Question

How do you find the exact value of $\sin \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right)$ ?

Answer 1

Hint : We can solve this by substituting $\theta = \arccos \left( { - \dfrac{2}{3}} \right)$ . Here arc means inverse function. We know that the Pythagoras relation between sine and cosine is ${\sin ^2}\theta + {\cos ^2}\theta = 1$ . Using this we can find sine value. After substituting the theta value in that we will get the required answer. Also remember that if we have $\cos (\arccos (\theta ))$ the arc cosine and cosine will get cancelled and give only $\theta$ .

Complete step-by-step answer :
Given, $\sin \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right)$ .
Now put $\theta = \arccos \left( { - \dfrac{2}{3}} \right)$ in the given problem. We have,
$\sin \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right) = \sin (\theta )$
Now from Pythagoras relation we have
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Rearranging we have
${\sin ^2}\theta = 1 - {\cos ^2}\theta$
Taking square root on both side we have,
$\sin \theta = \sqrt {1 - {{\cos }^2}\theta }$
Now substitute the value of theta,
$\Rightarrow \sin \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right) = \sqrt {1 - {{\cos }^2}\left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right)}$
$= \sqrt {1 - {{\cos }^2}\left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right)}$
We can rewrite it as
$= \sqrt {1 - {{\left( {\cos \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right)} \right)}^2}}$ .
Arccosine and cosine will cancel out we have,
$= \sqrt {1 - {{\left( { - \dfrac{2}{3}} \right)}^2}}$
$= \sqrt {1 - \dfrac{4}{9}}$
Taking LCM and simplifying we have,
$= \sqrt {\dfrac{{9 - 4}}{9}}$
$= \sqrt {\dfrac{5}{9}}$
$= \dfrac{{\sqrt 5 }}{3}$
Thus, we have
$\Rightarrow \sin \left( {\arccos \left( { - \dfrac{2}{3}} \right)} \right) = \dfrac{{\sqrt 5 }}{3}$ .
This is the required answer.
So, the correct answer is “$\dfrac{{\sqrt 5 }}{3}$”.

Note : We can do this without substituting $\theta = \arccos \left( { - \dfrac{2}{3}} \right)$ . But it is a little bit difficult compared to what we have done. Also know the relation between cosecant and cotangent. Note the difference between $\cos 2\theta$ and ${\cos ^2}\theta$ . Both are different. That is ${\csc ^2}\theta - {\cot ^2}\theta = 1$ . The relation between secant and tangent. That is ${\sec ^2}\theta - {\tan ^2}\theta = 1$ . We use these identities depending on the problem.