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Question: Find the principal value of \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] ....

Find the principal value of sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) .

Explanation

Solution

The principal value for this trigonometric function is value of angle for which , the sine of the angle will be in the range [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] , since . Now , to find the value of that angle we will assume an angle and then equate it with the inverse function to find the required value .

Complete step by step answer:
Given : sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)
Now , assume an angle say θ\theta such that it lies in the range of [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] which will provide the principal value of sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) .
Now equating θ\theta with sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) , we get
sin⁡−1(−12)=θ{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \theta
Now since θ\theta lies in the range [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] , therefore we can write the above equation as ,
sin⁡θ=−12\sin \theta = \dfrac{{ - 1}}{2} ,
Now , we can write 12\dfrac{1}{2} as sin⁡π6\sin \dfrac{\pi }{6} , since the value of sin⁡π6=12\sin \dfrac{\pi }{6} = \dfrac{1}{2} . Therefore , we get
sin⁡θ=sin⁡−π6\sin \theta = \sin \dfrac{{ - \pi }}{6}
Now , since both LHS and RHS sine terms are equal therefore , the angles are also equal , therefore we get ,
θ=−π6\theta = \dfrac{{ - \pi }}{6} .
Therefore , we get the value of the required angle as −π6\dfrac{{ - \pi }}{6} .

Therefore , which implies that the principal value of sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) is −π6\dfrac{{ - \pi }}{6}.

Note: Alternative Method:This method is short and easy . Use this method in MCQs type questions .
Given: sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)
Now , directly writing 12\dfrac{1}{2} as sin⁡π6\sin \dfrac{\pi }{6} , we get
=sin⁡−1(sin⁡(−π6))= {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{ - \pi }}{6}} \right)} \right)
Since , sin⁡\sin lies in the range [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] , therefore we can write above equation as ,
=−π6= \dfrac{{ - \pi }}{6}
Hence proved. Also here we get answers directly without substituting which includes skipping steps so, use this method when required. Trigonometric function is not cancelled out with its inverse; it is just like equating the angles as the angles in range of sin⁡x\sin x and domain of sin⁡x\sin x .