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Question: Find the principal value of \({{\sin }^{-1}}\left( \sin 3 \right)\)....

Find the principal value of sin⁡−1(sin⁡3){{\sin }^{-1}}\left( \sin 3 \right).

Explanation

Solution

Hint: Convert the angle 3 which is given in radian into degrees by multiplying it with 57.3. Check whether the obtained product is in the range -90 degrees to 90 degrees. If it is not, then subtract π\pi from three to get it in the required range. To balance the angle we have to add π\pi . Now, use the property sin⁡(π+θ)=−sin⁡θ\sin \left( \pi +\theta \right)=-\sin \theta and then the property sin⁡−1(sin⁡x)=x{{\sin }^{-1}}\left( \sin x \right)=x to get the answer.

Complete step-by-step answer:

Since, none of the six trigonometric functions are one-to-one, they are restricted in order to have inverse functions. Therefore, the ranges of the inverse functions are proper subsets of the domains of the original functions.
Now let us come to the question. We have to find the principal value of sin⁡−1(sin⁡3){{\sin }^{-1}}\left( \sin 3 \right).
Here, 3 is in radian. To convert radian into degrees, we multiply it by 57.3 because 1 radian = 57.3 degrees.
Therefore, 3 radian = 3×57.3=171.93\times 57.3=171.9 degrees.
We know that, the range of sin⁡−1x{{\sin }^{-1}}x is between −π2-\dfrac{\pi }{2} and π2\dfrac{\pi }{2} including these two values. So, we have to change the angle 3 such that it lies between −π2-\dfrac{\pi }{2} and π2\dfrac{\pi }{2}.
So, subtracting π\pi from 3 in the function, we get,
sin⁡−1(sin⁡(3−π)){{\sin }^{-1}}\left( \sin (3-\pi ) \right)
To balance this expression we have to add π\pi . Therefore, expression becomes
sin⁡−1(sin⁡(π+(3−π))){{\sin }^{-1}}\left( \sin (\pi +(3-\pi )) \right)
Using, sin⁡(π+θ)=−sin⁡θ\sin \left( \pi +\theta \right)=-\sin \theta , we get,
sin⁡−1(−sin⁡(3−π)){{\sin }^{-1}}\left( -\sin \left( 3-\pi \right) \right)
Using formula: sin⁡−1(−x)=−sin⁡−1x{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x, we get,
−sin⁡−1(sin⁡(3−π))-{{\sin }^{-1}}\left( \sin \left( 3-\pi \right) \right)
Here, (3−π)\left( 3-\pi \right) lies between −π2-\dfrac{\pi }{2} and π2\dfrac{\pi }{2}. Therefore, using the formula: sin⁡−1(sin⁡x)=x{{\sin }^{-1}}\left( \sin x \right)=x, we get,
−sin⁡−1(sin⁡(3−π)) =−(3−π) =π−3 \begin{aligned} & -{{\sin }^{-1}}\left( \sin \left( 3-\pi \right) \right) \\\ & =-\left( 3-\pi \right) \\\ & =\pi -3 \\\ \end{aligned}
Hence, the principal value of sin⁡−1(sin⁡3){{\sin }^{-1}}\left( \sin 3 \right) is π−3\pi -3.

Note: One may note that when the value of angle is given in radian which is a number, then it is difficult to directly check whether it lies in the range of a given function or not. So first we have changed it in degrees so that we can check the range. Once it is checked that it does not lie in range, we have subtracted π\pi from the angle and to balance, added π\pi to it. Some basic identities are then applied to get the answer.