Given tan x = 2.
tan(x/8) is the root of f(x), an 8th degree monic polynomial with integer coefficients.
What is the value of f(1)?
For brevity, let tan(z) = t.
Start with the identity tan(2z) = 2t / (1-t^2).
From that tan(4z) = (4t-4t^3) / (1-6t^2+t^4)
Then from that tan(8z) = (8t-56t^3+56t^5-8t^7) / (1-28t^2+70t^4-28t^6+t^8).
From the problem z=x/8, then t=tan(x/8) and tan(8z)=tan(x)=2.
Then (8t-56t^3+56t^5-8t^7) / (1-28t^2+70t^4-28t^6+t^8) = 2.
Which simplifies to f(t) = 1-4t-28t^2+28t^3+70t^4-28t^5-28t^6+4t^7+t^8 = 0
tan(x/8) is a root of this polynomial. When evaluated at t=1 then the value is 1-4-28+28+70-28-28+4+1 = 16.
Solution
