6
$\begingroup$

My question is:

If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+\tan^2\frac{5\pi}{11}=Y$ then find $5X^2-Y$.

I couldn't find any way to simplify it. Please help. Thanks.

$\endgroup$
  • $\begingroup$ Calculator shows that $5X^2=55$, $Y=55$, so the answer is $0$. $\endgroup$ – Mythomorphic Jun 4 '15 at 9:41
  • $\begingroup$ Is there any proper procudure to solve this? $\endgroup$ – Pratyush Jun 4 '15 at 9:43
  • $\begingroup$ I don't know...I'm trying $\endgroup$ – Mythomorphic Jun 4 '15 at 9:44
  • $\begingroup$ May help , math.stackexchange.com/questions/11246/… $\endgroup$ – Mann Jun 4 '15 at 9:46
  • 4
    $\begingroup$ In general, $$\prod_{k=1}^n \tan\bigg(\frac{k\pi}{2n+1}\bigg) ~=~ \sqrt{2n+1}~,$$ and $$\sum_{k=1}^n\tan^2\bigg(\frac{k\pi}{2n+1}\bigg) ~=~ n(2n+1).$$ $\endgroup$ – Lucian Jun 4 '15 at 9:59
10
$\begingroup$

The main building block of our solution will be the formula \begin{align*}\prod_{k=0}^{N-1}\left(x-e^{\frac{2k i\pi}{N}}\right)=x^N-1.\tag{0} \end{align*} It will be convenient to rewrite (0) for odd $N=2n+1$ in the form \begin{align*} \prod_{k=1}^{n}\left[x^2+1-2x\cos\frac{\pi k}{2n+1}\right]=\frac{x^{2n+1}-1}{x-1}. \tag{1} \end{align*} Replacing therein $x\leftrightarrow -x$ and multiplying the result by (1), we may also write \begin{align*} \prod_{k=1}^{n}\left[\left(x^2-1\right)^2+4x^2\sin^2\frac{\pi k}{2n+1}\right]=\frac{1-x^{4n+2}}{1-x^2}. \tag{2} \end{align*}

  1. Setting in (1) $x=-i$, we get $$\left(2i\right)^n\prod_{k=1}^n\cos\frac{\pi k}{2n+1}=\frac{i^{2n+1}-1}{i-1} \qquad \Longrightarrow\qquad \prod_{k=1}^n2\cos\frac{\pi k}{2n+1}=1.$$

  2. Setting in (2) $x=1$ and computing the corresponding limit on the right, we get $$\prod_{k=1}^n2\sin\frac{\pi k}{2n+1}=\left[\lim_{x\to 1}\frac{1-x^{4n+2}}{1-x^2}\right]^{\frac12}=\sqrt{2n+1}.$$

  3. Combining the two results yields $$\boxed{\quad\prod_{k=1}^n\tan\frac{\pi k}{2n+1}=\sqrt{2n+1}\quad}$$ and to find $X$, it suffices to set $n=5$.

  4. To find $Y$, let us rewrite (1) in the form (set $x=-e^{i\gamma}$) $$\prod_{k=1}^n \left[2\cos\gamma+2\cos\frac{\pi k}{2n+1}\right]=\frac{\cos\frac{\left(2n+1\right)\gamma}{2}}{\cos\frac{\gamma}{2}}$$ Taking the logarithm and differentiating twice with respect to $\gamma$, we find $$\sum_{k=1}^{n}\frac{1}{\left(\cos\gamma+\cos\frac{\pi k}{2n+1}\right)^2} =-\frac{1}{\sin\gamma}\frac{\partial}{\partial \gamma}\left(\frac{1}{\sin\gamma}\frac{\partial}{\partial \gamma}\ln \frac{\cos\frac{\left(2n+1\right)\gamma}{2}}{\cos\frac{\gamma}{2}}\right).\tag{3}$$

  5. Computing the right side of (3) and setting therein $\gamma=\frac{\pi}{2}$, we finally arrive at $$\sum_{k=1}^{n}\frac{1}{\cos^2\frac{\pi k}{2n+1}}=2n(n+1)\qquad \Longrightarrow\quad \boxed{\quad\sum_{k=1}^{n}\tan^2\frac{\pi k}{2n+1}=n(2n+1)\qquad}$$ This yields $Y=55$.
| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Like Roots of a polynomial whose coefficients are ratios of binomial coefficients,

$$\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$

If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is ay integer

$\implies x=\dfrac{r\pi}{2n+1}$ where $r\equiv-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$

So, the roots of $\displaystyle\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots+(-1)^{n-1}\binom{2n+1}{2n-1}\tan^{2n-1}x+(-1)^n\tan^{2n+1}x=0$ $\displaystyle\iff\tan^{2n+1}x-\binom{2n+1}2\tan^{2n-1}x+\cdots+(-1)^n(2n+1)\tan x=0$

are $\tan x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$

So, the roots of $\displaystyle\tan^{2n}x-\binom{2n+1}2\tan^{2n-2}x+\cdots+(-1)^n(2n+1)=0$

are $\tan x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$

As $\tan(-A)=-\tan A,$

the roots of $\displaystyle t^nx-\binom{2n+1}2t^{n-1}x+\cdots+(-1)^n(2n+1)=0\ \ \ \ (1)$

are $\tan^2x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv1,2,\cdots,n\pmod{2n+1}$

Using Vieta's formula on $(1),$ $$\sum_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=\dfrac{\binom{2n+1}2}1=\cdots$$

and $$(-1)^n\prod_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=(-1)^n(2n+1)\iff\prod_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=(2n+1)$$

Now, $\tan\dfrac{r\pi}{2n+1}>0$ for $\dfrac\pi2>\dfrac{r\pi}{2n+1}>0\iff2n+1>2r>0$

$$\implies\prod_{r=1}^n\tan\dfrac{r\pi}{2n+1}=\sqrt{2n+1}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.