Question

How do you solve and find the value of \[\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right)\]?

Answer 1

Hint: We will use concepts of trigonometry and their properties to solve this problem. First we will assume the inverse term as variable and then we use standard identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] to solve this problem.

Complete step by step answer:
The given question is \[\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right)\]
Take an assumption that, \[{\cos ^{ - 1}}\left( {\dfrac{3}{4}} \right) = \theta \]
Which implies that, \[\cos \theta = \dfrac{3}{4}\]
So, we can write this as \[\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right) = \sin \theta \]
Now, recall the trigonometric identity, \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
And from this, we can conclude that, \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]
So, this implies that, \[\sin \theta = \pm \sqrt {1 - {{\cos }^2}\theta } \]
We know the value of \[\cos \theta \]. So, we will substitute in this.
\[ \Rightarrow \sin \theta = \pm \sqrt {1 - {{\left( {\dfrac{3}{4}} \right)}^2}} \]
\[ \Rightarrow \sin \theta = \pm \sqrt {1 - \left( {\dfrac{9}{{16}}} \right)} \]
So, we can simplify this as,
\[ \Rightarrow \sin \theta = \pm \sqrt {\dfrac{{16 - 9}}{{16}}} \]
\[ \Rightarrow \sin \theta = \pm \sqrt {\dfrac{7}{{16}}} = \pm \dfrac{{\sqrt 7 }}{4}\]
So, like this, we can solve this problem, and finally we can find the result as \[\sin \theta = \pm \dfrac{{\sqrt 7 }}{4}\].

Note:
 Remember the identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] which is a standard identity in trigonometry. The result we got consists of both positive and negative values. We have to consider both the values.
\[{\sin ^{ - 1}}x\] and \[{\cos ^{ - 1}}x\] are defined only when \[ - 1 \leqslant x \leqslant 1\].
And also remember the identity \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\].