Question

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

Answer 1

Hint:The principal value for this trigonometric function is value of angle for which , the sine of the angle will be in the range \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\] , 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 : \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\]
Now , assume an angle say \[\theta \] such that it lies in the range of \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\] which will provide the principal value of \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] .
Now equating \[\theta \] with \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] , we get
\[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \theta \]
Now since \[\theta \] lies in the range \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\] , therefore we can write the above equation as ,
\[\sin \theta = \dfrac{{ - 1}}{2}\] ,
Now , we can write \[\dfrac{1}{2}\] as \[\sin \dfrac{\pi }{6}\] , since the value of \[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\] . Therefore , we get
\[\sin \theta = \sin \dfrac{{ - \pi }}{6}\]
Now , since both LHS and RHS sine terms are equal therefore , the angles are also equal , therefore we get ,
\[\theta = \dfrac{{ - \pi }}{6}\] .
Therefore , we get the value of the required angle as \[\dfrac{{ - \pi }}{6}\] .

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

Note:Alternative Method:This method is short and easy . Use this method in MCQs type questions .
Given: \[{\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\]
Now , directly writing \[\dfrac{1}{2}\] as \[\sin \dfrac{\pi }{6}\] , we get
\[ = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{ - \pi }}{6}} \right)} \right)\]
Since , \[\sin \] lies in the range \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\] , therefore we can write above equation as ,
\[ = \dfrac{{ - \pi }}{6}\]
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 \[\sin x\] and domain of \[\sin x\] .