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Too Small of an Angle (Posted on 2018-11-11) Difficulty: 3 of 5
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)?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 2 of 2 |
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.

  Posted by Brian Smith on 2018-12-16 21:50:19
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