shouldn't these equations be identical?

[allegansveritatem]

Why are these following two product to sum identities different? I would think that the commutative property of multiplication would ensure their being the same --I mean, of the same form:


My text book presents these as the first two of four product to sum identities, but does not explain why these have a different form, i.e., one sine on the right is added and one sine is subtracted.

 

[lev888]

Why are these following two product to sum identities different? I would think that the commutative property of multiplication would ensure their being the same --I mean, of the same form:
View attachment 18749

My text book presents these as the first two of four product to sum identities, but does not explain why these have a different form, i.e., one sine on the right is added and one sine is subtracted.

The commutative property does not apply. The arguments (u and v) are different.

 

[Steven G]

expand sin (u+v) and sin(u-v).

Then for (1) add the above two equations together and then divide by 2. What do you get?
Now for (2) subtract the two and then divide by 2. What do you get?

 

[Dr.Peterson]

Why are these following two product to sum identities different? I would think that the commutative property of multiplication would ensure their being the same --I mean, of the same form:
View attachment 18749

My text book presents these as the first two of four product to sum identities, but does not explain why these have a different form, i.e., one sine on the right is added and one sine is subtracted.

To get from the first to the second, you interchange u and v, right? Do so, and then use the fact that sin(-x) = -sin(x).

 

[allegansveritatem]

The commutative property does not apply. The arguments (u and v) are different.

yes, the arguments are different...but are they not just, so to speak, generic variables? I mean, the u and v of the second product, what have they to do with the first product?

 

[allegansveritatem]

expand sin (u+v) and sin(u-v).

Then for (1) add the above two equations together and then divide by 2. What do you get?
Now for (2) subtract the two and then divide by 2. What do you get?

I will try this and post back tomorrow. Thanks

 

[allegansveritatem]

To get from the first to the second, you interchange u and v, right? Do so, and then use the fact that sin(-x) = -sin(x).

well, this is part of the problem: The v is both positive and negative in the first case (in the sum)...but what would that have to do with anything? I mean, it seems to me like a whole new game with the second product. And there is no minus sign anywhere to be seen on the left side. There is something I am not seeing here...does it have anything to do with the double angle formula for sines?l I mean sin2A=sinAcosA? The crux of my dilemma is: I can't see why switching variables would have any bearing on the matter,other, of course, than that the numeric value would be different...but it seems the answer would be arrived at in the same way with the same sinage. Why wouldn't the commutative property assure that?

 

[topsquark]

You have sin(u - v). When you switch the u and v you get sin(v - u) = -sin(u - v).

And again, the commutative property does not apply when the arguments change. [math]f(u) g(v) = g(v) f(u)[/math] but [math]f(u) g(v) \neq g(u) f(v)[/math]. The two arguments are different.

-Dan

 

[Dr.Peterson]

well, this is part of the problem: The v is both positive and negative in the first case (in the sum)...but what would that have to do with anything? I mean, it seems to me like a whole new game with the second product. And there is no minus sign anywhere to be seen on the left side. There is something I am not seeing here...does it have anything to do with the double angle formula for sines?l I mean sin2A=sinAcosA? The crux of my dilemma is: I can't see why switching variables would have any bearing on the matter,other, of course, than that the numeric value would be different...but it seems the answer would be arrived at in the same way with the same sinage. Why wouldn't the commutative property assure that?

The way to see it is to do it, not just talk about it in general terms. This is true in much of learning; I tell students to write out the details of what they are thinking, and then they can see for themselves either that they are wrong, or why they are right.

So I'll do it for you, though you should have done this yourself to get full benefit:

Given that

[MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH],​

if we swap the names of u and v, we get

[MATH]\sin(v)\cos(u) = \frac{1}{2}(\sin(v+u)+\sin(v-u))[/MATH].​

(This is the very same fact, just with different letters in each place.)

Applying the commutative property, we can change order on the left, and in the arguments on the right:

[MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)+\sin(-u+v))[/MATH].​

But since, as has been point out, [MATH]\sin(-u + v) = \sin(-(u-v)) = -\sin(u-v)[/MATH], we have

[MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)-\sin(u-v))[/MATH].​

 

[lev888]

yes, the arguments are different...but are they not just, so to speak, generic variables? I mean, the u and v of the second product, what have they to do with the first product?

We have:
A. [MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH]B. [MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)-\sin(u-v))[/MATH]
Let's change the order on the left and apply sin(-x) = -sin(x) on the right in B.
A. [MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH]B. [MATH]\sin(v)\cos(u) = \frac{1}{2}(\sin(u+v)+\sin(v-u))[/MATH]
Do you see that they ARE the same?

 

[allegansveritatem]

You have sin(u - v). When you switch the u and v you get sin(v - u) = -sin(u - v).

