MHB Evaluate a floor function involving trigonometric functions

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The discussion revolves around evaluating the expression involving trigonometric functions: $$\left\lfloor{\tan^4 \frac{3\pi}{7}+\tan^4 \frac{2\pi}{7}+2\left(\tan^2 \frac{3\pi}{7}+\tan^2 \frac{2\pi}{7}\right)}\right\rfloor$$. Participants suggest that the answer is 412 and explore various trigonometric identities to simplify the problem. Key identities discussed include relationships involving cosine and tangent products, which may help in deriving the solution. A perfect square trinomial pattern is identified, leading to a potential simplification using secant functions. The conversation emphasizes the complexity of the problem and the need for deeper exploration of trigonometric properties.
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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?:)
 
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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?
 
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...
 
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