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Question: What does equal \( \arcsin \left( {\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right) \) equals?...

What does equal arcsin⁡(sin⁡(−π2))\arcsin \left( {\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right) equals?

Explanation

Solution

Hint : In order to solve the function given or equation, we need to simplify the brackets one by one. Simplifying with the inner bracket we get an odd function, solving it, we get a value and next left with the outer bracket, further simplifying it with some known facts, and we get our results.

Complete step-by-step answer :
We are given with the function arcsin⁡(sin⁡(−π2))\arcsin \left( {\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right) .
Solving with the inner bracket we can see that there is an odd function of sine and we know that sin⁡(−x)=−sin⁡x\sin \left( { - x} \right) = - \sin x , for an odd function.
Using this for our inner bracket we get that sin⁡(−π2)=−sin⁡π2\sin \left( { - \dfrac{\pi }{2}} \right) = - \sin \dfrac{\pi }{2} that implies arcsin⁡(sin⁡(−π2))=arcsin⁡(−sin⁡π2)\arcsin \left( {\sin \left( { - \dfrac{\pi }{2}} \right)} \right) = \arcsin \left( { - \sin \dfrac{\pi }{2}} \right) .
From trigonometric values we know that sin⁡π2=1\sin \dfrac{\pi }{2} = 1 that implies −sin⁡π2=−1- \sin \dfrac{\pi }{2} = - 1 .
Substituting it in our equation, we get that:
arcsin⁡(−sin⁡π2)=arcsin⁡(−1)\arcsin \left( { - \sin \dfrac{\pi }{2}} \right) = \arcsin \left( { - 1} \right) . ………..(i)
Now, let y=sin⁡(−x)y = \sin \left( { - x} \right) .
Multiplying both the sides by sin⁡−1{\sin ^{ - 1}} and, we get sin⁡−1y=sin−1(sin⁡(−x))=−x{\sin ^{ - 1}}y = si{n^{ - 1}}\left( {\sin \left( { - x} \right)} \right) = - x .
Similarly, if x=π2x = \dfrac{\pi }{2} , substitute in y=sin⁡(−x)y = \sin \left( { - x} \right) and we get:

y=sin⁡(−π2) y=−sin⁡π2=−1 =>−sin⁡π2=−1   y = \sin \left( { - \dfrac{\pi }{2}} \right) \\\ y = - \sin \dfrac{\pi }{2} = - 1 \\\ = > - \sin \dfrac{\pi }{2} = - 1 \;

Multiplying both the sides by sin⁡−1{\sin ^{ - 1}} :

\-1=−sin⁡π2 sin⁡−1(−1)=sin⁡−1(sin⁡(−π2)) =>sin⁡−1(−1)=−π2   \- 1 = - \sin \dfrac{\pi }{2} \\\ {\sin ^{ - 1}}\left( { - 1} \right) = {\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{2}} \right)} \right) \\\ = > {\sin ^{ - 1}}\left( { - 1} \right) = - \dfrac{\pi }{2} \;

As, we know, arcsin⁡\arcsin is nothing but sin⁡−1{\sin ^{ - 1}} .
Substitute this term in (i) and we get:
arcsin⁡(−sin⁡π2)=arcsin⁡(−1)=sin⁡−1(−1) sin⁡−1(−1)=−π2   \arcsin \left( { - \sin \dfrac{\pi }{2}} \right) = \arcsin \left( { - 1} \right) = {\sin ^{ - 1}}\left( { - 1} \right) \\\ {\sin ^{ - 1}}\left( { - 1} \right) = - \dfrac{\pi }{2} \;
Therefore, arcsin⁡(sin⁡(−π2))\arcsin \left( {\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right) equals −π2- \dfrac{\pi }{2} .
So, the correct answer is “−π2- \dfrac{\pi }{2}”.

Note : arcsin⁡(x)=sin⁡−1x\arcsin \left( x \right) = {\sin ^{ - 1}}x , arcsin is nothing but sine inverse.
sin⁡(x)\sin (x) is an odd function that means sin⁡(−x)=−sin⁡x\sin \left( { - x} \right) = - \sin x .
Since, we know that the range of sine function is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) , then according to that the range of arcsin function would be (−1,1)\left( { - 1,1} \right) as sin⁡π2=1\sin \dfrac{\pi }{2} = 1 .