And again, the commutative property does not apply when the arguments change. [math]f(u) g(v) = g(v) f(u)[/math] but [math]f(u) g(v) \neq g(u) f(v)[/math]. The two arguments are different.

-Dan

yes, I see that...I suppose what I am saying is this: Suppose you come upon a cosxsiny problem with no reference to a sinxcosy problem...why couldn't you take the former and present it thus sinycosx and then apply the first product to sum formula? The commutative property would seem to say YES. No?

 

[allegansveritatem]

The way to see it is to do it, not just talk about it in general terms. This is true in much of learning; I tell students to write out the details of what they are thinking, and then they can see for themselves either that they are wrong, or why they are right.

So I'll do it for you, though you should have done this yourself to get full benefit:

Given that

[MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH],​

if we swap the names of u and v, we get

[MATH]\sin(v)\cos(u) = \frac{1}{2}(\sin(v+u)+\sin(v-u))[/MATH].​

(This is the very same fact, just with different letters in each place.)

Applying the commutative property, we can change order on the left, and in the arguments on the right:

[MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)+\sin(-u+v))[/MATH].​

But since, as has been point out, [MATH]\sin(-u + v) = \sin(-(u-v)) = -\sin(u-v)[/MATH], we have

[MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)-\sin(u-v))[/MATH].​

I see what you are saying and I actually did this today:

I guess then that these product to sum formulas are based on identities....they are based on trigonometric logic rather than arithmetical (?) logic.

 

[allegansveritatem]

We have:
A. [MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH]B. [MATH]\cos(u)\sin(v) = \frac{1}{2}(\sin(u+v)-\sin(u-v))[/MATH]
Let's change the order on the left and apply sin(-x) = -sin(x) on the right in B.
A. [MATH]\sin(u)\cos(v) = \frac{1}{2}(\sin(u+v)+\sin(u-v))[/MATH]B. [MATH]\sin(v)\cos(u) = \frac{1}{2}(\sin(u+v)+\sin(v-u))[/MATH]
Do you see that they ARE the same?

yes, I see.We are dealing with trigonometry and I have been applying commutative properties as though it were an arithmetic problem.

 

[Dr.Peterson]

yes, I see that...I suppose what I am saying is this: Suppose you come upon a cosxsiny problem with no reference to a sinxcosy problem...why couldn't you take the former and present it thus sinycosx and then apply the first product to sum formula? The commutative property would seem to say YES. No?

But that's essentially what I did! I applied the first formula, for sin(u)cos(v), to cos(u)sin(v), which is the same as sin(v)cos(u), by replacing each u with v and each v with u. That led to the sign difference.

yes, I see.We are dealing with trigonometry and I have been applying commutative properties as though it were an arithmetic problem.

Well, the commutative property does apply perfectly well to trig, as long as you apply it correctly.

It does tell us that cos(u)sin(v) = sin(v)cos(u), where we are just switching the order of two quantities cos(u) and sin(v) that are being multiplied; but it doesn't tell us that cos(u)sin(v) = cos(v)sin(u), where the two quantities being multiplied are not the same numbers.

I see what you are saying and I actually did this today:
View attachment 18769
I guess then that these product to sum formulas are based on identities....they are based on trigonometric logic rather than arithmetical (?) logic.

This is a different point than we have been discussing, but, yes, these formulas are derived from the angle-addition-and-subtraction identities (to the extent that I only memorize the latter, not the former).

 

[Steven G]

Even in trigonometry you can use the commutative property of arithemetic!

 

[JeffM]

I am going to try a completely different tack.

First, it is NOT TRUE that even all arithmetic operations are commutative, e.g. subtraction is not commutative. So any analogy between aritmetic operations and trigonomtric functions with respect to commutivity starts from a weak base.

Second, it is ALWAYS TRUE that multiplication of real numbers is commutative, whether or not the numbers are generated from trigonometric functions. No one is denying that.

[MATH]sin(a) * cos(b) = cos(b) * sin(a).[/MATH]
Third, let's try proof rather than analogy.

[MATH]\text {ASSUME, for any pair of real numbers } a \text { and } b \text { that}\\ sin(a) * cos(b) = sin(b) * cos(a).\\ \therefore sin \left ( \dfrac{\pi}{4} \right ) * cos(0) = sin(0) * cos \left ( \dfrac{\pi}{4} \right ) \implies\\ \dfrac{\sqrt{2}}{2} * 1 = 0 * \dfrac{\sqrt{2}}{2} \implies\\ \dfrac{\sqrt{2}}{2} = 0 \implies\\ \sqrt{2} = 0 \implies\\ 2 = 0, \text { which is false.}[/MATH]Therefore, the original assumption was false. Therefore, it is true that

[MATH]\exists \text { at least one pair of real numbers } a \text { and } b \text { such that}\\ sin(a) * cos(b) \ne sin(b) * cos(a).[/MATH]Proof by contradiction can resolve many questions.