What is the connection between the Law of Cosines and Euclid's teachings?

In summary, the conversation discusses the value of ##\cos{90^{\circ}}## and various methods to determine its value, such as using the unit circle, Pythagorean theorem, and the law of cosines. It also mentions the concept of a terminal arm and how it relates to the cosine function. The conversation ends with a discussion on how the cosine function can be defined using the law of cosines and a polarization identity.
  • #1
SSG-E
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TL;DR Summary
In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.
I
Untitled.png
 
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  • #2
##\cos{90^{\circ}} = 0##
 
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  • #3
We can have geometric a proof, but how about this one:
$$
\sin^2 \psi + \cos^2 \psi = 1$$
$$if~\psi = 90^{\circ} ~and~accepting ~\sin 90^{\circ} = 1$$

$$1 +\cos^2 90^{\circ} = 1$$
$$\cos^2 90^{\circ} = 0 $$
$$\cos 90^{\circ}=0
$$
 
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  • #4
@Adesh perhaps, but that would rely on knowing ##\sin{90^{\circ}} = 1## and we're back where we started.

There are a few ways of determining the values of trigonometric functions. A useful tool is the so-called unit circle, which consists of a circle of radius 1 centred on the origin in the ##x##-##y## plane. The coordinates of any point on the circle at an angle ##t## to the positive ##x## axis are ##(\cos{t}, \sin{t})##, as shown:

1592752143895.png


If ##t = \frac{\pi}{2} = 90^{\circ}##, the point is at the top of the circle. The ##x##-coordinate is ##0##, and this is the cosine of the angle. It's also useful for determining the signs of trigonometric functions in the different quadrants (I've seen this called the "CAST Diagram" before).

Really you should memorise how ##\sin##, ##\cos## and ##\tan## behave at ##\theta = 0##, ##\theta = \frac{\pi}{6}##, ##\theta = \frac{\pi}{4}##, ##\theta = \frac{\pi}{3}## and ##\theta = \frac{\pi}{2}##. You can use the unit circle and trigonometric identities to deduce the values for lots more angles from them.
 
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  • #5
But how is its cosine equal to 0.
Not Correct.
 
  • #6
SSG-E said:
Summary:: In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.

IView attachment 264987
You can find the proof of Law Of Cosines in some Trigonometry books and some Calculus books. There is an old book by Anton which includes a proof.
 
  • #7
My favorite proofs are propositions 12, 13 Book II, Euclid's Elements.
 
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  • #8
SSG-E said:
Summary:: In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.

cos 90 = sin(90-90) = sin(0) = 0.

Does that now beg the question "how is sin(0) = 0" ?
 
  • #9
neilparker62 said:
cos 90 = sin(90-90) = sin(0) = 0.

Does that now beg the question "how is sin(0) = 0" ?

But you have turned the question into something that is much easier to answer, if you know or can derive the series expansions. Namely...

##\sin{0} = 0 - \frac{1}{3!}0^3 + \frac{1}{5!}0^5 + \dots = 0##
 
  • #10
Proof of Cosine Rule using Ptolemy's Thm

Proof of law of Cosines.png
 
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  • #11
mathwonk said:
My favorite proofs are propositions 12, 13 Book II, Euclid's Elements.

Proof of law of Cosines using Secant-Tangent Thm.png
 
  • #12
Another way, label the triangle like this:

IMG_1754.jpg


Pythagoras on the left triangle: ##x^2 + h^2 = b^2##
Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##

Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.
 
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  • #13
etotheipi said:
Another way, label the triangle like this:

View attachment 265361

Pythagoras on the left triangle: ##x^2 + h^2 = b^2##
Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##

Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.
😁 Oh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).
 
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  • #14
Adesh said:
😁 Oh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).

Yes my hand-writing is terrible, I never got my pen-license in primary school 😔
 
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  • #15
Draw an unit circle and add an additional line on the positive y-axis as the terminal arm. Let z denote the length of the terminal arm, we get: $$\cos(\theta)=\frac x z\implies \cos(90)=\frac 0 1 = 0$$
 
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  • #16
Leo Liu said:
Draw an unit circle and add an additional line on the positive y-axis as the terminal arm. Let z denote the length of the terminal arm, we get: $$\cos(x)=\frac x z\implies \cos(90)=\frac 0 1 = 0$$

What do you mean by a terminal arm?
 
  • #17
etotheipi said:
What do you mean by a terminal arm?
I'm pretty sure that he means the ray whose angle is measured relative to the reference direction, the positive x-axis.
 
