Question

What does `cos(arctan((3pi)/2))-2sin(arcsec(pi/4))` equal?

Answer 1

`2/sqrt(9pi^2+4)-2*sqrt(pi^2-16)/pi`

Explanation:

If we make `arctan((3pi)/2)=theta` then
`tan theta=(3pi)/2` => `sin theta/cos theta =(3pi)/2` => `sqrt(1-cos^2 theta)=(3pi)/2*cos theta` =>`1-cos^2 theta=(9pi^2)/4` => `cos^2 theta*(9pi^2+4)/4=1` => `cos theta=2/sqrt(9pi^2+4)` => `theta = arccos (2/sqrt(9pi^2+4))`

If we make `arcsec(pi/4)=phi` then
`sec phi =pi/4` => `cos phi=4/pi`
`sin phi=sqrt(1-cos^2 phi)=sqrt(1-16/pi^2)` => `sin phi=sqrt(pi^2-16)/pi` => `phi=arcsin(sqrt(pi^2-16)/pi)`

Using the results in the original expression
`cos theta-2sin phi=`
`cos (arccos(2/sqrt(9pi^2+4)))-2*sin (arcsin(sqrt(pi^2-16)/pi))=`
`=2/sqrt(9pi^2+4)-2*sqrt(pi^2-16)/pi`