proving $\csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8$

Question

How can I prove the following identity using complex variables $$ \begin{align*} 1) & \csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8 \\ 2) & \tan^2 \left( \frac{\pi}{16}\right) + \tan^2\left( \frac{3\pi}{16}\right) + \tan^2\left( \frac{5\pi}{16}\right)+ \tan^2\left( \frac{7\pi}{16}\right) = 28 \end{align*} $$ On earlier problem, I was given, $\displaystyle (z+a)^{2m}-(z-a)^{2m}=4maz \prod_{k=1}^{m-1} \left(z^2+a^2 \cot^2 \left(\frac{k\pi}{2m} \right )\right ) $ for integer $m>1$. I am not sure if I can use this is helpful. I am stumped please help.

Answer 1

$(1)$ Using this,

$\sin 7x=7s-56s^3+112s^5-64s^7$ where $s=\sin x$

If $\sin 7x=0,7x=n\pi$ where $n$ is any integer.

So, $x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$

So, the roots of $7s-56s^3+112s^5-64s^7=0$ are $\sin\frac{n\pi}7$ where $n=0,1,2,\cdots 5,6$

So, the roots of $64s^6-112s^4+56s^2-7=0$ are $\sin\frac{n\pi}7$ where $n=1,2,\cdots 5,6$

So, the roots of $64t^3-112t^2+56t-7=0$ are $\sin^2\frac{n\pi}7$ where $n=1,2,4$ or $3,5,6$

So, the equation whose roots are $\csc^2\frac{n\pi}7$ where $n=1,2,4$ or $3,5,6$ is $64\frac1{t^3}-112\frac1{t^2}+56\frac1t-7=0\iff 7t^3-56t^2+112t-7=0$

So, $\csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)$ is the sum of roots $=\frac{56}7=8$

$(2)$ $\cos2x=2c^2-1$ where $c=\cos x$

$\cos4x=2\cos^22x-1=2(2c^2-1)^2-1=8c^4-8c^2+1$

If $\cos4x=0,4x=(2m+1)\frac\pi2,x=\frac{(2m+1)\pi}8$ where $m=1,2,3,4$

So, the equation whose roots are $\cos\frac{(2m+1)\pi}8$ where $m=1,2,3,4$ is $8c^4-8c^2+1=0$

Now, as $$\cos2u=\frac{1-\tan^2u}{1+\tan^2u}\implies cos\frac{(2r+1)\pi}8=\frac{1-\tan^2\frac{(2r+1)\pi}{16}}{1+\tan^2\frac{(2r+1)\pi}{16}}$$

If $y=\tan^2\frac{(2r+1)\pi}{16},y=\frac{1-c}{1+c}\implies c=\frac{1-y}{1+y}$

So, $8c^4-8c^2+1=0$ becomes $$8\left(\frac{1-y}{1+y}\right)^4-8\left(\frac{1-y}{1+y}\right)^2+1=0$$ whose roots are $y=\tan^2\frac{(2r+1)\pi}{16}$ where $r=1,2,3,4$

or, $$8(y-1)^4-8(y-1)^2(y+1)^2+(y+1)^4=0$$

On simplification we get, $y^4-28y^3+52y^2-36y+1=0$

So, $\sum_{1\le r\le4}\tan^2\frac{(2r+1)\pi}{16}=\frac{28}1$

Answer 2

The relations can be generalized as $$\color{blue}{\sum_{k=1}^{n-1} \csc^2 \left( \frac{k \pi}{2n-1} \right ) = \frac{2}{3} n(n-1)} \tag{a}$$ $$\color{green}{\sum_{k=0}^{\lfloor \frac{n-1}{2}\rfloor } \tan^2 \left( \frac{(2k+1)\pi}{4n}\right ) = n (2n-1)} \tag{b}$$ Let us define a function by $$f(z) = (z+1)^{2n-1} - (z-1)^{2n-1} \tag{1}$$ Since the $(1)$ is a function in $2n-2$, there are $2n-2$ roots. So, let us calculate the roots of $(1)$ by setting $f(z)$ to zero. $$f(z) = (z+1)^{2n-1} - (z-1)^{2n-1} = 0 \\ \mathrm{or, } \left(\frac{z+a}{z-a}\right)^{2n-1} = 1 \tag{2}$$

