Sum Tangent $n^2$

Geometry Level 5

$\sum\limits_{n=1}^K \tan\left(\dfrac{n^2\pi}{12}\right) = 25$

Santa wants you to determine the value of $K$ to make the above equation true.

What is $K?$

The answer is 51.

2 solutions

Michael Huang
Nov 22, 2016

Finding the value of $\tan\left(\dfrac{n^2\pi}{12}\right)$

Case 1: $n$ is not a multiple of 2 or 3.
Since $n^ 2 \equiv 1 \pmod{4}$ and $n^2 \equiv 1 \pmod{3}$ , we get that $n^2 \equiv 1 \pmod{12}$ so $\tan\left(\dfrac{n^2\pi}{12}\right) = \tan\left(\dfrac{\pi}{12}\right)$ .

Using the tangent double angle formula $\tan(2x) = \dfrac{2\tan(x)}{1 - \tan^2(x)}$ , we set $x = \dfrac{\pi}{12}$ with $\tan(2x) = \dfrac{1}{\sqrt{3}}$ to conclude that

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

Solving for $\tan(x)$ , we obtain $\tan\left(\dfrac{\pi}{12}\right) = 2 - \sqrt{3}$ .

Case 2: $n$ is a multiple of 2 but not 3.
Since $n^2 \equiv 0 \pmod{4}$ and $n^2 \equiv 1 \pmod{3}$ , we get that $n^2 \equiv 4 \pmod{12}$ so $\tan\left(\dfrac{n^2\pi}{12}\right) = \tan\left(\dfrac{4\pi}{12}\right) = \tan \left( \dfrac{\pi}{3} \right) = \sqrt{3}$ .

Case 3: $n$ is a multiple of 3 but not 2.
Since $n^2 \equiv 1 \pmod{4}$ and $n^2 \equiv 0 \pmod{3}$ , we get that $n^2 \equiv 9 \pmod{12}$ so $\tan\left(\dfrac{n^2\pi}{12}\right) = \tan\left(\dfrac{9\pi}{12}\right) = \tan \left( \dfrac{3\pi}{4} \right) = -1$ .

Case 4: $n$ is a multiple of 2 and 3.
Since $n^2 \equiv 0 \pmod{4}$ and $n^2 \equiv 0 \pmod{3}$ , we get that $n^2 \equiv 0 \pmod{12}$ so $\tan\left(\dfrac{n^2\pi}{12}\right) = \tan 0 = 0$ .

Solution

Summarizing the above, we get that $\tan \frac{n^2 \pi } { 12 }$ is cyclic with period 6:
$\begin{array}{rlcrlr} & \text{First 6 terms} & & & \text{Next terms}\\ n = 1 &\rightarrow 2 - \sqrt{3} & & n = 7 &\rightarrow 2 - \sqrt{3}\\ n = 2 &\rightarrow \sqrt{3} & & n = 8 &\rightarrow \sqrt{3}\\ n = 3 &\rightarrow -1 & & n = 9 &\rightarrow -1\\ n = 4 &\rightarrow \sqrt{3} & & n = 10 &\rightarrow \sqrt{3}\\ n = 5 &\rightarrow 2 - \sqrt{3} & & n = 11 &\rightarrow 2 - \sqrt{3}\\ n = 6 &\rightarrow 0 && n = 12 &\rightarrow 0 & \\ \end{array}$

In each cycle of 6 terms, we get a total sum of $( 2 - \sqrt{3} ) + \sqrt{3} + (-1) + \sqrt{3} + 2 - \sqrt{3} + 0 = 3$ . Hence, we can conclude that the sum of the terms will be:

$\begin{array} { l l l } \text{ Number of terms} & \text{Sum } \\ 6k + 1 & 3k + 2 - \sqrt{3} \\ 6k+2 & 3k + 2 \\ 6k+3 & 3k+1 \\ 6k + 4 & 3k + 1 + \sqrt{3} \\ 6k+5 & 3k+3 \\ 6k+6 & 3k+3 \\ \end{array}$

Thus, to get a sum of $25 = 3 \times 8 + 1$ , we will need $k = 6 \times 8 +3 = 51$ terms.

I was fascinated to the see the periodicity in the values of $\tan \left(\frac{n^2 \pi}{ 12}\right)$ . I think the solution is well presented. Splitting into two sections of "Finding the values" and "Solution" makes it clear what exactly is happening in each section.

Pranshu Gaba - 4 years, 6 months ago

That is what I sometimes do when it comes to long solution. Since Calvin expected certain level of rigor in the solution and conciseness of the terminology, he displayed the formal version of mine, which is amazing. I wish I can figure that out as well as transforming deep mathematical concepts into best explanation. :)

The original solution was that the $\tan$ values at $6n, (1 + 6n), \cdots, (5 + 6n)$ for each nonnegative $n$ 's have the certain values. Instead, the newer version provides the deeper information with number theory and some abstract algebra. Before I joined Brilliant site, I wasn't into rigorous material and learning as I made through some courses that do not involve deep proofs. Since I find this site to be very useful and intriguing, I began to learn from other people, even if I already know the solutions. Honestly, I'm the solver more than being the writer. However, I like writing solutions as if I were the lecturer.

While thinking about other trigonometry ideas, I know that $\sin$ 's, $\cos$ 's and $\tan$ 's have certain periodicity because of their natures and number theory. The problem would be surprising if we were to determine the general pattern of the form, say: $\tan\left(\frac{n^k \pi}{z}\right)$ where $k$ and $z$ are strictly positive integers. I would like to see someone or anyone work this out.

