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.