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Question: Calculate \({\sin^{-1}}\sin2\) ?...

Calculate sin⁡−1sin⁡2{\sin^{-1}}\sin2 ?

Explanation

Solution

sin⁡x\sin x is a trigonometric function whose domain is real number and range is [−1,1]\left[ { - 1,1} \right] .
Trigonometric function sin⁡x\sin x is positive in the I and II quadrant.
sin⁡−1x{\sin ^{ - 1}}x is an inverse trigonometric function whose domain is [−1,1]\left[ { - 1,1} \right] and range is [−π2,π2]\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⁡−1sin⁡2{\sin ^{ - 1}}\sin 2. Let us consider the given value 22 is in degree. Now the value of sin⁡2∘\sin {2^ \circ } lies in between 00 to 11 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⁡−1sin⁡2∘=2∘{\sin ^{ - 1}}\sin {2^ \circ } = {2^ \circ }.

Now we consider that the given value 22 is in radian. Now we cannot directly cancel out the inverse trigonometric function and, trigonometric function as the obtained value 22 does not lie in the range [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right], so first we add and subtract π\pi in the value and rewrite the given expression, we get
sin⁡−1sin⁡2=sin⁡−1sin⁡(π−(π−2)){\sin ^{ - 1}}\sin 2 = {\sin ^{ - 1}}\sin \left( {\pi - \left( {\pi - 2} \right)} \right)

Now the angle π−(π−2)\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⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta so the given expression can be written as
⇒sin⁡−1sin⁡(π−2)\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 [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] so
⇒sin⁡−1sin⁡(π−2)=π−2\Rightarrow {\sin ^{ - 1}}\sin \left( {\pi - 2} \right) = \pi - 2

Hence the value of sin⁡−1sin⁡(2)=π−2{\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.