Let, $\omega$ be the $(2n-1)^{th}$ root of unity, then we get $$\frac{z+1}{z-1} = \omega^k \, , 1\le k \le 2n-2 \tag{3}$$

Note that $\omega^{2m-1} = 1$ is not the root of $(1)$. Solving $(3)$, we get, $$z = - \left( \frac{ 1 + \omega^k}{1-\omega^k} \right )\, , 1\le k \le 2n-2 \tag{4}$$

Upon simplification, we get \begin{align*} z &= -\left ( \frac{ 1 + \cos \left( \frac{2\pi k}{2n-1} \right )+i\sin \left( \frac{2\pi k}{2n-1} \right )}{ 1 - \cos \left( \frac{2\pi k}{2n-1} \right )-i\sin \left( \frac{2\pi k}{2n-1} \right )} \right )\\ &= -\, i \cot\left( \frac{\pi k}{2n-1} \right ) \, , 1\le k \le 2n-2 \tag{5} \end{align*}

Therefore the roots of $(1)$ are given by $(5)$. Again using fundamental theorem of algebra, $(1)$ can be written as, \begin{align*} f(z) &= 2 \binom{2n-1}{1} \prod_{k=1}^{2n-2} \left(z+ i \cot\left( \frac{\pi k}{2n-1} \right ) \right ) \\ &= 2 (2n-1) \prod_{k=1}^{n-1} \left(z+ i\cot\left( \frac{\pi k}{2n-1} \right ) \right )\prod_{k=n}^{2n-2} \left(z- i \cot\left( \frac{ \pi (2n - 1 - k)}{2n-1} \right ) \right )\\ &= 2 (2n-1) \prod_{k=1}^{n-1} \left(z+ i \cot\left( \frac{\pi k}{2n-1} \right ) \right )\prod_{k=1}^{n-1} \left(z- i \cot\left( \frac{ \pi k }{2n-1} \right ) \right ) \\ &= 2 (2n-1) \prod_{k=1}^{n-1} \left(z^2+ \cot^2\left( \frac{\pi k}{2n-1} \right ) \right ) \tag{6} \end{align*}

Binomially expanding the $(1)$, we get \begin{align*} 2(2n-1) \prod_{k=1}^{n-1} \left(z^2+ \cot^2\left( \frac{\pi k}{2n-1} \right ) \right ) &= \sum_{k=0}^{2n-1} \binom{2n-1}{k} z^{2n-1-k} (1 - (-1)^k)\\ &= 2 \sum_{k=0}^{n-1} \binom{2n-1}{2k+1} z^{2n-2k-2} \tag{7}\\ \end{align*}

Comparing coefficients of $z^{2n-4}$ on $(7)$ we get, $$2 (2n-1) \sum_{k=1}^{n-1} \cot ^2 \left( \frac{ \pi k}{2n-1 } \right ) = 2 \binom{2n-1}{2n-4}$$

Or, which gives the result. $$ \boxed{ \large \color{red} {\sum_{k=1}^{n-1} \cot ^2 \left( \frac{ \pi k}{2n-1 } \right ) = \frac{(2n-2)(2n-3)}{6}}} \tag{8}$$

The equation $(8)$ gives $$\color{blue}{\sum_{k=1}^{n-1} \csc^2 \left( \frac{k \pi}{2n-1} \right ) = \frac{2}{3} n(n-1)} \tag{9}$$

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Similarly, we find $$\displaystyle (z+1)^{2n}-(z-1)^{2n}=4nz \prod_{k=1}^{n-1} \left(z^2+ \cot^2 \left(\frac{k\pi}{2n} \right )\right ) \tag{10}$$

Comparing the coefficients of $z^{2n-3}$ we find the relation that $$2 \binom{2n}{2n-3} = 4n \sum_{k=1}^{n-1} \cot^{2} \left( \frac{\pi k}{2n} \right ) \\ \mathrm{or, } \sum_{k=1}^{n-1} \cot^{2} \left( \frac{\pi k}{2n} \right ) = \frac{(2n-1)(2n-2)}{6} \tag{11} $$

Changing $n\to 2n $ and subtracting $(11)$ we get, $$\boxed{\color{green}{\sum_{k=0}^{\lfloor \frac{n-1}{2}\rfloor } \cot^2 \left( \frac{(2k+1)\pi}{4n}\right ) = \sum_{k=1}^{2n-1}\cot^2\left( \frac{k \pi}{4n} \right ) - \sum_{k=1}^{n-1}\cot^2\left( \frac{k \pi}{2n} \right ) = n (2n-1)}} \tag{12}$$

By changing $z \to \frac{\pi}{2}-z$, we get the required result.

