Neat idea - I suppose R is just a place holder and disappears in the algebra that follows...

, into the law of cosines you'll
>get an identity, although it takes some effort to obtain it.
>This means that starting with
>
>a² + b² + 2ab cos(α + β)

Did you mean "a² + b² - 2ab cos(α + β)" ?

>
>you must be able to derive c².

That's neat also, but I wanted to start with the law of sines and "derive" the law of cosines. Actually what I had in mind was something that starts like this:

a / sinA + b / sinB = 2c / sinC

and then by algebraic techniques and non-pythagorean trig identities end with:

c² = a² + b² - 2ab cosC

I think if this is possible it would be a rather monumental task. But as I have read elsewhere on this site, I am morally obligated not to try something impossible.

>>a² + b² + 2ab cos(α + β)
>
>Did you mean "a² + b² - 2ab cos(α + β)" ?

No. The usual thing to write would be -2ab cosγ. But γ = 180° - α - β so that

cosγ = -cos(α + β).

>>you must be able to derive c².
>
>That's neat also, but I wanted to start with the law of
>sines and "derive" the law of cosines.

This is how it would happen if you start with

a² + b² + 2ab cos(α + β),

substitute the values from the law of sines and use the trig identities for the sum of angles, never the PT identity.

>Actually what I had
>in mind was something that starts like this:
>
>a / sinA + b / sinB = 2c / sinC
>
>and then by algebraic techniques and non-pythagorean trig
>identities end with:
>
>c² = a² + b² - 2ab cosC
>
>I think if this is possible it would be a rather monumental
>task. But as I have read elsewhere on this site, I am
>morally obligated not to try something impossible.

You have to try, or do you mean that somebody else should?

>You have to try, or do you mean that somebody else should?

I was just wondering if someone had already proven this impossible, in which case I would not try. But the gist of your statement tells me no one has ever tried this, so try I will.

>R is the circumradius of the triangle.

Ahh, one learns something new everyday.

>Actually what I had
>in mind was something that starts like this:
>
>a / sinA + b / sinB = 2c / sinC
>
>and then by algebraic techniques and non-pythagorean trig
>identities end with:
>
>c² = a² + b² - 2ab cosC
>
>I think if this is possible it would be a rather monumental
>task. But as I have read elsewhere on this site, I am
>morally obligated not to try something impossible.

I should also add that I thought of using the supplements on the sine function as well so that we would have:

a / sin(B+C) + b / sin(A+C) = 2c / sinC

Using the sum of angles identity and then squaring both sides makes for some messy algebra indeed.

Am I going down an endless path perhaps?

John, I am afraid to tell. Generally speaking in mathematics, if you can get from here to there in one way, there is at least one other way to accomplish the feat. I have done this once starting from the right hand side of the law of cosines. Whether what you suggest is a dead end I do not know.

One thing I can say is that I would not complicate matters passing to the supplementary angles and still maintain all three angles. Also, I would get rid of the denominators before squaring. Finally, I am pretty sure that you do not want to work with both the angles and the sides simultaneously, especially because the law of sines lets one express sides through angles, and vice versa.

Did not observe that, although this does happen occasionally.

It must take effort I cannot produce. I did what you suggested and I have:

(sinA cosB)^2 + (cosA sinB)^2 = (sinA)^2 + sin(B)^2 - 2(sinA sinB)^2

I believe I will give up this venture... But thanks for helping me with it.

sin²(A+B) = sin²(A) + sin²(B) + 2sin(A)sin(B)cos(A+B)

The right-hand side equals

sin²(A) + sin²(B) + 2sin(A)sin(B)(cos(A)cos(B) - sin(A)sin(B))

which is

sin²(A) + sin²(B) - 2sin²(A)sin²(B) + 2sin(A)sin(B)cos(A)cos(B).

The left-hand side equals

(sin(A)cos(B) + sin(B)cos(A))² = sin²(A)cos²(B) + sin²(B)cos²(A) + 2sin(A)cos(B)sin(B)cos(A)

These will be equal iff

sin²(A)cos²(B) + sin²(B)cos²(A) = sin²(A) + sin²(B) - 2sin²(A)sin²(B)

Now

sin²(A)cos²(B) = sin²(A)(1 - sin²(B)) and
sin²(B)cos²(A) = sin²(B)(1 - sin²(A)) so that

sin²(A)cos²(B) + sin²(B)cos²(A) =
sin²(A) + sin²(B) - 2sin²(A)sin²(B)

which is exactly what you need.

