Question

How do you evaluate `cos(Arctan 2)`?

Answer 1

Set `x=arctan2` hence `tanx=2` square both sides to get

`tan^2x=4=>tan^2x+1=5=>1/cos^2x=5`

Hence `cosx=1/sqrt5` or `x=arccos(1/sqrt5)`

But `x=arctan2` hence

`arctan2=arccos(1/sqrt5)=>cos(arctan2)=cos(arccos(1/sqrt5))=1/sqrt5`

Finally `cos(arctan2)=1/sqrt5`