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

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

Explanation

Solution

The arcsin is the inverse of the trigonometric representation of sine. The range of arcsine is from negative of π2\dfrac{\pi }{2} to positive of π2\dfrac{\pi }{2}. The domain of arcsin is from negative 1 to positive 1. It is a bijective function, which means it will be invertible. This property will be very useful in this question.

Complete step by step solution:
According to the question, we have to find the value of sin⁡−1(−1){\sin ^{ - 1}}( - 1),
So, let sin⁡−1(−1){\sin ^{ - 1}}( - 1)=x
Now,
Let us draw a triangle with sin⁡x\sin x= −1 - 1
In this right-angled triangle,

We have considered the angle between base and Hypotenuse as x.
The angle between the base and the height =90∘= 90^{\circ}
Now taking sine function both sides we will get,
sin⁡(sin⁡−1(−1))=sin⁡(x)\sin ({\sin ^{ - 1}}( - 1)) = \sin (x)
And as we know that sin⁡(sin⁡−1θ)=θ\sin ({\sin ^{ - 1}}\theta ) = \theta , so in above equation θ=(−1)\theta = ( - 1)
Hence, the equation changes to sin⁡(x)=(−1)\sin (x) = ( - 1)
So, as we know the value of sin⁡θ\sin \theta for different values of θ\theta , we know that when θ\theta = 270° or (-90°), sin⁡θ\sin \theta will give the value −1 - 1.
Hence now we know that for getting a result equal to a negative one, we must give input as θ\theta = 270° or (-90°),

So, our answer x=3π2x = \dfrac{{3\pi }}{2} or x=−π2x = - \dfrac{\pi }{2}.

Note:
The arcsin is the inverse of the tan trigonometric function. It is a bijective function, that is, it gives a unique value of every unique input and for every output one single input is considered. Bijective also means it is invertible, that is why we changed arsine to sin easily in our solution. This property is valid only for its respective domain, which is from a negative one to positive one.