An application of the law of cosines?

In summary, you have shown that Néroman's formula for the ratio of the musical intervals is $6\pm\sqrt{36 + 64(1 + \tan^2i)}}{8\sqrt{1+\tan^2i}}$.
  • #1
Kobzar
11
0
Hello, everybody:

I am a philologist who is fond of mathematics, but who unfortunately has just an elementary high school knowledge of them. I am translating La leçon de Platon, by Dom Néroman (La Bégude de Mazenc, Arma Artis, 2002), which deals with music theory and mathematics in the works of Plato. The problem which brings me here is not about translation, but about mathematics. I am attaching a file with my problem and what I have tried so far to solve it.

Thank you very much in advance and best greetings.
 

Attachments

  • Law of cosines?.pdf
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  • #2
In fact, your answer is essentially the same as Néroman's. You have shown that $$\begin{aligned}\frac{PM}{PH} &= \frac{6\pm\sqrt{36 + 64(1 + \tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{100 + 64\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4(25 + 16\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4}\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm2\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}}. \end{aligned}$$ Now divide top and bottom of the fraction by $2$ to get $$\frac{PM}{PH} = \frac{3\pm\sqrt{25 + 16\tan^2i}}{4\sqrt{1+\tan^2i}}.$$ That is exactly Néroman's formula apart from two things. One of these is the ambiguous $\pm$ sign. That is resolved by the fact that the $+$ sign gives the ratio $\frac{PM}{PH}$, and the $-$ sign gives the ratio $\frac{Pm}{PH}$. The other anomaly is the $PH$ (or $h$) that occurs in Néroman's formula. That must be a mistake on his part. The formula should be either $$\tfrac{PM}{PH} = \tfrac1{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr)$$ or $$PM = \tfrac{PH}{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr).$$ But he appears to have mistakenly put the $PH$ on both sides of the formula.
 
  • #3
Opalg said:
In fact, your answer is essentially the same as Néroman's. You have shown that $$\begin{aligned}\frac{PM}{PH} &= \frac{6\pm\sqrt{36 + 64(1 + \tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{100 + 64\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4(25 + 16\tan^2i)}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm\sqrt{4}\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}} \\ &= \frac{6\pm2\sqrt{25 + 16\tan^2i}}{8\sqrt{1+\tan^2i}}. \end{aligned}$$ Now divide top and bottom of the fraction by $2$ to get $$\frac{PM}{PH} = \frac{3\pm\sqrt{25 + 16\tan^2i}}{4\sqrt{1+\tan^2i}}.$$ That is exactly Néroman's formula apart from two things. One of these is the ambiguous $\pm$ sign. That is resolved by the fact that the $+$ sign gives the ratio $\frac{PM}{PH}$, and the $-$ sign gives the ratio $\frac{Pm}{PH}$. The other anomaly is the $PH$ (or $h$) that occurs in Néroman's formula. That must be a mistake on his part. The formula should be either $$\tfrac{PM}{PH} = \tfrac1{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr)$$ or $$PM = \tfrac{PH}{4\sqrt{1+\tan^2i}}\bigl(3 + \sqrt{25 + 16\tan^2i}\bigr).$$ But he appears to have mistakenly put the $PH$ on both sides of the formula.
Thank you very much!
 

Related to An application of the law of cosines?

1. What is the law of cosines?

The law of cosines is a mathematical formula used to find the length of a side or angle in a triangle when the lengths of the other sides and angles are known. It is also known as the cosine rule and is used for solving problems in trigonometry and geometry.

2. How is the law of cosines applied?

The law of cosines is applied by plugging in the known values for the sides and angles of a triangle into the formula: c² = a² + b² - 2abcos(C), where c is the unknown side, a and b are the known sides, and C is the angle opposite of side c.

3. What types of problems can be solved using the law of cosines?

The law of cosines can be used to solve problems involving angles, sides, and areas of triangles. It is particularly useful for solving problems involving non-right triangles, as it can be used to find missing sides and angles.

4. Can the law of cosines be used for any type of triangle?

Yes, the law of cosines can be used for any type of triangle, including acute, obtuse, and right triangles. However, it is most commonly used for solving problems involving non-right triangles.

5. What are some real-life applications of the law of cosines?

The law of cosines has many real-life applications, such as in navigation and surveying, where it is used to find distances and angles between objects. It is also used in physics and engineering for calculating forces and vectors in two and three-dimensional systems.

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