A trigonometric map is any one of sin, cos, tan, arcsin, arccos and arctan. Show that given any positive rational number x, one can find a finite sequence of trigonometric maps which take 0 to x. [So we need to show that we can always find a sequence of trigonometric maps ti so that: x1 = t0(0), x2 = t1(x1), ... , xn = tn-1(xn-1), x = tn(xn).]
Solution
We have cos2t + sin2t = 1. Hence cos t = cos t/(√(cos2t + sin2t)) = 1/(√(1 + tan2t)). So if we put tan t = √x, then cos tan-1√x = 1/√(1 + x). We also have cos(p/2 - x) = sin x and tan(p/2 - x) = 1/tan x. So tan cos-1 sin tan-1 x = 1/x (1). Hence also tan cos-1 sin tan-1 cos tan-1 √x = √(x + 1) (2). These two relations solve the problem.
Using (2) and iterating we can get √n for any positive integer n. Hence in particular we can get n for any positive integer n. Now suppose we want m/n with m and n relatively prime. We show that m/n can be achieved by induction on n. We have just dealt with the case n = 1. Suppose we have dealt with all a/b with b < n. If m > n, then we can write m = qn + r with 0 < r < n and use (2) to reduce the problem to getting r/n. If m < n, then put r = m. Now use (1) to reduce the problem to n/r, which is solved by induction.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002