Question

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

Answer 1

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 ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$.