Question

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

Answer 1

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.