How do you evaluate #sin^-1(sin((19pi)/10))#?

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Jim H Share
Jul 26, 2017

Answer:

#-pi/10#

Explanation:

#sin^-1(x)# is a number (or angle), #t# in #[-pi/2,pi/2]#
with #sin(t) = x#

So,

#sin^-1(sin((19pi)/10))# is a number (or angle), #t# in #[-pi/2,pi/2]#
with #sin(t) = sin((19pi)/10)#

#(19pi)/10# is almost #2pi#, so it is a fourth quadrant angele.

The reference angle for #(19pi)/10# is #pi/10#.

So #t = -pi/10#

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