Question

How do you evaluate `cos(2Arctan(3/4))`?

Answer 1

`cos(2arctan(3/4))=7/25.`

Explanation:

Let `arctan(3/4)=theta`, sothat, by defn. of `arctan` function, `tantheta=3/4`, and, `theta in (-pi/2,pi/2)=(-pi/2,0)uu{0}uu(0,pi/2).`

Note that, `tantheta=3/4 >0, theta !in (-pi/2,0]`, but, `theta in (0,pi/2).`

Now, reqd. value `=cos(2arctan(3/4))=cos2theta=(1-tan^2theta)/(1+tan^2theta)={1-(3/4)^2}/{1+(3/4)^2}=(16-9)/(16+9)=7/25.`