Temesvári Matematikai Lap // REVISTA MATEMATICĂ (a elevilor) din TIMIȘOARA - nr 7/1971

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SUPLIMENTUL cu EXERCIȚII al GMB N0 1/2024

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Identities for some $\frac{k\pi}{7}$ and $\frac{k\pi}{14}$ angles

           The problem has been discussed here before. 

          I showed that $\left \{ cos\frac{\pi}{7},\;-cos\frac{2\pi}{7},\;cos\frac{3\pi}{7} \right \}=\left \{ sin\frac{5\pi}{14},\; -sin \frac{3\pi}{14},\; \sin\frac{\pi}{14} \right \}$ 

are the roots of the equation  $8x^3-4x^2-4x+1=0$ (see Remark 2, relations (5)-(8) ). And identities are expressions of  Vieta's relationships for this equation.

          For 3.i), let $A=cos\frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}$. 

Let's calculate  $2sin\frac{\pi}{7} \cdot A=2sin\frac{\pi}{7}\cdot cos\frac{\pi}{7}-2sin\frac{\pi}{7}\cdot cos \frac{2\pi}{7}+2sin \frac{\pi}{7}\cdot cos \frac{5\pi}{7};$

with the formula of the double angle and the sum product formulas we get

$$2sin\frac{\pi}{7} \cdot A= sin\frac{2\pi}{7}-\left (sin \frac{3\pi}{7}-sin\frac{\pi}{7} \right )+ \left ( sin \frac{4\pi}{7}-sin\frac{2\pi}{7} \right ).$$

But the angles $\frac{4\pi}{7}$ and $\frac{3\pi}{7}$ are supplementary, so $sin \frac{4\pi}{7}=sin\frac{3\pi}{7}$, and we get $2sin\frac{\pi}{7} \cdot A= sin\frac{\pi}{7}$ from which 3.i) results.

          For 3.ii), let's note that $\frac{\pi}{2}-\frac{\pi}{7}=\frac{5\pi}{14},\;\frac {\pi}{2}-\frac{2\pi}{7}=\frac{3\pi}{14},\;\frac{\pi}{2}-\frac{3\pi}{7}=\frac{\pi}{14}$ 

and with the complement formula this is obtained from 3.i).

          For 3.iii), let $B=8\cdot cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7} \cdot cos \frac{3\pi}{7}.$

 We will calculate $sin\frac{\pi}{7}\cdot B$ applying several times the formula of the double angle and the formulas of the supplements.

$$sin\frac{\pi}{7}\cdot B=4\cdot \left ( 2sin \frac{\pi}{7} \cdot cos\frac{\pi}{7}\right )\cdot cos\frac{2\pi}{7}\cdot cos\frac{3\pi}{7}=4\cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \cdot cos \frac{3\pi}{7}=$$

$$=2\cdot \left ( 2 \cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \right ) \cdot cos \frac{3\pi}{7}=2\cdot sin\frac{4\pi}{7}\cdot cos\frac{3\pi}{7}=2\cdot sin\frac{3\pi}{7}\cdot cos \frac{3\pi}{7}=$$

$$=sin\frac{6\pi}{7}=sin\frac{\pi}{7}$$

hence $B=1.$

          Now 3.iv) is obtained from 3.iii) with complement's formulae.

$\blacksquare$

GAZETA MATEMATICĂ Seria B N0 1/2024

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En trigonometrisk identitet af Euklid // A Trigonometric Identity of Euclid

 (Daneză)

          An identity of Euclid

edit

Euclid showed in Book XIII, Proposition 10 of his Elements that the area of the square on the side of a regular pentagon inscribed in a circle is equal to the sum of the areas of the squares on the sides of the regular hexagon and the regular decagon inscribed in the same circle. In the language of modern trigonometry, this says:

$${\displaystyle {\sin }^2{18}^{\circ }+{\sin }^2{30}^{\circ }={\sin }^2{36}^{\circ }.}$$

Ptolemy used this proposition to compute some angles in his table of chords in Book I, chapter 11 of Almagest.

