Does A Trigonometric Table Help?

Relevant wiki: Proving Trigonometric Identites

We have the multi-angle formula $\tan(nt)=\frac{\Im(1+i\tan t)^n}{\Re(1+i\tan t)^n}$ , a rational function $\frac{p(x)}{q(x)}$ in $x=\tan t$ . For odd $n$ , $p(x)=nx+...\pm x^n$ . The roots of $\frac{p(x)}{x}$ are $x_k=\tan\left(\frac{k\pi}{n}\right)$ for $k=1,..,n-1$ so that $\prod_{k=1}^{n-1}|\tan\left(\frac{k\pi}{n}\right)|=n$ by Viète. Now $\prod_{k=1}^{(n-1)/2}\tan\left(\frac{k\pi}{n}\right)=\sqrt{n}$ by symmetry. For $n=81$ it's $\boxed{9}$ .

Claim : $\displaystyle \prod_{k=1}^n \tan \left( \dfrac{ k \pi}{2n+1} \right) = \sqrt{2n+1}$ .

Proof : The crux of this proof will be the formula

$\prod_{k=0}^{N-1} \left( x- e^{2k i \pi /N} \right) = x^N - 1 .$

It will be convenient to rewrite for odd $N=2n+1$ in the form

$\prod_{k=1}^n \left [ x^2 + 1 - 2x \cos \left( \dfrac{\pi k}{2n+1} \right) \right ] =\dfrac{x^{2n+1} - 1}{x-1} \qquad \qquad (1)$

Replacing $x \leftrightarrow -x$ and multiplying the result by $(1)$ , we may also write

$\prod_{k=1}^n \left [ (x^2-1)^2 + 4x^2 \sin^2 \left( \dfrac{\pi k}{2n+1} \right) \right ] = \dfrac{1-x^{4n+2}}{1-x^2} \qquad \qquad (2)$

Set $x=i$ for $(1)$ , we get

$(2i)^n \prod_{k=1}^n \cos \left( \dfrac{\pi k }{2n+1} \right) = \dfrac{i^{2n+1} - 1}{i-1} \Rightarrow \prod_{k=1}^n 2 \cos \left( \dfrac{\pi k}{2n+1} \right) = 1 \qquad \qquad (3)$

Set $x=1$ into $(2)$ and completing the corresponding limit on the right, we get

$\prod_{k=1}^n 2\sin \left( \dfrac{\pi k }{2n+1} \right) = \left( \lim_{x\to 1 } \dfrac{1-x^{4n+2}}{1-x^2} \right)^{1/2} = \sqrt{2n+1 } \qquad \qquad (4)$

By dividing $(4)$ by $(3)$ , the results follows. $\blacksquare$

In this case, $n=40$ , hence our answer is $\sqrt{2n+1} = \sqrt{2\cdot40+1} = \boxed9$ .