How do you evaluate #sin^-1(-sqrt2/2)# without a calculator?

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Answer 1 · 2016-11-30T10:40:03

#sin^(-1)(-sqrt(2)/2) = -pi/4#

This is a known value for #sin(x)#, you should remember that:

#sin(pi/4) = sqrt(2)/2#

and that:

#sin(-x) = -sin(x)#

Although many other angles have the same value for sine, namely:

#alpha = -pi/4+2kpi#
and
#alpha = -(3pi)/4 +2kpi#

conventionally #arcsin(x)# is defined in the interval #[-pi/2,pi/2]#
so you pick the value:

#x=-pi/4#