AN EQUATION THAT, SURPRISING, IS NOT GRADE SEVEN ...

 In the book https://ogeometrie-cip.blogspot.com/2021/02/getting-to-know-new-problem-book.html

see PROBLEM 28, at page 115.

           I have not found a solution to this problem.

           I had to look at the solution given in the book. On page 120 it says just that...

(Long after that I found out about the author's solution: see below, in the contribution of Silviu Boga)

 
       I posted the problem on several forums, let's see what consequences it will have...

 Posted on Facebook, 2/19/2021 https://www.facebook.com/petre.ciobanu.5621/posts/708103799874909 

 
 Posted on AOPS, 2/20/2021 https://artofproblemsolving.com/community/c4h2458976_equation_for_sinpi14

 

 

SOLUTION CiP

Answer CiP 

 The roots of the polynomial are    $sin\frac{\pi}{14},\;\;-sin\frac{3\pi}{14}\;\;and\;\;sin\frac{5\pi}{14}$.

Or, complementarily expressed    $cos\frac{\pi}{7},\;\;-cos\frac{2\pi}{7}\;\;and\;\;cos\frac{3\pi}{7}$.

           

           Solution

                     

          According to de Moivre's formula

$(cos \alpha +\imath \cdot sin \alpha)^{7}=cos 7 \alpha+\imath \cdot sin 7\alpha,\;(cos \alpha-\imath \cdot sin \alpha)^{7}=cos 7 \alpha-\imath \cdot sin 7\alpha$

express $sin7\alpha=\frac{(cos\alpha+\imath \cdot sin \alpha)^{7}-(cos\alpha-\imath \cdot sin\alpha)^{7}}{2\imath}$. We develop $(cos \alpha \pm \imath \cdot sin\alpha)^{7}$ with Newton's binomial theorem , considering that $\imath ^{2}=\imath ^{6}=-1,\;\imath ^{3}=\imath ^{7}=-\imath,$

$\imath ^{4}=1,\;\imath ^{5}=\imath$, and we obtain  $\sin7\alpha=$

$=\frac{1}{2\imath}\cdot [cos^{7}\alpha+7\imath\; cos^{6}\alpha sin \alpha-21 cos^{5}\alpha sin^{2}\alpha-35\imath\; cos^{4}\alpha sin^{3} \alpha +35cos ^{3} \alpha sin^{4} \alpha +21 \imath\; cos^{2} \alpha sin^{5} \alpha -7 cos \alpha sin^{6} \alpha-$

$-\imath \; sin^{7} \alpha -(cos^{7}\alpha-7\imath\; cos^{6}\alpha sin \alpha -21 cos^{5}\alpha sin^{2}\alpha-35 \imath\; cos^{4}\alpha sin^{3}\alpha +35 cos^{3}\alpha sin^{4} \alpha -21 \imath \;cos^{2}\alpha sin^{5} \alpha -$

$-7 cos \alpha sin^{6} \alpha +\imath \; sin^{7} \alpha )]$. 

At the result    $sin 7\alpha=7cos^{6} sin \alpha -35 cos^{4} \alpha sin^{3} \alpha +21 cos^{2}\alpha sin^{5} \alpha -sin^{7} \alpha$

we replace further $cos^{2}\alpha=1-sin^{2}\alpha, \;\;cos^{4} \alpha =1-2sin^{2}\alpha +sin^{4}\alpha,\;$

$\;cos^{6}\alpha=1-3sin^{2}\alpha +3sin^{4}\alpha-sin^{6} \alpha$, and after other calculations we obtain 

(1)          $sin 7\alpha =-64sin^{7}\alpha+112sin^{5}\alpha-56sin^{3}\alpha +7sin \alpha$.

(See Bromwich's formulas (24) in Weisstein, Eric W. "Multiple-Angle Formulas.")

