Question

How do you find `cos(sin^-1(sqrt2/2)+cos^-1(3/5))`?

Answer 1

`cos(sin^(-1)(sqrt(2)/2) + cos^(-1)(3/5)) = -sqrt(2)/10`

Explanation:

Let `alpha = sin^(-1)(sqrt(2)/2)` and `beta = cos^(-1)(3/5)`

Then:

`sin alpha = sqrt(2)/2`

`cos alpha = sqrt(1-sin^2 alpha) = sqrt(1-1/2) = sqrt(1/2) = sqrt(2)/2`

`sin beta = sqrt(1-cos^2 beta) = sqrt(1-3^2/5^2) = sqrt(1-9/25) = sqrt(16/25) = 4/5`

`cos beta = 3/5`

Alternatively you could pick out these values from the `1:1:sqrt(2)` and `3:4:5` right angled triangles...

`alpha = pi/4`, `sin alpha = sin beta = 1/sqrt(2) = sqrt(2)/2`

`beta = B`, `sin beta = 4/5`, `cos beta = 3/5`

Then using the formula for `cos` of a sum of angles we have:

`cos (alpha+beta) = cos alpha cos beta - sin alpha sin beta`

`color(white)(cos (alpha+beta)) = sqrt(2)/2 3/5 - sqrt(2)/2 4/5`

`color(white)(cos (alpha+beta)) = -sqrt(2)/10`