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Question: How do you find the exact value of \( {\sin ^{ - 1}}( - 0.5) \) ?...

How do you find the exact value of sin⁡−1(−0.5){\sin ^{ - 1}}( - 0.5) ?

Explanation

Solution

Hint : This question is related to the topic of inverse trigonometric function. In this type of question we need to find the principal value of sin⁡−1{\sin ^{ - 1}} function. Since in this question we are not given any branch, we mean the principal value branch of that function. To solve this question you need to know the definition of sin⁡−1{\sin ^{ - 1}} function

Complete step-by-step answer :
To solve this question we will first assume that sin⁡−1(−0.5)=y{\sin ^{ - 1}}( - 0.5) = y
And since we know that domain [−1,1][ - 1,1] and range [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] of sin⁡−1{\sin ^{ - 1}} function from its definition.
Since in this question we have not given the branch on which we have to find sin⁡−1{\sin ^{ - 1}} function value to find so we take principal value branch to find sin⁡−1(−0.5){\sin ^{ - 1}}( - 0.5) value.
We have above assumed sin⁡−1(−0.5)=y{\sin ^{ - 1}}( - 0.5) = y and we can write (−0.5)=1/2( - 0.5) = 1/2 . So we have
sin⁡−1(−0.5)=sin⁡−1[12]=y ⇒sin⁡−1[12]=y   {\sin ^{ - 1}}( - 0.5) = {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \\\ \Rightarrow {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \;
Then sin⁡y=12\sin y = \dfrac{1}{2}
As, we know the range of the principal value branch of
sin⁡−1{\sin ^{ - 1}} is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and also we know
sin⁡(π6)=12\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}
Therefore, the principal value of sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) is π6\dfrac{\pi }{6} .
So the exact value of sin⁡−1(−0.5){\sin ^{ - 1}}( - 0.5) is π6\dfrac{\pi }{6} .
So, the correct answer is “ π6\dfrac{\pi }{6} ”.

Note : The principal value of an inverse trigonometric function is that value which lies in the range of the principal branch of that inverse trigonometric function.
While solving this kind of question do not confuse between sin⁡−1x{\sin ^{ - 1}}x and (sin⁡x)−1{(\sin x)^{ - 1}} because both are not same even (sin⁡x)−1{(\sin x)^{ - 1}} means (sin⁡x)−1=1sin⁡x{(\sin x)^{ - 1}} = \dfrac{1}{{\sin x}} .Similarly, for other trigonometric inverse function. If this type of question branch is not specified always take it be the principal value branch of the corresponding inverse trigonometric function. These are some of common errors done by students while solving these questions.