
Calculate ${\sin^{-1}}\sin2$ ?
Answer
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Hint: $\sin x$ is a trigonometric function whose domain is real number and range is $\left[ { - 1,1} \right]$ .
Trigonometric function $\sin x$ is positive in the I and II quadrant.
${\sin ^{ - 1}}x$ is an inverse trigonometric function whose domain is $\left[ { - 1,1} \right]$ and range is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Complete step by step solution:
We have to evaluate the value of the inverse trigonometric function ${\sin ^{ - 1}}\sin 2$. Let us consider the given value $2$ is in degree. Now the value of $\sin {2^ \circ }$ lies in between $0$ to $1$ which lies in the range of inverse trigonometric function, so we can cancel out the trigonometric and inverse trigonometric function. Hence the value of ${\sin ^{ - 1}}\sin {2^ \circ } = {2^ \circ }$.
Now we consider that the given value $2$ is in radian. Now we cannot directly cancel out the inverse trigonometric function and, trigonometric function as the obtained value $2$ does not lie in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, so first we add and subtract $\pi $ in the value and rewrite the given expression, we get
${\sin ^{ - 1}}\sin 2 = {\sin ^{ - 1}}\sin \left( {\pi - \left( {\pi - 2} \right)} \right)$
Now the angle $\pi - \left( {\pi - 2} \right)$ lies in the second quadrant and we know that in the second quadrant is positive and we also know that $\sin \left( {\pi - \theta } \right) = \sin \theta $ so the given expression can be written as
$ \Rightarrow {\sin ^{ - 1}}\sin \left( {\pi - 2} \right)$
Now we can cancel out the inverse trigonometric function and trigonometric function as the obtained result lies in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ so
$ \Rightarrow {\sin ^{ - 1}}\sin \left( {\pi - 2} \right) = \pi - 2$
Hence the value of ${\sin ^{ - 1}}\sin \left( 2 \right) = \pi - 2$
Note: While evaluating the value of an inverse trigonometric function, remember that value must lie in the range of the inverse trigonometric function. If the given value is not specified then, consider the given value in degree as well as in radian also. If the obtained value does not lie in the range then, convert the given expression such that its result does not change and the obtained result comes in the range.
Trigonometric function $\sin x$ is positive in the I and II quadrant.
${\sin ^{ - 1}}x$ is an inverse trigonometric function whose domain is $\left[ { - 1,1} \right]$ and range is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Complete step by step solution:
We have to evaluate the value of the inverse trigonometric function ${\sin ^{ - 1}}\sin 2$. Let us consider the given value $2$ is in degree. Now the value of $\sin {2^ \circ }$ lies in between $0$ to $1$ which lies in the range of inverse trigonometric function, so we can cancel out the trigonometric and inverse trigonometric function. Hence the value of ${\sin ^{ - 1}}\sin {2^ \circ } = {2^ \circ }$.
Now we consider that the given value $2$ is in radian. Now we cannot directly cancel out the inverse trigonometric function and, trigonometric function as the obtained value $2$ does not lie in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, so first we add and subtract $\pi $ in the value and rewrite the given expression, we get
${\sin ^{ - 1}}\sin 2 = {\sin ^{ - 1}}\sin \left( {\pi - \left( {\pi - 2} \right)} \right)$
Now the angle $\pi - \left( {\pi - 2} \right)$ lies in the second quadrant and we know that in the second quadrant is positive and we also know that $\sin \left( {\pi - \theta } \right) = \sin \theta $ so the given expression can be written as
$ \Rightarrow {\sin ^{ - 1}}\sin \left( {\pi - 2} \right)$
Now we can cancel out the inverse trigonometric function and trigonometric function as the obtained result lies in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ so
$ \Rightarrow {\sin ^{ - 1}}\sin \left( {\pi - 2} \right) = \pi - 2$
Hence the value of ${\sin ^{ - 1}}\sin \left( 2 \right) = \pi - 2$
Note: While evaluating the value of an inverse trigonometric function, remember that value must lie in the range of the inverse trigonometric function. If the given value is not specified then, consider the given value in degree as well as in radian also. If the obtained value does not lie in the range then, convert the given expression such that its result does not change and the obtained result comes in the range.
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