Click on the image and use the password : ogeometrie .The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.
Click on the image and use the password : ogeometrie .The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.
The triangle has angle measures $2^{\circ}\;,\;\;89^{\circ}\;,\;\;89^{\circ}$
Solution CiP
If $x^{\circ},\;y^{\circ},\;z^{\circ}$ are the angles of this triangle, then :
$x^{\circ}\;,\;y^{\circ},\;z^{\circ}\;<90^{\circ}\;\;and\;\;x^{\circ}+y^{\circ}+z^{\circ}=180^{\circ} \tag{1}$
Being prime numbers, if they were all odd then their sum would be odd. We contradict (1). So one of the numbers, let's say $z^{\circ}$ is $2^{\circ}$. Then the second condition in (1) becomes :
$x^{\circ}+y^{\circ}=178^{\circ} \tag{2}$
An obvious solution for (2) , which meets the requirements of the problem, is $x^{\circ}=y^{\circ}=89^{\circ}$. This triangle is isosceles.
$x^{\circ}\underset{(1)}{<}89^{\circ}\Rightarrow y\underset{(2)}{=}178^{\circ}-x^{\circ}>89^{\circ}$. And we can still have $y^{\circ}=90^{\circ}$ , which violates all the requirements.
$\blacksquare$
In the GMB 7/2005, mentioned here, the solution to problem 25223 (published in GMB/1/2005, page 40) appears, which is the source of this post. (With nostalgia, I found on page 299 the name of my college colleague Constantin BUȘE...deceased in the meantime)
" 25223. To solve in $\mathbb{C}$ the equation : $3x^3-3x+1=0.$
[author : ] Gh. Szöllősy, Sighetul Marmației"
ANSWER
$x_1=\frac{2}{\sqrt{3}}\cos\frac{5\pi}{18}\;,\;x_2=\frac{2}{\sqrt{3}}\cos \frac{7\pi}{18}\;,\;x_3=\frac{2}{\sqrt{3}}\cos\frac{17\pi}{18}$
Solution
We are looking for solutions of the kind $x_t=a\cdot \cos t,\;a>0,\;t\in [0,\pi]$. Then
$3x_t^3-3x_t+1=0\Leftrightarrow 3a^3\cos^3x_t-3a\cos x_t+1=0 \tag{1}$
Because $\cos 3t=4\cos^3t-3\cos t$ we are looking to determine $a$ such that $\frac{3a^3}{4}=\frac{3a}{3}$ which leads to $a=\frac{2}{\sqrt{3}}.$ Thus we have
$\frac{2}{\sqrt{3}}=\frac{3a^3\cos^3x_t-3a\cos x_t}{4\cos^3 x_t-3\cos x_t}\underset{(1)}{=}\frac{-1}{\cos 3t}$
From $\cos 3t=-\frac{\sqrt{3}}{2}$ we obtain $3t \in \left \{\pm \frac{5\pi}{6}+2k\pi\right \}\cap [0,\pi]$, so $t\in \left \{\frac{5\pi}{18},\;\frac{7\pi}{18},\;\frac{17\pi}{18}\right\}$ hence the Answer.
$\blacksquare$
REMARK CiP Putting $x=\frac{2}{\sqrt{3}}\cdot y$ we see that $y$ are solutions of the equation
$3\cdot \frac{8}{3\sqrt{3}}y^3-3\cdot \frac{2}{\sqrt{3}}y+1=0\;\Leftrightarrow\;8y^3-6y+\sqrt{3}=0 \tag{2}$
that is $y_{1-3}\in \left \{\cos\frac{5\pi}{18},\;\cos\frac{7\pi}{18},\;\cos\frac{17\pi}{18}=-\cos\frac{\pi}{18}\right \}.$ We therefore have, from Vieta's Formulas :
$\cos\frac{\pi}{18}-\cos\frac{5\pi}{18}-\cos\frac{7\pi}{18}=0$
$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18}-\cos \frac{5\pi}{18}\cdot \cos \frac {7\pi}{18}+\cos \frac{\pi}{18}\cdot \cos \frac{7\pi}{18}=-\frac{3}{4}$
$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18} \cdot \cos \frac{7\pi}{18}=\frac{\sqrt{3}}{8}$
<end REM>
I wrote about the book here.
