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Evaluate: $$\tan^2 \frac{\pi}{18}+\tan^2 \frac{5\pi}{18}+\tan^2 \frac{7\pi}{18}$$

Tried in various ways but to no avail ,

$$\tan\left(\frac{\pi}{2}-\frac{2\pi}{18}\right) = \tan^2 \frac{7\pi}{18}$$,
$$\tan\left(\frac{\pi}{2}-\frac{4\pi}{18}\right) = \tan^2 \frac{5\pi}{18}$$, $$\tan\left(\frac{\pi}{2}-\frac{8\pi}{18}\right) = \tan^2 \frac{\pi}{18}$$

Xander Henderson
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Longa Erick
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  • What identities did you try using? Is this part of a larger question that might have some relevant context? – ConMan Jul 03 '23 at 00:27
  • No , they only asked me to calculate the value of that , that is the whole question – Longa Erick Jul 03 '23 at 00:39
  • Well, Wolfram Alpha says the answer is 9, so the hard part is going to be proving that. I assume you have some set of trig identities that would provide a hint, but I don't know exactly how to approach this. – ConMan Jul 03 '23 at 00:56
  • I guess we can use the fomula $tanxtanytanz=tanx+tany+tanz$. – Itoz Darbien Jul 03 '23 at 01:05
  • But that only works if the angles add up to $\pi$? – Longa Erick Jul 03 '23 at 01:34
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    The following pair of identities, referenced in an older answer, may be useful: $$\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{\left(n-1\right)\left(2n - 1\right)}{3},\\sum _{k=1}^{m}\tan^2 \frac {k\pi } {2m+1}=m(2m+1)$$ Specifically, consider the cases $n=9$ and $m=4$ respectively. – Semiclassical Jul 03 '23 at 01:38
  • I tried to improve the question, as I can . – lone student Jul 03 '23 at 04:07
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    @lonestudent: Compare Guidelines for context edits and rewrites on meta, in particular “Do not expand the question with so much context that it changes fundamentally.” and “What sort of context can be edited into a question?” – Martin R Jul 03 '23 at 04:55
  • Well, my intention is not to harm the site, but to contribute to it. This edit is rare. I clearly didn't think this was homework. Added publicly observable and reasonable context to the question. If the moderators suspend me for this reason, I'll leave this comment in advance. Thanks . – lone student Jul 03 '23 at 05:05
  • @lonestudent Please do not use edits to hijack someone else's question. – Xander Henderson Aug 08 '23 at 17:11
  • @XanderHenderson What is "hijack" ? Google translate obviously failed . By saving the question from being PSQ, being closed, excessive downvotes, deletion, I have ensured that it will useful for future users. Moreover, the context I've added doesn't include attempts instead of OP, also it doesn't blatantly violate any policy. The context I am providing is the publicly observable context. Finally, please note that not all moderators around the world may agree with that action . – lone student Aug 08 '23 at 23:32
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    @lonestudent I do not see any comments in which the original asker of the question indicates that your edits are in line with their own thinking. Editing a question to add material which is not clearly the author's intent is a violation of the consensus standards of the site. Again, please read https://math.meta.stackexchange.com/q/32402/ . – Xander Henderson Aug 09 '23 at 00:13
  • With respect to the word "hijack", the relevant definition is (2) from https://www.merriam-webster.com/dictionary/hijack: "to take or take control of (something) as if by hijacking", as related to (1b) "to commandeer (a vehicle in transit)". Your edits take control of the question away from the original asker. Personally, I would prefer questions like this deleted (as this is essentially a homework problem), but the compromise that we have reached on this site is that we will tolerate homework questions if the askers of those questions provide context. – Xander Henderson Aug 09 '23 at 00:17
  • It is the asker's responsibility to provide that context. – Xander Henderson Aug 09 '23 at 00:17

4 Answers4

2

In this answer, we show that :