         We put in the formula (1) $\alpha=\frac{\pi}{14}$. Since $sin 7\cdot \frac{\pi}{14}=sin \frac{\pi}{2}=1$, we obtain that $x=sin \frac{\pi}{14}$ is the root of the seventh degree equation

(2)                     $64 \cdot x^{7}-112 \cdot x^{5}+56\cdot x^{3}-7\cdot x+1=0$.

           IF WE ARE LUCKY ENOUGH, we can factorize the polynomial on the left, and find the form equivalent to the equation (2)

$(x+1)\cdot (8\cdot x^{3}-4\cdot x^{2} -4 \cdot x+1)^{2}=0$.

 
          From the above line, it is clear to whom $x=sin\frac{\pi}{14}$ is the root.

$\blacksquare$ 

  ..................................................................................................................................

                REMARK 1

         We write formula (1) in the form

(3)          $sin7\alpha-1=-(sin\alpha+1)(8sin^{3}\alpha-4sin^{2}\alpha-4sin\alpha+1)^{2}$.

The values for which $sin7\alpha-1=0$ are $\alpha_{k}=\frac{\pi}{14}+k\cdot \frac{2\pi}{7},\;\;k \in \mathbb{Z}$. So

(4)          $(sin \alpha_{k}+1)(8sin^{3}\alpha_{k}-4sin^{2}\alpha_{k}-4sin \alpha_{k}+1)^{2}=0$.

        We analyze only cases $k\in \left \{0,\;1,...,6\;\right \}$ because $\alpha_{k+7m}\equiv \alpha_{k}\;\;(mod\;2\pi)$.

     The value $\alpha_{5}=\frac{\pi}{14}+\frac{10\pi}{7}=\frac{21\pi}{14}=\frac{3\pi}{2}$ corresponds to the case $sin\alpha_{5}=-1$, which makes zero the first factor in the  relation  (4). Further,

$sin\alpha_{0}=sin\alpha_{3}$, because $\alpha_{0}+\alpha_{3}=\frac{\pi}{14}+\frac{13\pi}{14}=\pi$,

$sin\alpha_{1}=sin\alpha_{2}$, because $\alpha_{1}+\alpha_{2}=\frac{5\pi}{14}+\frac{9\pi}{14}=\pi$,

$sin\alpha_{4}=sin\alpha_{6}$, because $\alpha_{4}+\alpha_{6}=\frac{17\pi}{14}+\frac{25\pi}{14}=3\pi$.

     So the given polynomial $8\cdot x^{3}-4\cdot x^{2}-4\cdot x+1$ has three different roots $sin\alpha_{0}=\sin\frac{\pi}{14},\;sin\alpha_{1}=sin\frac{5\pi}{14},\;sin\alpha_{4}=sin\frac{17\pi}{14}=sin(\pi+\frac{3\pi}{14})=-sin\frac{3\pi}{14}$. We get the first answer.

Observing that $\frac{\pi}{2}-\alpha_{0}=\frac{3\pi}{7},\;\frac{\pi}{2}-\alpha_{1}=\frac{\pi}{7},\;\frac{\pi}{2}-\frac{3\pi}{14}=\frac{2\pi}{7}$, we can still express the answer $x_{1}=cos\frac{\pi}{7},\;x_{2}=-cos\frac{2\pi}{7},\;x_{3}=cos\frac{3\pi}{7}$.

(End Rem #1)

                REMARK 2

           Writing, for the three roots found, some of Viete's relationships, we have identities

 (5)          $sin\frac{\pi}{14}-sin\frac{3\pi}{14}+sin\frac{5\pi}{14}=\frac{1}{2}$,

(6)          $cos\frac{\pi}{7}-cos\frac{2\pi}{7}+cos\frac{3\pi}{7}=\frac{1}{2}$,

(7)          $sin\frac{\pi}{14}\cdot sin\frac{3\pi}{14}\cdot sin\frac{5\pi}{14}=\frac{1}{8}$,

(8)          $cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7}\cdot cso\frac{3\pi}{7}=\frac{1}{8}$.