It's not for nothing that I wrote somewhere that THIS BOOK IS FULL OF SCRATCHES. (Maybe that's why the author withdrew Part I from sale)
We are now debating Problem 62***, page 157 (solved on pages 516-519).
" 62***. Calculate $P_n\; Q_n\;and\; R_n$, where $n\in \mathbb{N},\;n\geqslant 3,\;a\in\mathbb{R}^*\;:$
$$a)\;\;\;P_n=\prod_{k=0}^{n-1} \cos \left (a+\frac{k\pi}{n}\right )\;;$$
$$b)\;\;Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\;;$$
$$c)\;\;R_n=\prod_{k=0}^{n-1}\tan \left (a+\frac{k\pi}{n}\right )."$$
ANSWER
$a)\;\;P_{2n}=\frac{(-1)^n\sin 2na}{2^{2n-1}}\;\;;\;\;P_{2n+1}=\frac{(-1)^n\cos (2n+1)a}{2^{2n}}$
$b)\;\;Q_n=\frac{\sin na}{2^{n-1}}$
$c)\;\;R_{2n}=-1\;\;;\;\;R_{2n+1}=(-1)^n\tan (2n+1)a$
Solution CiP
We will start with b). A solution that, at some point, requires a square root extraction is found in TURTOIU Fanică - Probleme de Trigonometrie, Ed. Tehnică, București, 1979 (pages 86-87, Problem 2.38 - received by the kindness of Liviu PODGORNEI). This is a personal attempt.
Let $z_k=\cos\left (a+\frac{k\pi}{n}\right )+\imath \sin \left (a+\frac{k\pi}{n}\right )\;,\;\;0\leqslant k\leqslant n-1.$ We have
$|z_k|=1=z_k\cdot \bar z_k$ so $\bar z_k=\cos \left (a+\frac{k\pi}{n}\right )-\imath \sin \left (a+\frac{k\pi}{n}\right )=\frac{1}{z_k}.$ In addition
$z_k^{2n}=\cos 2n \left (a+\frac{k\pi}{n}\right )+\imath \sin 2n\left (a+\frac{k\pi}{n}\right )=\cos 2na+\imath \sin 2na \tag{1}$
Then
$\sin \left (a+\frac{k\pi}{n}\right )=\frac{z_k-\bar z_k}{2\imath}=\frac{z_k-\frac{1}{z_k}}{2\imath}=\frac{z_k^2-1}{2\imath z_k} \tag{2}$
From (1) it follows that the equation
$z^{2n}=\cos 2na+\imath \sin 2na \tag{3}$
has the roots $\pm z_k\;,\;\;k\in \{0,\;1,\dots ,\;n-1 \}.$ Denoting
$f(z)=z^{2n}-(\cos na +\imath \sin na) \tag{4}$
we have $f(z)=(z-z_0)(z-z_1)\dots (z-z_{n-1})(z+z_0)(z+z_1)\dots (z+z_{n-1})\;$, or
$$f(z)=\prod_{k=0}^{n-1}(z^2-z_k^2) \tag{5}$$
From here we go further $$\prod_{k=0}^{n-1}(z_k^2-1)=(-1)^n\prod_{k=0}^{n-1}(1-z_k^2)=(-1)^nf(1)\overset{(4)}{=}(-1)^n(1-\cos 2na-\sin 2na)=$$
$$=(-1)^n(2\sin^2na-2\imath \sin na \cos na)=(-1)^n(-2\imath^2\sin^2 na-2\imath \sin na \cos na)$$
so
$$\prod_{k=0}^{n-1}(z_k^2-1)=-2\imath (-1)^n\sin na(\cos na+\imath \sin na) \tag{6}$$
On the other hand
$$\prod_{k=0}^{n-1}z_k=\prod_{k=0}^{n-1}\left [(\cos a+\imath \sin a)\left (\cos \frac{k\pi}{n}+\imath \sin\frac{k\pi}{n}\right )\right]=$$
$$=(\cos a+\imath \sin a)^n\prod_{k=0}^{n-1}\left (\cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )=$$
$$=(\cos na+\imath \sin na)\left [\cos \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}+\imath \sin \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}\right ]=$$
$=(\cos na+\imath \sin na)\left [\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}\right ]$
because $\sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}.$ Analyzing the cases $n=4m,\;4m+1,\;4m+2,\;4m+3$ it is immediately seen that $\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}=-\imath^{n+1}$ , and then the previous calculation gives us
$$\prod_{k=0}^{n-1}z_k=-\imath^{n+1}(\cos na+\imath\sin na) \tag{7}$$
Now we can finish
$$Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\overset{(2)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2-1)}{2^n\imath^n\prod_{k=0}^{n-1}z_k}\overset{(6)}{\underset{(7)}{=}}\frac{-2\imath(-1)^n\sin na(\cos na+\imath\sin na)}{2^n\imath^n(-\imath^{n+1})(\cos na+\imath \sin na)}=$$
$=\frac{\imath \cdot (-1)^n\sin na}{2^{n-1}\imath^{2n}\cdot \imath}=\frac{(-1)^n \sin na}{2^{n-1}(-1)^n}=\frac{\sin na}{2^{n-1}}.$ We got the answer of b).