$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{\tan^2\frac{\pi}{18}+\tan^2\frac{5\pi}{18}+\tan^2\frac{7\pi}{18}=9}\end{align} $$

and

$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{\tan\frac{\pi}{18}\tan\frac{5\pi}{18}\tan\frac{7\pi}{18}=\frac {\sqrt 3}{3}}\end{align} $$

You can observe that,

$$ \begin{align}\left|\tan \frac{\pi}{6}\right|=\left|\tan \frac{5\pi}{6}\right|=\left|\tan \frac{7\pi}{6}\right|\end{align} $$

or

$$ \begin{align}\tan^2 \frac{\pi}{6}=\tan^2 \frac{5\pi}{6}=\tan^2 \frac{7\pi}{6}\end{align} $$

Then, let us recall the $\tan (3x)$ formula in terms of $\tan x$ :

$$\tan (3x)=\frac {3\tan x-\tan^3x}{1-3\tan^2x}$$

By squaring both side of the identity and substituting $\color{#c00}{\tan x=y}$ you have :

$$\tan^2(3x)=\frac {9y^2-6y^4+y^6}{1-6y^2+9y^4}$$

By setting $x_1=\dfrac {\pi}{18}$ and $x_2=\dfrac {5\pi}{18}$ and $x_3=\dfrac {7\pi}{18}$, then substituting $\color{#c00}{y^2=u}$, we see that $\tan^2(3x)=\dfrac 13$, where

$$ \begin{align}u_1&=\tan^2 \frac {\pi}{18}\\ u_2&=\tan^2 \frac {5\pi}{18}\\ u_3&=\tan^2 \frac {7 \pi}{18}\end{align} $$

which leads to the following :

$$ \begin{align}&\frac {9u-6u^2+u^3}{1-6u+9u^2}=\frac 13\\\\ \iff &3u^3-27u^2+33u-1=0\end{align} $$

Finally, by applying the Vieta's formulas , you reach the following result :

$$u_1+u_2+u_3=\dfrac {27}{3}=9\thinspace .$$

which is equivalent to :

$$ \begin{align}\tan^2\frac{\pi}{18}+\tan^2\frac{5\pi}{18}+\tan^2\frac{7\pi}{18}=9\end{align} $$

and since

$$ \begin{align}\tan\frac{\pi}{18}\tan\frac{5\pi}{18}\tan\frac{7\pi}{18}>0\end{align} $$

then again by Vieta's formulas we have :

$$ \begin{align}u_1u_2u_3&=\frac 13\\ \implies \sqrt {u_1u_2u_3}&=\frac {\sqrt 3}{3}\end{align} $$

which means that,

$$ \begin{align}\tan\frac{\pi}{18}\tan\frac{5\pi}{18}\tan\frac{7\pi}{18}=\frac {\sqrt 3}{3}\thinspace .\end{align} $$

Note that, you can also prove that some other identities . For instance, try to show that :

$$ \begin{align}\bbox[5px,border:2px solid #C0A000]{\tan^4\frac{\pi}{18}+\tan^4\frac{5\pi}{18}+\tan^4\frac{7\pi}{18}=59}\end{align} $$

lone student
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  • It uses triple angle identities for tangent function, and square it. Can this be generalized? – MathFail Jul 03 '23 at 06:28
  • @MathFail Generalization in which sense, exactly? To construct any possible explicit formula for $\tan^2a+\tan^2b+\tan^2c$ ? Or $$\begin{align}\tan^n\frac{\pi}{18}+\tan^n\frac{5\pi}{18}+\tan^n\frac{7\pi}{18}\end{align}$$ I like to find formulas but, I have no motivation to expand the answer but see .e.g.(which is trivial) $$\begin{align}\bbox[5px,border:2px solid #C0A000]{\tan^4\frac{\pi}{18}+\tan^4\frac{5\pi}{18}+\tan^4\frac{7\pi}{18}=59}\end{align} $$ By the way, I still see, my first boxed and last lines formula involve some typos (missing squaring) . Editing can make people tired . – lone student Jul 03 '23 at 09:04
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For example, for $(\cos(θ) + i\sin(θ))^6$

