MHB Evaluate a floor function involving trigonometric functions

[anemone]

Evaluate $$\left\lfloor{\tan^4 \frac{3\pi}{7}+\tan^4 \frac{2\pi}{7}+2\left(\tan^2 \frac{3\pi}{7}+\tan^2 \frac{2\pi}{7}\right)}\right\rfloor$$.

Hi MHB,

I don't know how to solve the above problem, as I have exhausted all possible methods that I could think of, and I firmly believe there got to be an easy way to crack it because this is a competition problem...any help, please

 

[Ackbach]

Re: Evaluate a floor function involves of trigonometric functions

Hmm. That is, indeed, a tough one. The answer, according to my calculator, is $412$, but how to get that? I'm thinking esoteric trig identities are the way to go. Here's one that might be useful:
$$\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right)=\frac18.$$
Here's another:
$$\prod_{k=1}^{m}\tan\left(\frac{k\pi}{2m+1}\right)=\sqrt{2m+1}.$$
Fleshing this out for your case yields
$$\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right)=\sqrt{7}.$$
We can combine these two together to get
$$\sin\left(\frac{\pi}{7}\right)\sin\left(\frac{2\pi}{7}\right)\sin\left(\frac{3\pi}{7}\right)=\frac{\sqrt{7}}{8}.$$
Hmm. Squaring some of your expressions looks like we might be able to do something here.

I also noticed that there's a perfect square trinomial pattern hidden in your original expression:
\begin{align*}
&\left\lfloor\tan^4\left(\frac{3\pi}{7}\right)+\tan^4\left(\frac{2\pi}{7}\right)+2\left(\tan^2\left(\frac{3\pi}{7}\right)+\tan^2\left(\frac{2\pi}{7}\right)\right)\right\rfloor \\
=&\left\lfloor\left(\tan^2\left(\frac{3\pi}{7}\right)+1\right)^{\!2}+\left(\tan^2\left(\frac{2\pi}{7}\right)+1\right)^{\!2}
-2\right\rfloor \\
=&\left\lfloor \sec^4\left(\frac{3\pi}{7}\right)+\sec^4\left(\frac{2\pi}{7}\right)-2\right\rfloor.\end{align*}

I'm not sure where to go from here; does this give you any ideas?

 

[anemone]

Thanks so much Ackbach for your reply!

I will think of it based on your observations and hopefully I can crack it soon and when I have done so, I sure will post back...it may take a while as I am very, very busy these days...