calculate tangent of 1e100

How do you calculate trig functions with very large arguments?

For example, in "Surely you're joking, Mr Feynman", Richard tells how he claimed to be able to solve within 60 seconds and within 10 percent any problem that can be stated in 10 seconds, but is stumped when asked to calculate tan(10^100).

MATLAB gives tan(1e100) = -0.4116 whereas google returns -0.6895. Which one is more accurate?

The MATLAB algorithm seems to not be based on simple remainders mod 2*pi, since tan(mod(1e100,2*pi)) returns 0.

[Seth DeLand]

Using Symbolic Toolbox I got:

>> vpa('tan(1e100)',100)

ans =

0.4012319619908143541857543436532949583238702611292440683194415381168718098221191211467267309749320831

Answer 1

[Tom]

Given that

eps(1e100) = 1.9427e+84

it won't be that accurate.

For what it's worth, sin(1e100)/cos(1e100) gives the same answer.

(DuckDuckGo says that Wolfram Alpha says it's 0.4012319619908143541857543436532949583238702611292440)