#cos ( 2 arctan(12/5) ) #
I enumerated here all the forms like
#cos arctan (a/b) = pm b/sqrt{a^2+b^2}#
These are easy to thing about as sides of a right triangle. #arctan(a/b)# is the angle whose opposite is #a# and adjacent is #b# so a hypotenuse of #sqrt{a^2+b^2}.#
Before we can apply this we have to first solve
#tan ( 2 arctan(12/5) )#
That's the double angle formula for tangent:
# tan 2 theta = {2 tan theta}/{1 - tan^2 theta}#
#tan ( 2 arctan(12/5) ) #
#= {2 (12/5) }/{1 - (12/5)^2} #
# = {2(12)(5)}/{5^2 - 12^2} #
# = -120/119 #
Now we can apply
#cos ( 2 arctan(12/5) ) = cos arctan ({-120}/{ 119}) = pm 119/sqrt{120^2+119^2} = pm 119/169
#
If we want the principal value, #arctan(12/5)# is first quadrant, more than #45^circ.# So its double is second quadrant, negative cosine.
#cos ( 2 text{Arc}text{tan}(12/5) ) =-119/169#
