The answer is: ##arcsin(sin(7pi/6))=-pi/6##

The answer is: ##arcsin(sin(7pi/6))=-pi/6##.

The range of a function ##arcsin(x)## is, by definition , ##-pi/2<=arcsin(x)<=pi/2## It means that we have to find an angle ##alpha## that lies between ##-pi/2## and ##pi/2## and whose ##sin(alpha)## equals to a ##sin(7pi/6)##.

From trigonometry we know that ##sin(phi+pi)=-sin(phi)## for any angle ##phi##. This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle ##phi## with the X-axis (counterclockwise from the X-axis to a radius). We also know that sine is an odd function, that is ##sin(-phi)=-sin(phi)##.

We will use both properties as follows: ##sin(7pi/6)=sin(pi/6+pi)=-sin(pi/6)=sin(-pi/6)##

As we see, the angle ##alpha=-pi/6## fits our conditions. It is in the range from ##-pi/2## to ##pi/2## and its sine equals to ##sin(7pi/6)##. Therefore, it's a correct answer to a problem.