Question

Find the principal value of ${\sin ^{ - 1}}\left( {\tan \dfrac{{5\pi }}{4}} \right)$.

Answer 1

The range of ${\sin ^{ - 1}}x$ is between $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. By using tangent function, find the value of $\left( {\tan \dfrac{{5\pi }}{4}} \right)$. Now substitute this back into our given expression and simplify it to get the principal value.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. The principal value of ${\sin ^{ - 1}}x$ for $x > 0$ , is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason ${\sin ^{ - 1}}x$ is also denoted by $\operatorname{arcsinx}$.
The principal value of ${\sin ^{ - 1}}x$ branches to,
${\sin ^{ - 1}}x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.
Hence the principal value of the given function will be between the range$\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ .
Now we have been given the function, ${\sin ^{ - 1}}\left( {\tan \dfrac{{5\pi }}{4}} \right)$.
Now we need to find the value of $\left( {\tan \dfrac{{5\pi }}{4}} \right)$. We can also write it as,
$\tan \dfrac{{5\pi }}{4} = \tan \left( {\pi + \dfrac{\pi }{4}} \right)$
By using basic trigonometric ratios, we can write the tangent function as,
$\tan (\pi + \theta ) = \tan \theta$, by comparing both the equations we can find that,

$$\tan \left( {\pi + \dfrac{\pi }{4}} \right) = \tan \dfrac{\pi }{4} \ \\\ \tan \dfrac{\pi }{4} = \tan {45^ \circ } = 1 \ \\\$$

Now let us substitute the value of $\left( {\tan \dfrac{{5\pi }}{4}} \right)$back in our given expression. Thus it becomes,
${\sin ^{ - 1}}\left( {\tan \dfrac{{5\pi }}{4}} \right) = {\sin ^{ - 1}}1$
From the trigonometric table we can write the value of 1 as,
$\sin \dfrac{\pi }{2} = 1$
Hence put $\sin \dfrac{\pi }{2}$in the place of 1.

$${\sin ^{ - 1}}1 = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{2}} \right) \ \\\ {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{2}} \right) = \dfrac{\pi }{2} \ \\\$$

Thus we got the principal value of the expression as $\dfrac{\pi }{2}$.
$\therefore {\sin ^{ - 1}}\left( {\tan \dfrac{{5\pi }}{4}} \right) = \dfrac{\pi }{2}$.

Note: To solve a question like these you should be familiar with the domain and range of the sine functions as well as the domain and range of the inverse sine functions. For us the range of inverse sine function is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and the domain of inverse function of sine is $[ - 1,1]$. Students should remember the important trigonometric ratios and standard angles to solve these types of questions.