Question

How do you calculate \[\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right)\]?

Answer 1

Hint:To solve this question, we need to use the concept of inverse of the trigonometric functions. Inverse trigonometric functions are defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, anti trigonometric functions or cyclometric functions. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios which we will do in this question.

Formula used:
$\cot \theta = \dfrac{1}{{\tan \theta }}$
\[{\text{cose}}{{\text{c}}^2}\theta = 1 + {\cot ^2}\theta \]
$\sin \theta = \dfrac{{\text{1}}}{{{\text{cosec}}\theta }}$

Complete step by step answer:
We are asked to calculate the value of the term \[\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right)}
\right)\].
For this, first we will consider the term \[{\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)\].
Let us take \[{\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) = \theta \]
We know that the range of inverse of tangent function is from $ - \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$
which is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
Therefore we can say that the angle $\theta $ belongs to this range $\left( { - \dfrac{\pi }{2},\dfrac{\pi
}{2}} \right)$.
\[
{\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) = \theta \\
\Rightarrow \tan \theta = \dfrac{3}{4} \\
\]
Here, the value of $\tan \theta $ is greater than zero. This means that the angle $\theta $ does not belong to the range $\left( { - \dfrac{\pi }{2},0} \right)$.
$ \Rightarrow \theta \in \left( {0,\dfrac{\pi }{2}} \right)$
Now, our desired value \[\sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right)\] becomes \[\sin
\theta \].
We know that cotangent is the reciprocal of the tangent function.

We have \[\tan \theta = \dfrac{3}{4}\]
\[
\cot \theta = \dfrac{1}{{\tan \theta }} \\
\Rightarrow \cot \theta = \dfrac{4}{3} \\
\]
Now we will find the value of \[{\text{cosec}}\theta \] by applying the formula
\[
{\text{cose}}{{\text{c}}^2}\theta = 1 + {\cot ^2}\theta = 1 + {\left( {\dfrac{4}{3}} \right)^2} = 1 +
\dfrac{{16}}{9} = \dfrac{{25}}{9} \\
\Rightarrow {\text{cosec}}\theta = \pm \dfrac{5}{3} \\
\]
We know that \[{\text{cosec}}\theta \]is the reciprocal of \[\sin \theta \].
$ \Rightarrow \sin \theta = \dfrac{{\text{1}}}{{{\text{cosec}}\theta }} = \pm \dfrac{3}{5}$
But, we have seen that the angle $\theta \in \left( {0,\dfrac{\pi }{2}} \right)$
Therefore the value of \[\sin \theta \]must be greater than zero.
$ \Rightarrow \sin \theta = \dfrac{3}{5}$
We know that \[{\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) = \theta \]
\[ \Rightarrow \sin \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right) = \dfrac{3}{5}\]
Hence, our required value is $\dfrac{3}{5}$.

Note:
In this question, the most important part is to consider the range of the inverse tangent function.
Because, without considering it, we cannot obtain the single value of the function. For example, here we have seen that the range of inverse tangent function is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
But as the given ratio $\dfrac{3}{4}$is greater than zero, we have determined that the angle $\theta
\in \left( {0,\dfrac{\pi }{2}} \right)$. Due to which we can know that the angle will be in the first quadrant and accordingly we have determined the value of \[\sin \theta \].