It is obvious that if $z = 1$ , then we have a harmonic pattern with integers. What if suppose we have $z \geq 2$ ? Would there exist the certain pattern if $z$ is rational? :)

Michael Huang - 4 years, 6 months ago

Great problem! It is quite clear that anything of the form $\tan(n^k \pi/z)$ will be periodic, for $k,z$ positive integers, with the period, no greater than $z$ , but what is more interesting is the actual value of the period, which is determined by the distribution of the $k$ power residue class among $\{0,1,\cdots,\ z-1\}$ . Similar inferences can be made about the period and other related patterns if the distribution of the residue class for the system $q\cdot n^k \equiv r\pmod p$ is investigated when $z=p/q$ , a positive rational.

Samrat Mukhopadhyay - 4 years, 5 months ago

Very interesting problem! I was initially surprised to see that there was a pattern in $\tan \frac{ n^2 \pi } { 12 }$ .

I've cleaned up the solution in 2 ways
1. Explaining why the pattern exists, and how we can characterize it
2. Figuring out the sum for the number of terms, which allow for us to solve for the answer explicitly / rigorously.

Keep it up!

Calvin Lin Staff - 4 years, 6 months ago

Nice and neat solution. This time, I have a solution with more formal approach. I am not an expert in explaining things, but better keep on trying! ^.^

Thanks!

Michael Huang - 4 years, 6 months ago

It takes time and effort to express yourself clearly in a solution, so that others who couldn't solve the problem can quickly understand what you did. Keep on practicing!

Calvin Lin Staff - 4 years, 6 months ago

Niranjan Khanderia
Dec 7, 2016

\(\color{orange}{Tan\frac x 2=\sqrt{\dfrac{1-Cosx}{1+Cosx}}.\\
Let x=\frac \pi 6.\ \ \therefore\ Tan\frac \pi {12}=\sqrt{\dfrac{1-Cos\frac \pi 6}{1+Cos\frac \pi 6}}.\\
On\ rationalizing \ and\ simplifying\ we\ get \ Tan\frac \pi {12}=2 - \sqrt3.}\\
\color{blue}{Let\ us\ study\ the\ behavior\ of \ Tan\frac{n^2*\pi}{12}.\\
\begin {array} {} n & Tan\frac{n^2*\pi}{12} &=Tan\frac{..\pi}{....} & =.......\\
1& Tan\dfrac{\pi}{12} &= Tan\dfrac{\pi}{12} &= 2 - \sqrt3. \\
2& Tan\dfrac{4*\pi}{12} &= Tan\dfrac{\pi}{3} &= \ \ \ \sqrt3. \\
3& Tan\dfrac{9*\pi}{12} &= Tan\dfrac{3\pi}{4} &=\ \ \ -1 \\
4& Tan\dfrac{16*\pi}{12} &= Tan\dfrac{\pi}{3} &=\ \ \ \sqrt3. \\
5& Tan\dfrac{25*\pi}{12} &= Tan\dfrac{\pi}{12} &= 2 - \sqrt3. \\
6& Tan\dfrac{36*\pi}{12} &= Tan\dfrac{0*\pi}{12} &=\ \ \ 0. \\
7& Tan\dfrac{49*\pi}{12} &= Tan\dfrac{\pi}{12} &= 2 - \sqrt3. \\
8& Tan\dfrac{64*\pi}{12} &= Tan\dfrac{\pi}{3} & =\sqrt3. \\
9& Tan\dfrac{81*\pi}{12} &= Tan\dfrac{3\pi}{4} & = - 1. \\
10& Tan\dfrac{100*\pi}{12} &= Tan\dfrac{\pi}{3} &= \sqrt3. \\
11& Tan\dfrac{121*\pi}{12} &= Tan\dfrac{\pi}{12} &= 2 - \sqrt3. \\
12& Tan\dfrac{144*\pi}{12} &= Tan\dfrac{0*\pi}{12} &= 0. \\
13& Tan\dfrac{169*\pi}{12} &= Tan\dfrac{\pi}{12} & =2 - \sqrt3. \\
\end {array} } \\
\color{pink}{As\ seen\ from\ above\ Tan\dfrac{n^2*\pi}{12}\ is\ a\ periodic\ function,\ with\ period\ of\ 6.\\
We\ also\ note\ that \ for\ n^2\equiv\ m\ (mod\ 6),\\
for\ m=1\ and\ 5\ Tan\dfrac{m*\pi}{12}=2 - \sqrt3, \\
for \ m=2\ and\ 4\ Tan\dfrac{m*\pi}{12}=\ \ \ \ \sqrt3, \\
for\ m=3\ \ \ \ \ Tan\dfrac{m*\pi}{12}= - 1 \\
for \ m=6\ \ \ \ \ Tan\dfrac{m*\pi}{12}= 0 \\
So\ each\ period \ adds\ \ 2*(2 - \sqrt3)+2*\sqrt3 -1+0=3. \\
\therefore\ \ to\ obtain\ 25=24+1\ we\ need\ \dfrac{24} 3=8 \ periods\ of \ 6.\\
But\ integer\ 1\ remain.\\
Two\ more\ terms\ would \ add\ 2\ and\ three\ more\ terms\ would\ add\ 2 - 1=1.\ This\ is\ OK.\\
So\ 8\ periods\ of \ 6 +3=48+3=51.\\
\therefore\ K=}\large\ \ \ \color{red}{51}.\)

Sum Tangent n 2 n^2 n 2

The answer is 51.

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Sum Tangent $n^2$