Remark CiP On page 518 the following stupid answer is given :
$$Q_n=\begin{cases}\frac{(-1)^n\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-even\\\frac{(-1)^{n+1}\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-odd\end{cases} \tag{Q}$$
............................................................................................................................
a) With the same notations as in point b), first we have
$\cos\left (a+\frac{k\pi}{n}\right )=\frac{z_k+\bar z_k}{2}=\frac{z_k^2+1}{2z_k} \tag{8}$
Then, starting from (5) we see that
$$f(\imath)=\prod_{k=0}^{n-1}(-1-z_k^2)=(-1)^n\prod_{k=0}^{n-1}(z_k^2+1)$$
so
$$\prod_{k=0}^{n-1}(z_k^2+1)=(-1)^nf(\imath) \tag{9}$$
Then we can write
$$P_n=\prod_{k=0}^{n-1}\cos \left (a+\frac{k\pi}{n}\right )\overset{(8)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2+1)}{2^n\prod_{k=0}^{n-1}z_k}\overset{(7)}{\underset{(9)}{=}}\frac{(-1)^nf(\imath)}{2^n(-\imath^{n+1})(\cos na+\imath \sin na)} \tag{10}$$
But $f(\imath)\underset{(4)}{=}(-1)^n-\cos 2na-\imath \sin 2na$ and we will treat the $n-even$ and $n-odd$ cases separately.
$\fbox{$n$-even}$
$f(\imath)=1-\cos na-\imath \sin na=2\sin^2 na-2\imath \sin na \cos na=$
$=-2\imath^2\sin^2 na-2\imath \sin na \cos na=-2\imath \sin na(\cos na+\imath \sin na)$
In (10) also appears $-\imath^{n+1}=-\imath^n\cdot \imath=(-\imath)(-1)^{\frac{n}{2}}$ and then we continue like this
$P_n=\frac{-2\imath \sin na(\cos na+\imath \sin na)}{2^n(-\imath)(-1)^{\frac{n}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n}{2}}\sin na}{2^{n-1}}.$
$\fbox{$n$-odd}$
$(-1)^nf(\imath)=-(-1-\cos 2na-\imath \sin 2na)=2\cos^2 na+2\imath \sin na \cos na=$
$=2\cos na(\cos na+\imath \sin na)$
and because $-\imath^{n+1}=-(-1)^{\frac{n+1}{2}}=-(-1)^{\frac{n-1}{2}+1}=-(-1)^{\frac{n-1}{2}}(-1)=(-1)^{\frac{n-1}{2}}$
we have $P_n\underset{(10)}{=}\frac{2\cos na(\cos na+\imath \sin na)}{2^n(-1)^{\frac{n-1}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n-1}{2}}\cos na}{2^{n-1}}.$ We got the answer of a).
Remark CiP Although in the solution I used COPILOT, who first told me incorrectly that the minus sign always appears in the formula, then came back and corrected the mistake, we must admit that in the book, on page 517, the correct answer is given. Marius DRĂGAN and Sorin RĂDULESCU are cited as the authors of the solution.