(End Rem #2)

$\bigstar$     $\bigstar$     $\bigstar$

 P.S.  Here ENDS the solitary adventure, starting from the attempt to solve the problem in the statement. I followed the path suggested by the indication in the book.With the price of a substantial number of scribbled sheets of paper, I am still glad that I solved more than was required:I determined ALL the roots of the given equation.

$\bigstar$      $\bigstar$      $\bigstar$

 

           During the days I worked on this problem, I was stimulated by the help of "fellow mathematicians" who sent me ideas or even other methods to solve this problem. I mention them below, trying not to miss any. My thanks to everyone.

 

Feb 20, 2021, 6:23 PM  AOPS- the user vanstraelen suggested the development of $sin7\alpha$ in terms of $sin\alpha$.  He gets the formula (1) which he writes in the form (3).

(It all happened 65 minutes after I posted the question. Others here did not answe.)

 

Now, about those on Facebook.

Andreea Schier [19, Feb at 23:03] put up a photo

 I kind of twisted my neck, then thought about rotating it

     She chooses, as was most normal, to calculate the value of the polynomial $8x^{3}-4x^{2}-4x+1$ for $x=sin\frac{\pi}{14}$.

     It goes without saying that he uses the formula $8\cdot sin^{3}\alpha =6\cdot sin\alpha -2 \cdot sin3\alpha$ (obtained from $sin 3\alpha=3\cdot sin \alpha-4 \cdot sin^{3} \alpha$). However, it mentions the formula $sin^{2} \alpha=\frac{1-cos 2 \alpha}{2}$ as well as a transition from the cosine to the sine of the complementary angle. An expression is reached in the sinuses of various multiples of $\frac{\pi}{14}$, but only at the first power:

$S=2\cdot sin\frac{\pi}{14}-2\cdot sin\frac{3\pi}{14}+2\cdot sin \frac{5\pi}{14}-1$.

 She proves that $S=0$, calculating for this $S \cdot sin\frac{\pi}{7}$ ( it transforms sine products into sums with the formula $2\cdot sin \alpha \cdot sin \beta=cos(\alpha-\beta)-cos(\alpha+\beta)$ and all terms are canceled).

      I notice that, by the same calculation as hers, a demonstration of the Viete type relationship (5) is obtained.

 $\bigstar$

          In her comment  Marcelina Popa  starts from the equation $sin 4\alpha=sin 3\alpha$. This equation has in the open interval $(\;0,\;\pi)$ only the solutions $\alpha_{k}=(2k+1)\cdot \frac{\pi}{7}\;\;k=0,\;1,\;2$.

Equivalently, the equation is written successively (remember that in the interval $(\;0,\;\pi)$)

$4\cdot sin\alpha \cdot cos\alpha\cdot cos2\alpha=sin\alpha\cdot (3-4sin^{2}\alpha)$

$\overset{sin\alpha \neq 0}{\Leftrightarrow} 4\cdot cos \alpha\cdot (2cos^{2}\alpha-1)=3-4(1-cos^{2}\alpha)$

$\Leftrightarrow \;8\cdot cos^{3}\alpha-4\cdot cos^{2}\alpha-4\cdot cos \alpha+1=0$.

As a result, the equation  $8x^{3}-4x^{2}-4x+1=0$ is verified for $cos\frac{\pi}{7},\;cos\frac{3\pi}{7}=sin\frac{\pi}{14},\;cos\frac{5\pi}{7}$.

$\bigstar$

          Another calculation-based answer gave Dumitru Pietreanu. He starts from equality

$sin(\frac{3\pi}{14}+\frac{4\pi}{14})=sin\frac{\pi}{2}$,

which he develops using known formulas, finding that $x=sin\frac{\pi}{14}$ is the root of a polynomial $Q(x)$ and this can be expressed as $Q(x)=P^{2}(x)\cdot (x+1)$ where $P(x)$ is the given polynomial.

