Question

How do you evaluate `cos(arctan(2) +arctan(3))`?

Answer 1

`cos(arctan(2)+arctan(3)) = -sqrt(2)/2`

Explanation:

Consider the following two right angled triangles:

Then:

`{ (tan(alpha) = 2), (tan(beta) = 3) :}`

Notice that:

`{ (sin(alpha) = 2/sqrt(5)), (cos(alpha) = 1/sqrt(5)), (sin(beta) = 3/sqrt(10)), (cos(beta) = 1/sqrt(10)) :}`

The sum of angles formula for `cos` tells us:

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

`=1/sqrt(5) 1/sqrt(10) - 2/sqrt(5) 3/sqrt(10)`

`=(1-6)/sqrt(50)=-5/(5sqrt(2))=-1/sqrt(2)=-sqrt(2)/2`

Answer 2

`cos(tan^-1 2+tan^-1 3)`

`=cos(tan^-1( (2+ 3)/(1-2*3)))`

`=cos(tan^-1( -1))`

`=cos(tan^-1( -tan(pi/4)))`

`=cos(tan^-1( tan(pi-pi/4)))`

`=cos(pi-pi/4)`

`=-cos(pi/4)`

`=-1/sqrt2`