Question

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

Answer 1

`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`