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Question: What is \({{\sin }^{-1}}\left( -1 \right)\) ?...

What is sin⁡−1(−1){{\sin }^{-1}}\left( -1 \right) ?

Explanation

Solution

We are asked to find the value of sin⁡−1(−1){{\sin }^{-1}}\left( -1 \right). We can do so by using the properties of inverse trigonometric functions and then comparing the value inside the inverse function to known values of normal trigonometric functions to arrive at the answer.

Complete step by step solution:
In the question, we have been asked to find the inverse sine transform of -1. We know that, by the property of sine inverse function, since it is an odd function, for any x∈[−1,1]x\in \left[ -1,1 \right]
sin⁡−1(−x)=−sin⁡−1(x){{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right)
Now, we can write sin⁡−1(−1){{\sin }^{-1}}\left( -1 \right) using the above property, as follows,
sin⁡−1(−1)=−sin⁡−1(1){{\sin }^{-1}}\left( -1 \right)=-{{\sin }^{-1}}\left( 1 \right)
Next, we will try to find the value of the angle for which sine function gives the value of 1. The sine function reaches 1 at the angle 90∘{{90}^{\circ }} or π2\dfrac{\pi }{2} radians. Hence we get the final result as,
−sin⁡−1(1)=−π2 sin⁡−1(−1)=−π2 \begin{aligned} & -{{\sin }^{-1}}\left( 1 \right)=-\dfrac{\pi }{2} \\\ & {{\sin }^{-1}}\left( -1 \right)=-\dfrac{\pi }{2} \\\ \end{aligned}

Hence, the value of sin⁡−1(−1){{\sin }^{-1}}\left( -1 \right) is found to be equal to −π2-\dfrac{\pi }{2}.

Note: Another way to solve this problem is by converting the sine inverse function to cosine inverse by using the inverse trigonometric law,
cos⁡−1θ=π2−sin⁡−1θ{{\cos }^{-1}}\theta =\dfrac{\pi }{2}-{{\sin }^{-1}}\theta .
We can write the function given to us using the odd function law as,
sin⁡−1(−1)=−sin⁡−1(1){{\sin }^{-1}}\left( -1 \right)=-{{\sin }^{-1}}\left( 1 \right)
Now, we can apply the conversion law of sine inverse to cosine inverse function. We get the expression as,
sin⁡−1(−1)=−(π2−cos⁡−1(1)){{\sin }^{-1}}\left( -1 \right)=-\left( \dfrac{\pi }{2}-{{\cos }^{-1}}\left( 1 \right) \right)
We need to find the angle for which cosine function gives 1 as the result. We know that at 0 degrees or radians cosine function gives 0, hence we obtain the answer as,
      sin⁡−1(−1)=−(π2−0) ⇒sin⁡−1(−1)=−π2 \begin{aligned} & \,\,\,\,\,\,{{\sin }^{-1}}\left( -1 \right)=-\left( \dfrac{\pi }{2}-0 \right) \\\ & \Rightarrow {{\sin }^{-1}}\left( -1 \right)=-\dfrac{\pi }{2} \\\ \end{aligned}
Hence, the value of sin⁡−1(−1){{\sin }^{-1}}\left( -1 \right) using this method is found to be equal to −π2-\dfrac{\pi }{2}.