
How do you find the exact value of $ {\sin ^{ - 1}}( - 0.5) $ ?
Answer
558.6k+ views
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}} $ 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}} $ function
Complete step-by-step answer:
To solve this question we will first assume that $ {\sin ^{ - 1}}( - 0.5) = y $
And since we know that domain $ [ - 1,1] $ and range $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ of $ {\sin ^{ - 1}} $ function from its definition.
Since in this question we have not given the branch on which we have to find $ {\sin ^{ - 1}} $ function value to find so we take principal value branch to find $ {\sin ^{ - 1}}( - 0.5) $ value.
We have above assumed $ {\sin ^{ - 1}}( - 0.5) = y $ and we can write $ ( - 0.5) = 1/2 $ . So we have
$
{\sin ^{ - 1}}( - 0.5) = {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \\
\Rightarrow {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \;
$
Then $ \sin y = \dfrac{1}{2} $
As, we know the range of the principal value branch of
$ {\sin ^{ - 1}} $ is $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and also we know
$ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} $
Therefore, the principal value of $ {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ is $ \dfrac{\pi }{6} $ .
So the exact value of $ {\sin ^{ - 1}}( - 0.5) $ is $ \dfrac{\pi }{6} $ .
So, the correct answer is “ $ \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 ^{ - 1}}x $ and $ {(\sin x)^{ - 1}} $ because both are not same even $ {(\sin x)^{ - 1}} $ means $ {(\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.
Complete step-by-step answer:
To solve this question we will first assume that $ {\sin ^{ - 1}}( - 0.5) = y $
And since we know that domain $ [ - 1,1] $ and range $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ of $ {\sin ^{ - 1}} $ function from its definition.
Since in this question we have not given the branch on which we have to find $ {\sin ^{ - 1}} $ function value to find so we take principal value branch to find $ {\sin ^{ - 1}}( - 0.5) $ value.
We have above assumed $ {\sin ^{ - 1}}( - 0.5) = y $ and we can write $ ( - 0.5) = 1/2 $ . So we have
$
{\sin ^{ - 1}}( - 0.5) = {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \\
\Rightarrow {\sin ^{ - 1}}\left[ {\dfrac{1}{2}} \right] = y \;
$
Then $ \sin y = \dfrac{1}{2} $
As, we know the range of the principal value branch of
$ {\sin ^{ - 1}} $ is $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and also we know
$ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} $
Therefore, the principal value of $ {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ is $ \dfrac{\pi }{6} $ .
So the exact value of $ {\sin ^{ - 1}}( - 0.5) $ is $ \dfrac{\pi }{6} $ .
So, the correct answer is “ $ \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 ^{ - 1}}x $ and $ {(\sin x)^{ - 1}} $ because both are not same even $ {(\sin x)^{ - 1}} $ means $ {(\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.
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