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  • #18
In some texts, the law of cosines or a polarization identity is used to define the cosine function:
$$\vec C = \vec A +\vec B$$

[itex] \begin{align*}
C^2 &= (\vec A+\vec B)\cdot (\vec A+\vec B) =A^2 +B^2 + 2AB\cos\theta \\
\cos\theta_{AB} &\equiv\frac{C^2-A^2-B^2}{2AB}
\end{align*} [/itex][itex] \begin{align*}
(\vec A+\vec B)\cdot (\vec A+\vec B) &=A^2 +B^2 + 2AB\cos\theta \\
(\vec A-\vec B)\cdot (\vec A-\vec B) &=A^2 +B^2 - 2AB\cos\theta \\
\cos\theta_{AB} &\equiv\frac{(\vec A+\vec B)\cdot (\vec A+\vec B) - (\vec A-\vec B)\cdot (\vec A-\vec B)}{4AB}
\end{align*} [/itex]
 
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  • #19
etotheipi said:
Another way, label the triangle like this:

1632175177519.png


Pythagoras on the left triangle: ##x^2 + h^2 = b^2##
Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##

Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.
##h=b\sin\hat{C}## and ##x=b\cos\hat{C}##. Hence:
$$c^2=b^2\sin^2\hat{C}+(a-b\cos\hat{C})^2=a^2 + b^2 - 2ab\cos{\hat{C}}$$, and as a 'bonus': $$\tan\hat{B}=\frac{b\sin\hat{C}}{a-b\cos\hat{C}}$$
 
  • #20
neilparker62 said:
##h=b\sin\hat{C}## and ##x=b\cos\hat{C}##. Hence:
$$c^2=b^2\sin^2\hat{C}+(a-b\cos\hat{C})^2=a^2 + b^2 - 2ab\cos{\hat{C}}$$, and as a 'bonus': $$\tan\hat{B}=\frac{b\sin\hat{C}}{a-b\cos\hat{C}}$$It
It is a great loss of archaeology that you do not work in that field.
 
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  • #21
Leo Liu said:
It is a great loss of archaeology that you do not work in that field.
thanks - cool profile pic by the way. Where is it from ?

1632316479317.png
 
  • #22
neilparker62 said:
thanks - cool profile pic by the way. Where is it from ?

View attachment 289504
I found it on a website for sharing wallpapers.
 

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  • #23
To me the cosine law became interesting and beatiful only after i learned from euclid, as an old man by then myself, that it is a natural complement to the pythagorean theorem, and can be expressed as areas of rectangles, no formulas needed. It just asks how much excess area does the square on the "hypotenuse" create when the angle opposite is greater than a right angle. And it answers in terms of the area of a simply given rectangle. The side of this rectangle then gives rise to the concept of a cosine. Since it was not taught in my primitive high school geometry course, I did not realize it was a geometry theorem, and only encountered it in a trig course, expressed in mysterious formulas, which did not speak as clearly to me. So it took me almost 50 years to understand this topic, for the simple reason that in the US, geometry is usually not taught from Euclid, as (at least I think) it should be.

Indeed as in post #18, this historical approach does to me fully justify using the deviation in the pythagorean formula, as a definition of the cosine.
 
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Related to What is the connection between the Law of Cosines and Euclid's teachings?

What is the Law of Cosines?

The Law of Cosines is a trigonometric formula that relates the lengths of the sides of a triangle to the cosine of one of its angles. It can be used to solve for missing side lengths or angles in a triangle.

When is the Law of Cosines used?

The Law of Cosines is used when solving triangles that do not have a right angle. It is particularly useful when you know the length of two sides and the measure of the included angle, but need to find the length of the third side or another angle.

What is the formula for the Law of Cosines?

The formula for the Law of Cosines is c² = a² + b² - 2ab cos(C), where c is the length of the side opposite the angle C, and a and b are the lengths of the other two sides.

How is the Law of Cosines derived?

The Law of Cosines can be derived from the Pythagorean Theorem and the Law of Sines. By substituting the formula for the sine of an angle into the Pythagorean Theorem, you can manipulate the equation to get the Law of Cosines.

What are some real-life applications of the Law of Cosines?

The Law of Cosines has many practical applications, such as in navigation and surveying. It can be used to find the distance between two points on a map, or to determine the height of a building or mountain. It is also used in physics and engineering to calculate forces and vectors.

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