How do you evaluate #arcsin(sqrt 2/2)#?

1 Answer
George C.
May 24, 2015

#sin (pi/4) = sqrt(2)/2# is the length of one side of the isoceles right angled triangle with sides #sqrt(2)/2#, #sqrt(2)/2# and #1#, which has internal angles #pi/4#, #pi/4# and #pi/2#.

(#pi/4# radians = #45^o# and #pi/2# radians = #90^o# if you prefer)

To show this is right angled, check with Pythagoras:

#(sqrt(2)/2)^2 + (sqrt(2)/2)^2#

#= sqrt(2)^2/2^2 + sqrt(2)^2/2^2#

#= 2/4 + 2/4 = 1/2+1/2 = 1 = 1^2#

So since #sin (pi/4) = sqrt(2)/2# and #pi/4# is in the

required range for #arcsin# viz #-pi/2 <= theta <= pi/2#, we find

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

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
67272 views around the world
You can reuse this answer
Creative Commons License