How do you solve and find the value of #sin^-1(1/sqrt2)#?

1 Answer
Mar 2, 2017

Answer:

The angle = #pi/2 = 45^@#

Explanation:

#sin^-1(1/sqrt(2))# says, give me the angle that has a sin of #(1/sqrt(2)) = sqrt(2)/2#

Either from a trig circle or from a #45^@-45^@-90^@# triangle, that angle would be either #pi/4 = 45^@# or #(3 pi)/4 = 135^@#. However, the #arcsin# #(sin^-1)# function has a limited domain #[-1, 1]#and range #[-pi/2, pi/2] = [-90^@, 90^@]#. This means #(3 pi)/4 = 135^@# is not a valid answer.

You can see this from the graph #f(x) = arcsin(x) = sin^-1(x)#:
graph{arcsin(x) [-4.933, 4.932, -2.466, 2.467]}