In theory you could just use the series expansion:
#cos(theta) = 1 - theta^2/(2!) + theta^4/(4!) - theta^6/(6!) +...#
This will converge faster for small values of #theta#, so it's better to use it in conjunction with some angle sum formulae:
#cos(alpha + beta) = cos(alpha)cos(beta)-sin(alpha)sin(beta)#
#sin(alpha + beta) = sin(alpha)cos(beta) + sin(beta)cos(alpha)#
and
#sin(theta) = theta - theta^3/(3!) + theta^5/(5!) - theta^7/(7!) +...#
So if you can find an angle #alpha# for which you know some trig values that is close to #theta#, then you can use these to construct a good approximation for #cos(theta)#
For your example, I am not sure whether #5.22# is in radians or degrees. I will assume radians and just try to find an answer to a few significant digits.
I will use #pi ~= 355/113#
#5.22 / pi ~= 113 * 5.22 / 355 = 113 * 522 / 35500 = 58986 / 35500 ~= 5/3#
#cos((5pi)/3) = cos(-pi/3) = cos(pi/3) = 0.5#
#sin((5pi)/3) = sin(-pi/3) = -sin(pi/3) = -sqrt(3)/2#
#~= -1.7320508/2 = -0.8660254#
Let #alpha = (5pi)/3# and #beta = 5.22 - (5pi)/3#
#beta = 5.22 - (5pi)/3 ~= 5.22 - (5*355)/(113*3) ~= 5.22 - 5.235988#
#=-0.0159882#
#cos(5.22) = cos(alpha+beta)#
#=cos(alpha)cos(beta) - sin(alpha)sin(beta)#
#~=0.5 cos(beta)+0.8660254 sin(beta)#
#cos(beta) ~= 1 - beta^2/(2!)#
#~=1 - 0.0002556/2 = 1 - 0.0001278 = 0.9998722#
#sin(beta) ~= beta - beta^3/(3!)#
#~=-0.0159882 + 0.0000007#
#=-0.0159875#
So
#cos(5.22) ~= 0.5 * 0.9998722 - 0.8660254 * 0.0159875#
#~= 0.49993610 - 0.01384558#
#=0.48609052#
Actually #cos(5.22) ~= 0.486090886#
so we were good to about #6# significant digits.
As you see, the process is a little tedious and fraught with possibilities for miscalculation.
