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$\textbf{Question:}$ Prove that $$\frac{1}{2}+\cos\frac{2\pi}{11}+\cos\frac{4\pi}{11}=\frac{1}{4\sin\frac{\pi}{22}}$$

$\textbf{My Attempt:}$

Let $z= e^{\frac{i\pi}{22}}$ then the given relation will simplify out to $$z^{18}-z^{16}+z^{14}-z^{12}+z^{10}-z^9-z^{8}+z^{6}-z^4=0$$ $$z^{14}-z^{12}+z^{10}-z^{8}+z^{6}-z^5-z^{4}+z^{2}-1=0$$ Now I am not getting how to proceed further by this approach so I switched to another approach i.e. multiply out the whole expression and then open it using product of trigonometric ratio rules as follows:

$$2\sin\frac{\pi}{22}+2\left(\sin\frac{5\pi}{22}-\sin\frac{3\pi}{22}\right)+2\left(\sin\frac{9\pi}{22}-\sin\frac{7\pi}{22}\right)=1$$ which also seems like a dead end to me, so if anyone has insights on how to proceed further please help me out...

ACB
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sparrow_2764
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    Perhaps start using this: https://math.stackexchange.com/q/3597300/42969 – Martin R Apr 04 '23 at 13:35
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    The first approach can be made to work. In terms of $z=e^{\pi i/11}$ you have $2\cos(2\pi/11)=z^2+z^{-2}$, $2\cos(4\pi/11)=z^4+z^{-4}$ and $2\sin(\pi/22)=z^5+z^{-5}$. – Jyrki Lahtonen Apr 04 '23 at 13:41

4 Answers4

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Let $z = e^{i\pi/22}$, then following your first approach,

$$\begin{align*} &&\frac 12 + \cos \frac{2\pi}{11} + \cos \frac{4\pi}{11} &= \frac{1}{4\sin \frac\pi{22}}\\ &\iff &\frac12\left(z^0 + z^4 + z^{-4} + z^8 + z^{-8}\right) &= \frac{2i}{4\left(z-z^{-1}\right)}\\ &\iff &\left(z^{-7} + z^{-3} + z^1 + z^5+ z^9\right)-\left(z^{-9} + z^{-5} + z^{-1} + z^3+ z^7\right) - i &= 0\\ &\iff &-z^{-9}\cdot\frac{1-\left(-z^{2}\right)^{11}}{1-\left(-z^{2}\right)} &= 0\\ &\iff &-z^{-9}\cdot\frac{1+z^{22}}{1+z^{2}} &= 0\\ &\iff &-z^{-9}\cdot\frac{0}{1+z^{2}} &= 0\\ \end{align*}$$

Similar to your first approach, by converting the left-hand side to a geometric sum,

$$\begin{align*} LHS &= \frac 12 + \cos \frac{2\pi}{11} + \cos \frac{4\pi}{11}\\ &= \frac12\left(z^0 + z^4 + z^{-4} + z^8 + z^{-8}\right)\\ &= \frac{z^{-8}}2\cdot\frac{\left(z^4\right)^5-1}{z^4-1}\\ &= \frac{1}2\cdot\frac{z^{12}-z^{-8}}{\left(z^{2}-1\right)\left(z^2+1\right)}\\ &= \frac{1}2\cdot\frac{z^{11}-z^{-11}z^2}{\left(z-z^{-1}\right)\left(z^2+1\right)}\\ &= \frac{1}2\cdot\frac{i}{z-z^{-1}}\\ &= \frac{1}{2\cdot2\sin\frac{\pi}{22}}\\ &= RHS \end{align*}$$

peterwhy
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$$\frac{1}{2}＋\cos \left(\frac {2\pi}{11} \right)＋\cos \left(\frac {4\pi}{11} \right)＝\frac {\sin \left(\textstyle\frac {5\pi}{11} \right)}{2\sin \left(\textstyle\frac {\pi}{11} \right)}$$

And $$4\sin \left(\frac {\pi}{22} \right)\sin \left(\frac {5\pi}{11} \right)＝2 \left(\cos \left(\frac {9\pi}{22} \right)-\cos \left(\frac {\pi}{2} \right) \right)＝2\cos \left(\frac {9\pi}{22} \right)＝2\sin \left(\frac {\pi}{2}-\frac {9\pi}{22} \right)＝2\sin \left(\frac {\pi}{11} \right)$$

This implies that

$$\frac {\sin \left(\textstyle\frac {5\pi}{11} \right)}{2\sin \left(\textstyle\frac {\pi}{11} \right)}＝\frac {1}{4\sin \left(\frac {\pi}{22} \right)}$$

Bumblebee
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Moustapha_M_I
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Both ways are fine. Let us complete the computations.

