How do you find the #arcsin(sin((7pi)/6))#?

1 Answer
Zor Shekhtman
Feb 10, 2015

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.

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