This was a bit much to type out, so I took a snip of it.
drive.google.com/file/d/1AUtOTEvfSw-cvv5_tCm9hk1NsADr0vZP/view?usp=sharing
you should be familiar with the relationships
cos (2x)
= cos^2 x - sin^2 x
or
= 1 - 2sin^2 x
or
= 2cos^2 x -1
then if cos u = 9/41
cos u = 1 - 2sin^2 (u/2)
9/41 = 1 - 2sin^2 (u/2)
2sin^2 (u/2) = 32/41
sin^2 (u/2) = 16/41
sin (u/2) = ± 4/√41
by the Pythagorean relationship cos^2 (u/2) = 25/41
cos (u/2) = ± 5/√41
going back to the original cos u = 9/41, then u must be in quadrants I or IV
so u/2 must be in I or II
In quad I, sin(u/2) = 4/√41 , cos(u/2) = 5/√41, tan(u/2)=4/5
in quad II sin(u/2) = 4/√41, cos(u/2) = -5/√41 , tan(u/2) = -4/5
for the tan(u/2) I used the fact that tanx = sinx/cosx