Rational trig identities?

[Albert Chan]

General case, by expanding tan(n*x), in terms of t = tan(x)

tan(n*x) = (t + tan((n-1)*x)) / (1 - t*tan((n-1)*x))

tan(2*x) = (2t) / (1 - t^2)
tan(3*x) = (3t - t^3) / (1 - 3t^2)
tan(4*x) = (4t - 4t^3) / (1 - 6t^2 + t^4)
tan(5*x) = (5t - 10t^3 + t^5) / (1 - 10t^2 + 5t^4)
...

Pattern appeared, (t+1)^n, odd powers goes on top, even powers bottom, alternate sign.

tan(n*x) = (n*t - binom(n,3)*t^3 + ...) / (1 - binom(n,2)*t^2 + ...)

Better with cotangent c = 1/t: (with x = atan(1/n), we have c = n)

tan(n*x) = (n*c^(n-1) - binom(n,3)*c^(n-3) + ...) / (c^n - binom(n,2)*c^(n-2) + ...)

Or, just let CAS do the work

CAS> a(n) := texpand(tan(n*x)) (tan(x)=1/n)
CAS> apply(a, range(1,9))

\(\displaystyle \left[
1,\frac{4}{3},\frac{13}{9}, \frac{240}{161},
\frac{719}{475}, \frac{42372}{27755},
\frac{92567}{60319}, \frac{14970816}{9722113}
\right]\)