Question

Find the principal values of ${\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$

Answer 1

Hint : To find the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ . We have to know some inverse trigonometric properties. Let the given function ${\sin ^{ - 1}}\left( y \right) = x$ , where y is given. Taking $\sin$ on both sides of the above equation we get $y = \sin x$ . Hence, we have to find the value of $x$ such that $\sin x = y$ and check $x$ lies in the range of inverse sine function.

Complete step-by-step answer :
The inverse functions ${\sin ^{ - 1}}x$ , ${\cos ^{ - 1}}x$ , ${\tan ^{ - 1}}x$ , ${\cot ^{ - 1}}x$ , $\cos e{c^{ - 1}}x$ , ${\sec ^{ - 1}}x$ are called inverse circular functions. For the function $y = \sin x$ , there are infinitely many angles $x$ which satisfy $\sin x = a$ , $- 1 \leqslant a \leqslant 1$ .Of these infinite set of values, there is one which lies in the interval $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ . This angle is called the Principal angle and denoted by ${\sin ^{ - 1}}a$ . The Principal value of an inverse function is that value of the general value which is numerically least. It may be positive or negative. When there are two values, one is positive and the other is negative such that they are numerically equal, then the principal value is the positive one.
Given ${\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$
Let ${\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = x$ ---(1)
Taking $\sin$ on both sides of the equation (1), we get
$\sin \left( {{{\sin }^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right) = \sin x$
$\Rightarrow \sin x = - \dfrac{1}{{\sqrt 2 }}$ ------(2)
We know that $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and also $\sin$ is an odd function. Then the equation (2)
$\Rightarrow \sin x = \sin \left( { - \dfrac{\pi }{4}} \right)$ ---(3)
Taking ${\sin ^{ - 1}}$ on both sides of the equation (3)
$\Rightarrow x = - \dfrac{\pi }{4}$ .
Since the range of ${\sin ^{ - 1}}$ lie in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $- \dfrac{\pi }{4} \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
Hence the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ is $- \dfrac{\pi }{4}$ .
So, the correct answer is “ $- \dfrac{\pi }{4}$ .”.

Note : Note that the inverse of the trigonometric function must be used to determine the measure of the angle. Also note that $\sin ( - x) = - \sin (x)$ , $\cos ( - x) = \cos (x)$ and $\tan ( - x) = - \tan (x)$ . Also $\sin x$ is a periodic function with period $2\pi$ . The inverse of the sine function is read sine inverse and is also called the arcsine relation.