Didn't you use a pythagorean relationship here? cos²(B) = 1 - sin²(B)

And even still I do not think I can produce c² from your original expression with cos (A+B)... I even tried inserting an expression from Mollweide's Formula for c^2 along with using half angle identities and could not get the two sides to be equal (although I did find some interesting facts along the way). I think I should do my homework on this level of algebra before continuing this project. But thanks again for clarifying. This is quite intense...

Again my plus signs were missing, but this is what I meant to write in the previous post... and is where I got stuck...

Exactly?...

Can't tell you that. But one thing is clear. The above derivation is from the law of sines towards the law of cosines, not in the other direction. So I do not know about equivalence.

To see that, consider a triangle with three right angles and then its half by drawing one of the altitudes. This will produce a triangle with angles 90-45-90 degrees. Two sides will remain the same but the base will be halved, whereas sin(45) = sqrt(2)/2. So the law of sines can't hold in both triangles.

The law of cosines is equivalent to the PT.

Just thoughts.

And I also suppose the answer to my original question is 'yes' as long as one is willing to accept the cos^2 = 1 - sin^2 rule.

Is this is "definition by convenience" as you have stated in another proof?

Generally speaking,

(a ⇒ c) & (b ⇒ c)

does not imply

a ⇔ b

>And I also suppose the answer to my original question is
>'yes' as long as one is willing to accept the cos^2 = 1 -
>sin^2 rule.

In view of my last response, what if the law of sines implies the PT?

>Is this is "definition by convenience" as you have stated in
>another proof?

No, here it was essential.

OK but what about (a ⇒ b) & (b ⇒ c)

a being the law of sines
b being the law of cosines
c being the pythagorean theorem

This would imply a ⇒ c but still not a ⇔ b.

ok I think the PT is a necessary ingredient in either direction now, what I am thinking is a ⇒ b and b ⇒ c and c ⇒ a. a the law of sines, b the PT and c the law of cosines. The most difficult of these algebraically being the last one.

forgot to subscribe...

Let ABC have sides a,b,c, and let CD be an altitude to AB. Then the base c can be written as AD + DB = a*cos(B) + b*cos(A), and the altitude can be written equivalently as a*sin(B) or b*sin(A) (note the equality of these is the law of sines), so the area can be written as (a*cos(B))*(b*sin(A)) + (b*cos(A))*(a*sin(B)) = ab(sin(A)cos(B) + cos(A)sin(B)). Note that lengths must be treated as signed, with a negative appearing when D is not between A and B.

But the area can also be written ab*sin(C), since a*sin(C) is the altitude to side b, or equivalently b*sin(C) is the altitude to side a. (By the way, this is also related to the law of sines, inasmuch as the easiest proof is to note that multiplying the product abc by the law in the form sin(A)/a = sin(B)/b = sin(C)/c produces bc*sin(A) = ca*sin(B) =ab*sin(C), which are all formulas for the triangle area, hence equal.) Now equating these two forms and cancelling the ab factor leaves sin(A)cos(B) + cos(A)sin(B) = sin(C), which is essentially the sum of angles formula for sine, since sin(C) = sin(pi-C) = sin(A+B).

Specializing to a right triangle with hypotenuse c forces cos(B) = sin(A) and cos(A) = sin(B) directly from their definitions. Therefore, sin(A)^2 + sin(B)^2 = sin(C)^2 = 1. This trivially implies the PT. The PT then implies the Law of Cosines (a nice example of a special case equivalent to the general case).

There are undoubtedly other methods, perhaps more direct.

--Stuart Anderson

a. The law of sines
b. The law of cosines
c. The PT
d. The addition formulas for sine

are equivalent and all follow from the additivity of the area.

Looks quite right.

The question is then, What machinery is needed to justify that identity?

https://en.wikipedia.org/wiki/Bhaskara_I's_sine_approximation_formula

Also, I have additionally found a page that explains a proof of the addition formula for sin without using the PT:

https://staff.jccc.net/swilson/trig/anglesumidentities.htm

(see the 'alternate proof' section).

Therefore is it possible to use the cofunction and sin angle sum formula to derive the PT?

In right triangle ABC, C the right angle we have:

a = c*cosB = c*sinA
b = c*sinB = c*cosA

Then a^2 + b^2 = (c*cosB)(c*sinA) + (c*sinB)(c*cosA) = c^2*sin(A+B) = c^2 * sinC = c^2 * 1 = c^2

Additionally, in using the law of sines a / sinA = b / sinB, I have found that using the sum of the squares of the cross products along with cofunction identit's we have:

(a*sinB)^2 + (b*sinA)^2 = (a*sinB)(a*cosA) + (b*sinA)(b*cosB) which if added to (a*sinA)^2 + (b*sinB)^2 produces a^2 + b^2 if C is 90 degrees.

