Too small of an angle!

We have one value of tan(x). Hence we'll get two values tan(x/2). Similarly 4 values tan(x/4) and ultimately 8 values of tan(x/8) Thus we have a degree 8 equation in tan(x/8) as variable whose roots are the eight values of tan(x/8). So now, replace tan(x/8) with 'x' and obtain similar equation in 'x' Lastly put x=1 and get the answer....

$\tan(x)=2$

$\Rightarrow \frac {2\tan(\frac {x}{2})}{1-\tan^2(\frac {x}{2})}=2$

$\Rightarrow \tan (\frac {x}{2})=1-\tan^2(\frac {x}{2})$ ......for simplicity..let $\tan(\frac {x}{4})=K$

$\Rightarrow \frac {2K}{1-K^2}=1-\frac {4K^2}{(1-K^2)^2}$

$\Rightarrow K^4+2K^3-6K^2-2K+1=0$

again..let $\tan (\frac {x}{8})=L$ ..we get $K=\frac {2L}{1-L^2}$

also....

$(\frac {2L}{1-L^2})^4+2(\frac {2L}{1-L^2})^3-6(\frac {2L}{1-L^2})^2-2(\frac {2L}{1-L^2})+1=0$

$\Rightarrow (2L)^4+2(2L)^3(1-L^2)-6(2L)^2(1-L^2)-2(2L)(1-L^2)^3+(1-L^2)^4=0$

Clearly we can see a MONIC POLYNOMIAL with $\tan (\frac {x}{8})$ as a ROOT

$\boxed{f(1)=16}$