          Solution 1 CiP

          We know the values $$sin 18^\circ=\frac{\sqrt{5}-1}{4},\;sin 30^\circ=\frac{1}{2},\;sin 36^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}.$$ 

The calculation is simple: $sin^2 18^\circ +sin^2 30^\circ =\left (\frac{\sqrt{5}-1}{4} \right )^2+\left (\frac{1}{2} \right )^2=\frac{5-2\sqrt{5}+1}{16}+\frac{1}{4}=\frac{6-2\sqrt{5}+4}{16}=\frac{10-2\sqrt{5}}{16}=sin^2 36^\circ .$

$\blacksquare$

          Solution 2 CiP

          First we show that

$cos 36^\circ -cos 72^\circ=\frac{1}{2}. \tag{1}$

For this, we denote $A:=cos 36^\circ -cos 72^\circ.$ We will calculate, with double-angle formula and product-to-sum identity

$$2sin 36^\circ \cdot A=2sin 36^\circ \cdot cos 36^\circ-2sin 36^\circ \cdot cos 72^\circ=sin (2\cdot 36^\circ)-(sin 108^\circ-sin36^\circ)=$$

$$=(sin 72^\circ-sin 108^\circ)+sin 36^\circ=sin 36^\circ,$$

(because of reflection in $90^\circ$ formulae, $sin108^\circ=sin(180^\circ-108^\circ)=sin 72^\circ)$, hence (1) follow.

     Now we calculate with power-reduction formula

$sin^2 36^\circ-sin^2 18^\circ=\frac{1-cos (2\cdot 36^\circ)}{2}-\frac{1-cos(2\cdot 18^\circ)}{2}=\frac{1}{2}(cos 36^\circ-cos 72^\circ)\;\underset{(1)}{==}\;\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}=sin^2 30^\circ.$

$\blacksquare$

          Remark CiP We also know $cos 36^\circ =\frac{\sqrt{5}+1}{4},\;cos 72^\circ =\frac {\sqrt{5}-1}{4}$ and (1) results easier.

<End Rem>

Barycentric coordinate on the Straight Line

          Let $A$ and $B$ be distinct points. 

     LEMMA  The following statements are equivalent:

                  (i) points $\;X,\;A,\;B\;$ are collinear and $\frac{\overline{XA}}{\overline{XB}}=\kappa$;

                 (ii) $\overrightarrow{OX}=\frac{\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}}{1-\kappa}$ for a certain point $O$;

                (iii)  $\overrightarrow{OX}=\frac{\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}}{1-\kappa}$ for any point $O$.

     Corollary  The following statements are equivalent:

          ($\alpha$)  points $\;X,\;A,\;B\;$ are collinear and $\frac{\overline{AX}}{\overline{AB}}=r$;

          ($\beta$)  $\overrightarrow{OX}=(1-r)\cdot \overrightarrow{OA}+r \cdot \overrightarrow{OB}$ for a certain/any point $O$.

 

     Remark  The formula from point ($\beta$) is still writte

$$X=(1-r)\cdot A+r \cdot B.$$

          Proof CiP

          (i)$\Rightarrow$(iii) As oriented segments, we have $\frac{\overline{XA}}{\overline{XB}}=\kappa\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Rightarrow$

$$\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa(\overrightarrow{OB}-\overrightarrow{OX})\;\Rightarrow\;(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}$$

where $O$ is an arbitrary point. We get the conclusion.

          (ii)$\Rightarrow$(i) $(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}\;\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa (\overrightarrow{OB}-\overrightarrow{OX}\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}$

so the vectors $\overrightarrow{XA}$ and $\overrightarrow{XB}$ are collinear and their orientation is according to the sign of $\kappa$, as the figure shows. Hence $\overline {XA}=\kappa \cdot \overline{XB}$.

          (iii)$\rightarrow$(ii)  obvious.

     The Lemma is proved. To prove the corollary we note that

$$\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Leftrightarrow\;\overrightarrow{AX}=-\kappa \cdot \overrightarrow{XB}=-\kappa(\overrightarrow{AB}-\overrightarrow{AX})\;\Leftrightarrow \;(1-\kappa)\overrightarrow{AX}=-\kappa \cdot \overrightarrow{AB}$$

hence $r=-\frac{\kappa}{1-\kappa}$.

$\blacksquare \blacksquare$

Was kann eine Maschine lernen und ein Mensch nicht?

                A young researcher thinks he has made a great discovery. A prestigious journal even published his research. Here are the links to the two articles:

https://www.researchgate.net/publication/334837030_Lecture_Notes_on_Machine_Learning_Maximum_Product_of_Numbers_of_Constant_Sum 

https://www.researchgate.net/publication/334848381_Lecture_Notes_on_Machine_Learning_Minimum_Sum_of_Numbers_of_Constant_Product

Unfortunately, neither the mentor of this young researcher nor the Referent of these articles noticed that the two theorems have been known for a long time. Some would say that they are consequences of the inequality of the arithmetic and geometric means. But I say that the inequality precedes the notion of a radical of order $n$.

            For these the inequality

$$\left ( \frac{a_1+a_2+\cdots +a_n}{n} \right )^n \geqslant a_1\cdot a_2 \cdots a_n$$

is sufficient, which I think, at least in the case of $n=2$, was known since the time of Euclid.

          Therefore, man still needs to learn, so that the machine does not overtake him !!