I thought the above paragraph was a little interesting. Not sure how to prove this to be c^2 but I am sure you and Stuart could shed some light on this....

Then the next order of business is to show (rather easily I think) that the PT produces the law of cosines for non-right triangles.

The challenge of course for me is to show that the law of cosines becomes the law of sines again... but I am sure there is way to do it. Just haven't had the time really....

Any thoughts?

How do I ensure they stay when I type them in here?

I did in the preview when I submitted the message. They must have made it through without my knowledge.

Sorry...

https://www.scribd.com/doc/36021278/Sin-90-Equals-1-Proof

If sine is the ratio of the opposite leg to the hypotenuse of a right triangle then sin90° could not possibly have a meaning, let alone a value because in Euclidean geometry there are no triangles with two right angles.

If sine is anything else then the question of its values is moot.

...even if we invoke a limit?

Again, just say aloud what is sine.

I suppose it depends on where you get your definition. I found one here that does not use the words "opposite leg to the hypotenuse."

https://wordnetweb.princeton.edu/perl/webwn?s=sine

OK, "sine is opposite side from the angle divided by the hypotenuse." But for a right angle the hypotenuse is the opposite side, so hypotenuse divided by hypotenuse = 1. No degenerate triangle with two right angles...just the quotient of two lengths that are equal.

Also, I was pondering how similar this might be to using the point (0,c) in finding the constant of integration in proof #90. There, did I not also make a degenerate triangle? I suppose it is OK to do it in that context but not in the case of the sin(90) which I think results in the same kind of triangle you are talking about.

Nonetheless, I have always accepted the definition I use above for sine. If I were to use this for cosine, there of course would be ambiguity but I do not think that applies to the argument.

I may of course be fighting a losing battle, but if not this is a fairly straightforward proof where sin(A+B) turns quickly into (sinA)^2 + (cosA)^2 = 1.

I appreciate that, but it's always the ratio of a leg to the hypotenuse. There is no point in hanging your hopes on an inaccurate definition.

The question may be that it might be worthwhile to expand sine first to angles that are not accute, 0 in particular, and then prove the addition formula. The expansion is by taking limits and imposing the period and symmetry in the origin. This done, you are free to deduce the PT from the addition formula.

In all likelihood this can be done with the principles similar to the ones annunciated in your calculus proof of the PT. If PT is not needed to define the deriavtive, it is certainly not needed to define continuity.

I believe I successfully did this in the paper I posted in the other thread. It is under III #10. I never really got any closure on this. It'seems that you have ignored it as trivial, but I maintain that one can extend the definition of sine as a continuous function whereas sin(90) = 1, and therefore sin(A+B)=sin(C) which leads to (sinA)^2 + (cosA)^2 = 1. A quite simple trigonometric proof I think.

I think also it bears repeating that using a 'degenerate triangle' in the sense of continuity is comparable to using the initial condition (0,c) in the calculus proof. What makes one proof more acceptable than the other?

I see the discussion has gone on while I was away briefly. As to the question of sin(pi/2)=1, the trouble you mention never occured to me, because I have for a long time thought of sine in the following way:

For an arbitrary nondegenerate triangle ABC, sin(A) is the ratio of the altitude BD to the side AB. (And it is immaterial whether one takes altitude BD or altitude CE, since ABD is similar to ACE.)

For a triangle right at C, this agrees exactly with the common definition, since the altitude then coincides with BC. For a triangle right at A, the altitude coincides with AB, leading trivially to sin(pi/2)=1. Sine is automatically defined for angle A > pi/2 as well. So this definition has the highly desirable properties of agreement with the prior definition and intuitively reasonable interpretation in the new cases to which it extends.

Adopting new definitions is, of course, optional, but this one has much to recommend it and no disadvantages that I can see. The benefit arises from the definition being stated for an arbitrary nondegenerate triangle, rather than for a right triangle. The "implied right triangle" ABD does become degenerate when A=pi/2, but ABC does not, so the case A=pi/2 is not exceptional under this definition.

I agree that without some kind of extended definition of sine, the assumption that sin(pi/2)=1 is unjustifiable. However, upon rereading my post #21, I think I see a way to avoid the trouble altogether. First, let me note that (to my embarassment) I have omitted the factor of 1/2 in the triangle area throughout the post. Oops.