.............................................................................................................................
c) $R_{2n}=\frac{Q_{2n}}{P_{2n}}=\frac{\frac{\sin 2na}{2^{2n-1}}}{\frac{(-1)^n\sin 2na}{2^{2n-1}}}=(-1)^n\;;$
$R_{2n+1}=\frac{Q_{2n+1}}{P_{2n+1}}=\frac{\frac{\sin (2n+1)a}{2^{2n}}}{\frac{(-1)^n\cos (2n+1)a}{2^{2n}}}=(-1)^n\tan(2n+1)a.$
$\blacksquare\;\blacksquare\;\blacksquare$
See the full article in GMB 7/2005, pages 289-293.
First identity :
(is formula (5) , page 290, in the cited article)
$$\prod_{k=0}^{n-1}\cos \frac{x+2k\pi}{n}=\frac{1}{2^{n-1}}\left [\cos\frac{n\pi}{2}+(-1)^{n+1}\cdot \cos x\right ] \tag{1}$$
For $x=\pi$ we obtain :
$\cos \frac{\pi}{n}\cdot \cos \frac{3\pi}{n} \dots \cos\frac{(2n-1)\pi}{n}=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos \pi \right]=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^n\right ]$
i.e. Exercise 9a).
For $x=0$ we obtain :
$\cos \frac{2\pi}{n}\cdot \cos \frac{4\pi}{n}\dots \cos \frac{2(n-1)\pi}{n}=\frac{1}{2^{n-1}} \left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos 0\right ]=\frac{1}{2^{n-1}}\left [ \cos \frac{n\pi}{2}-(-1)^n\right ]$
i.e. Exercise 9b).
In particular we have the equalities :
$\cos \frac{\pi}{5}\cdot \cos \frac{3\pi}{5}\cdot \cos \frac{5 \pi}{5}\cdot \cos \frac{7\pi}{5}\cdot \cos \frac{9\pi}{5}=\frac{1}{2^4}\left [ \cos\frac{5\pi}{2}-1\right ]=-\frac{1}{16}\;;$
$\cos \frac{2\pi}{5}\cdot \cos \frac{4\pi}{5}\cdot \cos \frac{6\pi}{5}\cdot \cos\frac{8\pi}{5}=\frac{1}{2^4}\left [\cos \frac{5\pi}{2}+1\right ]=\frac{1}{16}.$
(in construction)
(O încercare de regăsire a profesorului meu din Liceu, Vasile BIVOLARU. A plecat in GERMANIA, sotia, Ecaterina, a murit acolo - mi-a spus fiul lor, Cristian, pe Facebook)
Scanning, little by little, my collection of "Gazeta Matematica" magazines (aka GMB), I reached issue 6 from 1978 where the Topics from the 1977 Final Stage are published, the only one I participated in, in 10th grade.
(in consrtuction)
The problem was published in GMB 12 / 2004; I don't think it's here. It was solved in GMB 6 / 2005.
"C:2816. Prove the equality $\frac{\sin 110^{\circ}}{\sin 30^{\circ}}-\frac{\sin 60^{\circ}}{\sin 80^{\circ}}=1."$
{authors : Romanța GHIȚĂ and Ioan GHIȚĂ, Blaj
Solution CiP - taken from the cited source
The given equality is successively equivalent to :
$\sin 110^{\circ}\cdot \sin 80^{\circ}=\sin 30^{\circ}\cdot \sin 60^{\circ}+\sin 30^{\circ}\cdot \sin 80^{\circ}\;\Leftrightarrow$
$\Leftrightarrow\;\;2\sin 70^{\circ}\cdot \sin 80^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}\;\Leftrightarrow$
$\Leftrightarrow\;\;\cos 10^{\circ}-\cos 150^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}$
which is obvious, because $\cos 10^{\circ}=\sin 80^{\circ}\;\;and\;\; -\cos 150^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}.$
$\blacksquare$
REMARK CiP It would be interesting to study some trigonometric equations starting from the present equality, as I did in the post of September/7/ 2025 A forgotten trigonometric equation.