)$.

Now we also know that if:

$

\arcsin \left( x \right) = \theta \\

\Rightarrow - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}.........................\left( v \right) \\

$

So from the condition specified in (iv) we can take the positive value.

**Therefore thee value of$\cos \left( {\arcsin \left( { - \dfrac{2}{3}} \right)} \right) = \dfrac{{\sqrt 5 }}{3}$.**

**Additional Information:**

Some other properties useful for solving trigonometric questions:

Quadrant I: $0\; - \;\dfrac{\pi }{2}$ All values are positive.

Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.

Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.

Quadrant IV: $\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.

**Note:**

One of the basic property of $\arcsin x$is the range of the angle, which is given by:

$

\arcsin \left( x \right) = \theta \\

\Rightarrow - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2} \\

$

So the values only in that range have to be taken under consideration.

Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side.

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