      Indeed, after the first step $sin\frac{3\pi}{14}\cdot cos\frac{4\pi}{14}+cos\frac{3\pi}{14}\cdot sin\frac{4\pi}{14}=1$. If $x=sin\frac{\pi}{14}$, then $cos^{2}\frac{\pi}{14}=1-x^{2},\;sin\frac{2\pi}{14}=2x\cdot cos\frac{\pi}{14},\;cos\frac{2\pi}{14}=1-2x^{2},$

$\;sin\frac{3\pi}{14}=3x-4x^{3},\;cos\frac{3\pi}{14}=cos\frac{\pi}{14}\cdot (4cos^{2}\frac{\pi}{14}-3)=cos\frac{\pi}{14}\cdot [4(1-x^{2})-3]=...,$

$sin\frac{4\pi}{14}=4x\cdot cos \frac{\pi}{14}\cdot cos\frac{2\pi}{14}=...,\;cos\frac{4\pi}{14}=2cos^{2}\frac{2\pi}{14}-1=...$.

In the end it results $-64x^{7}+112x^{5}-56x^{3}+7x=1$, so just the formula (2).

      He also congratulated Marcel Tena for his contribution.

 

$\bigstar$

 
          The one who found the mentioned book from other sources, also identified the solutions (two) of the author, on page 296, is

          Silviu Boga

Sufleor la pag. 296 http://poincare.matf.bg.ac.rs/.../Polynomials_EJBarbeau.pdf

 

          The first solution exploits the fact that $cos \frac{4\pi}{7}+\imath \cdot sin \frac{4\pi}{7} \neq 1$ verifies the equation $z^{7}=1$ (they are all $cos (k\cdot  \frac{2\pi}{7})+\imath \cdot sin (k\cdot\frac{2\pi}{7}), \;k=0,\;1,...,6$.

  !!! There was a small mistake, probably transcription, where I circled in the picture: it should be written   "u"  instead of  "1".  

          The second solution is similar to ours. 

$\bigstar$

 

          Ghimisi Dumitrel used the complex number $\zeta=cos\frac{\pi}{14}+\imath\cdot sin\frac{\pi}{14}$. We have $\zeta^{7}=\imath$. But $\zeta^{7}-\imath=\zeta^{7}+\imath^{7}=(\zeta+\imath)(\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1)$

$\Rightarrow \;\zeta^{6}-\zeta^{5}\imath-\zeta^{4}+\zeta^{3}\imath+\zeta^{2}-\zeta\imath -1=0$.

 With $sin\frac{\pi}{14}=\frac{\zeta-\bar{\zeta}}{2}=\frac{\zeta^{2}-1}{2\imath z}$ is calculated $8sin^{3}\frac{\pi}{14}-4sin^{2}\frac{\pi}{14}-4sin\frac{\pi}{14}+1=$

$=8\cdot (\frac{\zeta^{2}-1}{2\imath z})^{3}-4\cdot (\frac{\zeta^{2}-1}{2\imath z})^{2}-4\cdot \frac{\zeta^{2}-1}{2\imath z}+1=-\frac{1}{\imath \zeta^{3}}\cdot [(\zeta^{2}-1)^{3}-\imath \zeta (\zeta^{2}-1)^{2}+2\zeta^{2}(\zeta^{2}-1)-\imath \zeta^{3}]=$

$=-\frac{1}{\imath \zeta^{3}}\cdot [\zeta^{6}-\imath \zeta^{5}-\zeta^{4}+\imath \zeta^{3}+\zeta^{2}-\imath \zeta-1]=0$.

 $\bigstar$

          Doru GHERASA  presented two methods of solving.

          In the first solution, he observes that $sin\frac{\pi}{14}=cos(\frac{\pi}{2}-\frac{\pi}{14})=cos\frac{6\pi}{14}$, so $\frac{\pi}{14}$ is a solution of the equation

(9)                     $sin \alpha =cos 6\alpha$.