Real part $$\cos(6θ) = 32\cos^6(θ) - 48 \cos^4(θ) + 18 \cos^2(θ)-1$$

Imaginary part $$\sin(6θ)=32\cos^5(θ)\sin(θ)-32\cos^3(θ)\sin(θ)+6\cos(θ)\sin(θ)$$

Let the polynomial $64x^3-96x^2+36x=3$, if $x=\cos^2(θ)$ then $$\cos(6θ)=\frac{1}{2}$$

$$6θ=2k\pi\pm\frac{\pi}{2}$$ $$θ=\frac{1}{18}(6k\pm1)\pi$$

The roots are $\cos^2(\frac{\pi}{18}), \cos^2(\frac{5\pi}{18}), \cos^2(\frac{7\pi}{18})$

The equation whose roots are the reciprocals of teh above roots is $$3x^3-36x^2+96x-64=0$$

And the sum of roots is $$\sec^2(\frac{\pi}{18})+\sec^2(\frac{5\pi}{18})+\sec^2(\frac{7\pi}{18})=12$$

Your case is similar

jimbo
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Note that: $\tan \frac{5\pi}{18}=\cot\frac{2\pi}{9}$ and $\tan \frac{7\pi}{18}=\cot\frac{\pi}{9}$. Then we have

$$\begin{align} S&=\frac{1}{\cos^2\frac{\pi}{18}}-1+\frac{1}{\sin^2\frac{2\pi}9}-1 +\frac{1}{\sin^2\frac{\pi}9} -1\\ \\ &=\frac{4\sin^2\frac{\pi}{18}}{4\sin^2\frac{\pi}{18}\cos^2\frac{\pi}{18}}+\frac{1}{\sin^2\frac{2\pi}9} +\frac{1}{\sin^2\frac{\pi}9} -3\\ \\ &=\frac{1+4\sin^2\frac{\pi}{18}}{\sin^2\frac{\pi}{9}}+\frac{1}{\sin^2\frac{2\pi}9}-3\\ \\ &=\frac{4\cos^2\frac{\pi}{9}\left(1+4\sin^2\frac{\pi}{18}\right)}{4\cos^2\frac{\pi}{9}\sin^2\frac{\pi}{9}}+\frac{1}{\sin^2\frac{2\pi}9}-3\\ \\ &=\frac{1+4\cos^2\frac{\pi}{9}\left(1+4\sin^2\frac{\pi}{18}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{1+2(1+\cos\frac{2\pi}{9})\left[1+2(1-\cos\frac{\pi}{9})\right]}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{7+6\cos\frac{2\pi}{9}-4\cos\frac{\pi}{9}-4\cos\frac{\pi}{9}\cos\frac{2\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{7+6\cos\frac{2\pi}{9}-4\cos\frac{\pi}{9}-2(\cos\frac{\pi}{3}+\cos\frac{\pi}{9})}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{6+6\cos\frac{2\pi}{9}-6\cos\frac{\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{12\cos^2\frac{\pi}{9}-6\cos\frac{\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{12\cos\frac{\pi}{9}\left(\cos\frac{\pi}{9}-\frac12\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{12\cos\frac{\pi}{9}\left(\cos\frac{\pi}{9}-\cos\frac{\pi}{3}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{24\cos\frac{\pi}{9}\left(\sin\frac{2\pi}{9}\sin\frac{\pi}{9}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=12-3\\ \\ &=9 \end{align}$$

MathFail
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Let $x=(\tan 10^{\circ})^2$. Then $\cos 20^{\circ}=\frac{1-x}{1+x}$ and from $\cos60^{\circ}=4(\cos20^{\circ})^3-3\cos20^{\circ}$ we get $3x^3-27x^2+33x-1=0.$ Similarly for the other two.

Bob Dobbs
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