(a) The complex number $$ z =\exp\frac{2\pi i}{44} $$ is a root of the cyclotomic polynomial $\Phi_{44}$ of order $44=2^2\cdot11$, which has degree $\varphi(44)=44\left(1-\frac 12\right)\left(1-\frac 1{11}\right)=20$, explicitly: $$ \Phi_{44}(x)= x^{20} - x^{18} + x^{16} - x^{14} + x^{12} - x^{10} + x^{8} - x^{6} + x^{4} - x^{2} + 1\ . $$ Note that $z^{11}=i$, and $z^{22}=-1$. Then $$ \begin{aligned} 2\cos\frac {2\pi}{11}&=2\cos\frac {4\cdot 2\pi}{44}=z^4+z^{-4}\ ,\\ 2\cos\frac {4\pi}{11}&=2\cos\frac {8\cdot 2\pi}{44}=z^8+z^{-8}\ ,\\ 2\sin\frac {2\pi}{44}&=\frac 1i(z-z^{-1}) =\frac 1{z^{11}}(z - z^{-1}) =\frac 1{z^{12}}(z^2 - 1) \ . \end{aligned} $$ Now note that the relation $0=\Phi_{44}(z)$ can be rewritten as: $$ 0=(z^2-1)(z^{18}+z^{14}+z^{10}+z^6+z^2)+1\ . $$ So we know the inverse of $(z^2-1)$. Collecting all data together: $$ \begin{aligned} \frac 1{\displaystyle 2\sin\frac {2\pi}{44}}=\frac{z^{12}}{z^2-1} &=-z^{12}(z^{18}+z^{14}+z^{10}+z^6+z^2) \\ & =z^8+z^4+1+z^{-4}+z^{-8} =1+2\cos\frac {2\pi}{11}+2\cos\frac {4\pi}{11} \ . \end{aligned} $$ We have used for instance $-z^{12}\cdot z^{18}=-z^{30}=z^{\pm 22}\cdot z^{30}=z^8$, and next terms follow with powers in arithmetic progression.

(b) Alternatively, use the relation $$ 2i\sin\frac{2k\pi}{44}=z^k-z^{-k} $$ to switch from the relation involving only sine-values to: $$ (z-z^{-1}) + (z^5-z^{-5}-z^3+z^{-3}) + (z^9-z^{-9}-z^7+z^{-7}) = i\ . $$ Multiply it by $z^9$ to get again a relation equivalent to $\Phi_{44}(z)=0$.

(On the left hand side we get an alternate sum of the powers $z^{18},z^{16},\dots,z^2,z^0$. On the right side $i\cdot z^9=z^{11}\cdot z^9=z^{20}$.)

dan_fulea
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Since no one else directly followed through on the comments that immediately followed the original posting:

To Prove: $~\displaystyle \frac{1}{2} + \cos(2\pi/11) + \cos(4\pi/11) = \frac{1}{4\sin(\pi/22)}.$

Let $~\alpha = \pi/11, ~z = e^{i\alpha}.$

Per the comment of Jyrki Lahtonen:

  • $2\cos(2\pi/11) = z^2 + z^{-2}.$

  • $2\cos(4\pi/11) = z^4 + z^{-4}.$

  • $2\sin(\pi/22) = 2\cos(10\pi/22) = 2\cos(5\pi/11) = z^5 + z^{-5}.$

Therefore, the original problem may be re-expressed as proving that:

$$\frac{1}{2} ~\left[ ~1 + \left(z^2 + z^{-2}\right) + \left(z^4 + z^{-4}\right) ~\right] = \frac{1}{2\left(z^5 + z^{-5}\right)}. \tag1 $$

The equation in (1) above is equivalent to showing that

$$1 = \left(z^5 + z^{-5}\right) \times ~\left[ ~1 + \left(z^2 + z^{-2}\right) + \left(z^4 + z^{-4}\right) ~\right] \iff $$

$$1 = \left(z^1 + z^{-1}\right) + \left(z^3 + z^{-3}\right) + \left(z^5 + z^{-5}\right) + \left(z^7 + z^{-7}\right) + \left(z^9 + z^{-9}\right) \iff $$

$$\frac{1}{2} = \cos(\alpha) + \cos(3\alpha) + \cos(5\alpha) + \cos(7\alpha) + \cos(9\alpha). \tag2 $$

So, the problem has been converted into showing that the equation in (2) above is true, when $~\alpha = (\pi/11).$

For $~\theta \in \Bbb{R}, ~n \in \Bbb{Z^+},~$
let $~\displaystyle f(\theta,n)~$ denote $~\displaystyle 1 + \sum_{k = 1}^n \cos(k\theta).$

Per the comment of Martin R, you have that

$$f(\theta,n) = \frac{1}{2} + \frac{\sin ~[ ~(2n + 1) ~\theta/2 ~]}{2\sin(\theta/2)}.$$

With $~\alpha = (\pi/11), ~$ let :

  • $T = \cos(\alpha) + \cos(3\alpha) + \cos(5\alpha) + \cos(7\alpha) + \cos(9\alpha).$

  • $R = f(\alpha,10).$

  • $S = f(2\alpha,5).$

Then

$$T = R - S$$

$$= \left\{ ~\frac{1}{2} + \frac{\sin ~[ ~21\pi/22 ~]}{2\sin(\pi/22)} ~\right\} - \left\{ ~\frac{1}{2} + \frac{\sin ~[ ~11\pi/11 ~]}{2\sin(\pi/11)} ~\right\}$$

$$= \left( ~\frac{1}{2} + \frac{1}{2} ~\right) - \left( ~\frac{1}{2} + 0 ~\right) = \frac{1}{2}.$$

user2661923
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