Now consider just the first paragraph of post #21. The area of a triangle is shown to be (ab/2)*(sin(A)cos(B) + cos(A)sin(B)). Let us immediately specialize to the case of a triangle right at C. Then its area is obviously ab/2, and also (from paragraph 3) cos(B)=sin(A) and cos(A)=sin(B). As before, equating the area formulas gives (sin(A)^2 + cos(A)^2 = 1, but this time without any assumption about sin(pi/2).

Is that any better?

--Stuart Anderson

I'll need to digest your definition of sine. It's clear why the standard one uses right triangles. The ratios are invariant wrt similarity which is only a function of one angle. Your definition gives the same result of course, but, I'd say, loses elegance by requiring one more step for a proof. The benefit is, however, obvious and weighty.

But now what did we get here? What you now showed is that the PT follows from the additivity of area (via the addition formula for sine - or, sort of). Does it follow from the law of sines?

I would like to go back to an earlier post where I mentioned that a huge mess turns into a^2 + b^2.

Using the law of sines a*sin(B) = b*sin(A) = c*sin(A)sin(B) and taking the liberty of sin(pi/2) = 1 and the addition formula for sine, I have the following identity:

a^2 + b^2 = 2c^2 * (sinA)^2 * (sinB)^2 + a^2 * (sinA)^2 + b^2 * (sinB)^2

Letting sin(A) = a/c and sin(B) = b/c we have:

a^2 + b^2 = 2c^2 * a^2 / c^2 * b^2 / c^2 + a^2 * a^2 / c^2 + b^2 * b^2 / c^2

After a bit of simplifcation we have:

c^2 = (a^2 + b^2)^ 2 / (a^2 + b^2) = a^2 + b^2

So we have the law of sines, the addition formula for sine, the cofunction identities, and the definitions for sine and cosine...together lead to PT.

This of course only if one is willing to accept the special case of the law of sines where sin(pi/2) = 1, since my original law of sines looks like this:

a / sin(A) = b / sin(B) = c

The bigger question I think I originally asked though is whether to use this version:

a / sin(A) = b / sin(B) = c / sin(C)

to produce:

c^2 = a^2 + b^2 - 2ab * cosC

I am being persuaded throughout this discussion that the only way to do that is through the PT.

However this one specifically makes use of the law of sines.

If we take the law of sines to apply to all triangles, including right ones, is an extended definition of sine necessary?

I think it is in my post 28, but not here, since in this formula sine applies to all angles 0 < C < 180.

Also, I do not think that a proof of the addition formula for sine is needed here, since due to an earlier post, it is possible to show this formula geometrically without recourse to the PT. Perhaps I would need to include that proof as well as comments above in my writeup.

https://www.cut-the-knot.org/proofs/sine_cosine.shtml#law

Here there is a diagram where you mention:

"The Law of Sines supplies the length of the remaining diagonal. The addition formula for sine is just a reformulation of Ptolemy's theorem."

Here what if AD passes through O? Then we have a rectangle, the pythagorean theorem and also sin(A+B)=1 since the diagonals of a rectangle are congruent, as well as A+B = 90 since the angle sum is one of the angles of a rectangle.

I could go on and on with justifications. I suppose the question is what is acceptable and what isn't in the proof, or even appropriate might be a better word.

Is it indeed true that the PT follows from the law of sines? If so, it is easy to show the law of cosines, although not without the PT (quoting "I'll be extremely curious to learn of any proof of the Cosine Rule completely independent of the Pythagorean Theorem.")

I suppose we could go on and on with this. But perhaps this thread is too long as it is.

I would like to echo your sentiment, "I'll be extremely curious to learn of any proof of the Cosine Rule completely independent of the Pythagorean Theorem."

No doubt my proof takes advantage of many properties of a right triangle: cofunction identities and trigonometric ratio definitions in particular I would not be able to use in such a proof of the Cosine Rule. But like I started this one, perhaps one could start by letting sinC not necessary equal 1 with:

(a sinB sinC)^2 + (b sinA sinC)^2 = 2 * (c sinA sinB)^2

This would make for a largely complex algebraic proof, and I am not quite sure if this is even a good way to start.

As to the second point, it'seems to me that the law of sines is still in use even in my improved, abbreviated proof. It is the law of sines which allows the area formula to be rewritten in such a way that the product ab factors out; without this, the derivation gets stuck at the first step. Perhaps there is another way to proceed, but I do not immediately see it.