Answer 3

I am going to answer the question 2 by grouping terms.

Firstly, we are going to evaluate the sum $\displaystyle \sum_{k=1}^{7} \tan ^{2}\left(\frac{k \pi}{16}\right). $

$\begin{aligned}\because \tan \frac{(8-k) \pi}{16} &=\tan \left(\frac{\pi}{2}-\frac{k \pi}{16}\right) =\frac{1}{\tan \frac{k\pi}{16}} \\\therefore \tan ^{2} \frac{k \pi}{16}+\tan ^{2} \frac{(8-k) \pi}{16}&=\tan ^{2} \frac{k \pi}{16}+\frac{1}{\tan ^{2} \frac{k \pi}{16}} \\&=\frac{\sin ^{4} \frac{k \pi}{16}+\cos ^{4} \frac{k \pi}{16}}{\cos ^{2} \frac{k \pi}{16} \cdot \sin ^{2} \frac{k \pi}{16}} \\&=\frac{1-2 \sin ^{2} \frac{k \pi}{16} \cos ^{2} \frac{k \pi}{16}}{\cos ^{2} \frac{k \pi}{16} \cdot \sin ^{2} \frac{k \pi}{16}} \\&=\frac{4}{\sin ^2\left(\frac{k \pi}{8}\right)}-2\end{aligned} \tag*{} $ $\begin{aligned} \sum_{k=1}^{7} \tan ^{2} \frac{k \pi}{16}=& \sum_{k=1}^{3}\left(\tan ^{2} \frac{k \pi}{16}+\tan ^{2} \frac{(8-k) \pi}{16}\right)+\tan ^{2} \frac{\pi}{4} \\=& \sum_{k=1}^{3}\left(\frac{4}{\sin ^{2} \frac{k \pi}{8}}-2\right)+1 \\=& 4 \sum_{k=1}^{3} \frac{1}{\sin ^{2} \frac{k \pi}{8}}-5\end{aligned} \tag*{} $

Now let’s evaluate the three terms one by one and get

$\displaystyle \sin ^{2}\left(\frac{\pi}{8}\right) =\frac{1-\cos \frac{\pi}{4}}{2} =\frac{1-\frac{\sqrt{2}}{2}}{2} =\frac{2-\sqrt{2}}{4} \tag*{} $ $\displaystyle \sin ^{2}\left(\frac{2 \pi}{8}\right)=\frac{1}{2} \tag*{} $ $\displaystyle \sin ^{2}\left(\frac{3 \pi}{8}\right) =\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{8}\right) =\cos ^{2} \frac{\pi}{8} =1-\frac{2-\sqrt{2}}{4} =\frac{2+\sqrt{2}}{4} \tag*{} $

Summing them up yields

$\displaystyle \sum_{k=1}^{7} \tan ^{2}\left(\frac{k \pi}{16}\right)=4\left(\frac{4}{2-\sqrt{2}}+2+\frac{4}{2+\sqrt{2}}\right)-5 =35 \tag*{} $ Moreover, $$\tan \left(\frac{2\pi}{16}\right)=\sqrt{2}-1 \Rightarrow \tan ^{2}\left(\frac{2\pi}{16}\right)=3-2 \sqrt{2} $$

$$\begin{aligned} \tan \left(\frac{6 \pi}{16}\right) &=\tan \left(\frac{\pi}{2}-\frac{\pi}{8}\right) =\frac{1}{\tan \frac{\pi}{8}} =\sqrt{2}+1 \end{aligned}\Rightarrow \tan ^{2}\left(\frac{6 \pi}{16}\right)=3+2 \sqrt{2}$$ Now we can conclude that

$$ \displaystyle \tan ^{2}\left(\frac{\pi}{16}\right)+\tan ^{2}\left(\frac{3 \pi}{16}\right)+\tan ^{2}\left(\frac{5 \pi}{16}\right)+\tan ^{2}\left(\frac{7 \pi}{16}\right)=35-(3-2\sqrt 2)-1-(3+2\sqrt 2)=28$$