 Equation (9) is successively transformed into $sin \alpha = 4cos^{3}2\alpha-3cos2\alpha$

$\Leftrightarrow\;sin\alpha=4(1-2sin^{2}\alpha)^{3}-3(1-2sin^{2}\alpha)\;\Leftrightarrow x=4(1-2x^{2})^{3}-3(1-2x^{2})$, so $sin\frac{\pi}{14}$ is a solution of equation

(10)                   $32x^{6}-48x^{4}+18x^{2}+x-1=0$.

He notices the factorization of (10) 

$(8x^{3}-4x^{2}-4x+1)(x+1)(4x^{2}-2x-1)=0$

 and, since $0<sin\frac{\pi}{14}<\frac{1}{2}$, it can be the root only for the first factor.

 

          In the second method, we start from  de Moivre's formula

$(a+\imath b)^{7}\overset{def}{=}(cos\frac{\pi}{14}+\imath sin \frac{\pi}{14})^{7}=cos\frac{\pi}{2}+\imath sin\frac{\pi}{2}=\;\imath$. 

The imaginary part of the equality, developing on the left side with Newton's binomial theorem,  involves

$7a^{6}b-35a^{4}b^{3}+21a^{2}b^{5}-b^{7}=1\;\;\;\overset{a^{2}=1-b^{2}}{\Rightarrow}\;\;\; 64b^{7}-112b^{5}+56b^{3}-7b+1=0$, similar to (2). Then mention that the last equation can be written

$(8b^{3}-4b^{2}-4b+1)^{2}(b+1)=0.$ 

 

$\bigstar$ 

          AT LAST, BUT NOT LEST, we thanks Mr Marcel ȚENA. I reported the problem to him for the first time. He also had that book. He is, among many other things, the author of the book "RADACINILE UNITATII".

In a discussion with him, I asked his opinion on which cyclotomic polynomial between $\Phi_{14}$ or $\Phi_{7}$ to use. From his answers I realized that it is possible with both.

          The solution now has only a few lines:

          The cyclotomic polynomial $\Phi_{14}$ has, by definition, one of its roots 

$\zeta=cos(3\cdot \frac{2\pi}{14})+\imath \cdot sin(3\cdot \frac{2\pi}{14})=cos\frac{3\pi}{7}+\imath \cdot sin\frac{3\pi}{7}$.

But $\Phi_{14}(X)=\Phi_{7}( -X)$ (Theorem 16, page 24) and 7 being prime number we have $\Phi_{7}(X)=X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1$ (Theorem 11, page 20). So $\zeta$ is a root of polynomial $X^{6}-X^{5}+X^{4}-X^{3}+X^{2}-X+1$.

      So $\zeta$ verifies the equation

$\zeta ^{6}-\zeta ^{5}+\zeta ^{4}-\zeta ^{3}+\zeta ^{2}-\zeta+1=0$

$\Leftrightarrow \;\;(\zeta ^{3}+\frac{1}{\zeta ^{3}})-(\zeta ^{2}+\frac{1}{\zeta ^{2}})+(\zeta+\frac{1}{\zeta})-1=0$.

 We replace above $\zeta +\frac{1}{\zeta}=2cos\frac{3\pi}{7}=sin(\frac{\pi}{2}-\frac{3\pi}{7})=sin \frac{\pi}{14}=2x$,

 $\zeta ^{2}+\frac{1}{\zeta ^{2}}=(\zeta +\frac{1}{\zeta})^{2}-2=(2x)^{2}-2=4x^{2}-2$,

 $\zeta ^{3}+\frac{1}{\zeta ^{3}}=(\zeta +\frac{1}{\zeta})^{3}-3(\zeta +\frac{1}{\zeta})=(2x)^{3}-3(2x)=8x^{3}-6x$

and we get the equation  $8x^{3}-4x^{2}-4x+1=0$.

$\blacksquare\;\blacksquare$