Finding equation in terms of variable $x$ whose roots are $\displaystyle \tan\left(\frac{(4n+1)\pi}{16}\right)$ for $n=0,1,2,3$
Try: $$\tan\bigg(n\theta\bigg)=\frac{\binom{n}{1}\tan\theta-\binom{n}{3}\tan^3\theta+\binom{n}{5}\tan^5\theta\cdots\cdots}{1-\binom{n}{2}\tan^2\theta+\binom{n}{4}\tan^4\theta\cdots}$$
Substitute $\displaystyle\bigg(\frac{(4n+1)\pi}{16}\bigg)=\theta$ and put $n=16$
We have $$\binom{16}{1}\tan \theta-\binom{16}{3}\tan^3\theta+\binom{16}{5}\tan^5\theta\cdots\cdots\cdots=0$$
Could some help me to solve it, thanks
Risky reusage of $n$
Put $n=4$ in $\tan n\theta$
where $4\theta=\dfrac{(4m+1)\pi}4, m=0,1,2,3$