However, the whole discussion has moved away from the original question of whether the law of cosines could be derived from the law of sines pure and simple. I never really said much on that point, so here are my thoughts: proofs occur within a context of axioms and laws of inference. The law of sines, by itself, does not let you derive anything at all; therefore, there must be some set of additional definitions and assumptions. What exactly to choose is the main question: choose too few and you will not succeed, too many and the proof will be trivial. I think the best formulation of the question would be: what other minimal assumptions and definitions are needed in order to make sense of the law of sines, and will these allow the derivation of the law of cosines; or if not, what is a miminal set of assumptions that will allow it?

--Stuart Anderson

I think I have completed the derivation from the law of cosines to the law of sines:

c^2 = a^2 + b^2 - 2ab cosC -> a / sin(A) = (a - 2b cosC) / sin(C - B) or b / sin(B) = (b - 2a cosC) / sin(C - A)

and

b^2 = a^2 + c^2 - 2ac cosB -> a / sin(A) = (a - 2c cosB) / sin(B - C) or c / sin(C) = (c - 2a cosB) / sin(B - A)

and

a^2 = b^2 + c^2 - 2bc cosA -> b / sin(B) = (b - 2c cosA) / sin(A - C) or c / sin(C) = (c - 2b cosA) / sin(A - B)

Now set equal any two fractions from an identical law of sines term, say c / sin(C) for example.

Here we'd have (c - 2a cosB) / sin(B - A) = (c - 2b cosA) / sin(A - B) and since sine is an odd function c - 2a cosB = -c + 2b cosA which produces an identity c = a cosB + b cosA.

Using any other fractions will result in similar identit's for b and a. And since the each version of the law of cosines can produce two different identities, can we not say that a / sinA = b / sinB = c / sinC?

It'seems from this site:

https://www.mae.ufl.edu/~uhk/TRIANGLE-FORMULAS.pdf

that Mollweide's formula can be derived from the law of sines. Is my argument then circular?

We talked about that. In itself it does not mean anything. Mollweide's formula could be derived in many other ways.

Thank you. Forgive my ignorance, but who do refer to as "we..." ? You and I?

So then I think the derivation is complete. The law of sines leads to the PT, which leads to the law of cosines, which leads to the law of sines.

Therefore all three are equivalent.

I suppose I am a little confused on where all this is going, or if we even accomplished anything here. I am aware of the following in all of the discourse since Stuart's message in #21:

1) We need an extended definition of sine that will allow us to use the definition sin(90) = 1. I believe the law of sines affords us this option as the right triangle is a single case of the law of sines where one angle is 90 and it's sine value is 1.

2) The law of sines, PT, and law of cosines being equivalent all rely on (1) as well as the use of the sine addition formula, where we assume that A + B = C in a right triangle. This either by the additivity of area (Stuart) or algebraically (mine).

3) I have still somewhat a lack of closure on whether all of these are equivalent... and really whether this whole thing is done or not.

Here is what I think I have done so far. You can verify if this is right or not:

a) The law of sines -> The PT
b) The PT -> the law of cosines (although I have not done a writeup for this it is fairly simple).
c) The law of cosines -> the law of sines. Here I am still a little unsure because of the complex web of implications that some equations have on others from my post #47. I have 9 different equations going on and they do not all imply one another (or do they?)

there is nothing very surprising in that there are several statements that circularly imply each other. Happens all the time. What may make this particular instance attractive is that the statements are named and popular.

It was different with the Calculus proof because of an implicit traditional assumption that somehow Calculus is a layer or two above the elementary math.

In my view, the curious part is that the PT implies the law of cosines (this is appears somewhere at the site), because one is a genralization of the other. Still, they are equivalent.

As to the properties of sine, my assumption is that, since the only thing needed to extend the sine from the right triangle is continuity, then certainly (with the experience of your calculus proof) it can be done. What it takes must be pretty straightforward. So yes, you can say that you have accomplished something.

OK, then I shall make an argument about the sine function being continous at 90 degrees and beyond. From there I'll write a proper paper on the whole thing:

1) The sine function is continuous and extends to 90 and beyond.
2) The law of sines holds for right triangles and implies the PT.
3) The PT implies the law of cosines.
4) The law of cosines implies the law of sines.
5) Therefore these three laws are equivalent.

Should make a rather large paper I think. Thanks for your feedback. I was starting to think that all this algebra had me being a babbling idiot, or a genius, or somewhere in between. Now at least I know where I lie in that continuum.

Again, most grateful for the collegiality.

Thanks again,

Perhaps someday...

So sorry for my rambling. But after a little contemplation, I believe I have answered my own question. See the attached. The equivalence is not 'direct' but nonetheless exists, unless there is an error in my work somewhere...