How to approach this

A

allegansveritatem

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Jan 10, 2018
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Here is the identity to be verified:
sinuvprob.PNG

I don't have a clue how to get LS to look like RS. Here is the best I could do:
sinuv1.PNG

This seems to be one of those brick-wall problems. How to make a dent in it? I'm sure once I find my way I will say: Now, why didn't I see that before?
 
I'd usually start by expressing everything in terms of the more familiar sine and cosine, rather than csc.

I'd also try to simplify each side, since both are equally complicated, and put my final answer together by reversing the process, rather than try to find a way to make one side look like the other.

Try doing what you do to simplify a complex fraction in general.
 
allegansveritatem said:
Here is the identity to be verified:
View attachment 19487

I don't have a clue how to get LS to look like RS. Here is the best I could do:
View attachment 19488

This seems to be one of those brick-wall problems. How to make a dent in it? I'm sure once I find my way I will say: Now, why didn't I see that before?
Click to expand...
If sin(u) is a, what is csc(u)?
Do the same for sin(v) = b.
Now you have an equality that uses just a's and b's - saves you some ink. Simplify each side.
 
Personally I like to work with the sine and cosine, and I hate fractions.. Moreover, it is implicit that neither sin(u) nor sin(v) is zero.

[MATH]p = \dfrac{sin(u) + sin(v)}{csc(u) + csc(v)} \implies\\ p\{csc(u) + csc(v)\} = sin(u) + sin(v) \implies\\ p \left ( \dfrac{1}{sin(u)} + \dfrac{1}{sin(v)} \right ) = sin(u) + sin(v) \implies\\ \dfrac{p\{(sin(u) + sin(v)\}}{sin(u)sin(v)} = sin(u) + sin(v) \implies \\ sin(u)sin(v)\{sin(u) + \sin(v)\} = p\{sin(u) + sin(v)\}.[/MATH]Now there are several ways to go. One way is to note that if the sum is not equal to zero

[MATH]p = sin(u)sin(v).[/MATH]
Now set up q to equal the other expression and try to reduce it to the same simplicity.
 
Last edited:
allegansveritatem said:
Here is the identity to be verified:
View attachment 19487

I don't have a clue how to get LS to look like RS. Here is the best I could do:
View attachment 19488

This seems to be one of those brick-wall problems. How to make a dent in it? I'm sure once I find my way I will say: Now, why didn't I see that before?
Click to expand...
One way to make a dent in it is NOT to multiply out as you did. You want a dent, then I would multiply both the numerator and denominator on the lhs by 1-sinusinv. At least now I have the correct numerator times something else. That is always a nice trick to do when you are completely stuck. I would however first turn the lhs into sines and cosines first.
 
Dr.Peterson said:
I'd usually start by expressing everything in terms of the more familiar sine and cosine, rather than csc.

I'd also try to simplify each side, since both are equally complicated, and put my final answer together by reversing the process, rather than try to find a way to make one side look like the other.

Try doing what you do to simplify a complex fraction in general.
Click to expand...
I suppose what you are saying is: Work with both sides until you get them to equal each other. What is that if not verifying?Thanks. I will proceed thus.
 
lev888 said:
If sin(u) is a, what is csc(u)?
Do the same for sin(v) = b.
Now you have an equality that uses just a's and b's - saves you some ink. Simplify each side.
Click to expand...
I hear you. I like it! a and b look a lot more friendly than sin(u) and sin(v). Thanks for the suggestion.
 
JeffM said:
Personally I like to work with the sine and cosine, and I hate fractions.. Moreover, it is implicit that neither sin(u) nor sin(v) is zero.

[MATH]p = \dfrac{sin(u) + sin(v)}{csc(u) + csc(v)} \implies\\ p\{csc(u) + csc(v)\} = sin(u) + sin(v) \implies\\ p \left ( \dfrac{1}{sin(u)} + \dfrac{1}{sin(v)} \right ) = sin(u) + sin(v) \implies\\ \dfrac{p\{(sin(u) + sin(v)\}}{sin(u)sin(v)} = sin(u) + sin(v) \implies \\ sin(u)sin(v)\{sin(u) + \sin(v)\} = p\{sin(u) + sin(v)\}.[/MATH]Now there are several ways to go. One way is to note that if the sum is not equal to zero

[MATH]p = sin(u)sin(v).[/MATH]
Now set up q to equal the other expression and try to reduce it to the same simplicity.
Click to expand...
yes, I have decided to work this little meanie from both sides. I will have to study what you have done here and will do so later in the day when my brain gets into the math groove. Thanks
 
allegansveritatem said:
I suppose what you are saying is: Work with both sides until you get them to equal each other. What is that if not verifying?Thanks. I will proceed thus.
Click to expand...
This is why I said, "put my final answer together by reversing the process". Many teachers object to working on both sides when verifying, and with some reason (because many students confuse proving with solving, and don't think about the logic that justifies what they are doing).

But if A is equivalent to C and B is equivalent to C, then A is equivalent to B. So you can just reverse the work for one side, and show that A = ... = C = ... = B. In this way, your initial simplification of each side can be taken as preparatory exploration, which then leads to a direct path from one side to the other.
 
Jomo said:
One way to make a dent in it is NOT to multiply out as you did. You want a dent, then I would multiply both the numerator and denominator on the lhs by 1-sinusinv. At least now I have the correct numerator times something else. That is always a nice trick to do when you are completely stuck. I would however first turn the lhs into sines and cosines first.
Click to expand...
I know that what I did makes very little sense. I guess I was tired. I know my attitude was something like: Well, let's just do things to this and see what happens.
 
Two points.

It is traditional to present trig identities as going from A to B rather than A going to C and B to C. The reason is that, unless you can also go from C to B and from C to A, the identity is false. Do you see why? So the proof is A to C to B and B to C to A. But when l a proof you can try almost anything.

Second, my assumption that a sum did not equal zero means that a proof based on that assumption cannot alone prove the identity. You must also prove the identity on the contrary assumption that the sum is zero. Frequently, however, that zero case is almost trivial to prove.
 
JeffM said:
Personally I like to work with the sine and cosine, and I hate fractions.. Moreover, it is implicit that neither sin(u) nor sin(v) is zero.

[MATH]p = \dfrac{sin(u) + sin(v)}{csc(u) + csc(v)} \implies\\ p\{csc(u) + csc(v)\} = sin(u) + sin(v) \implies\\ p \left ( \dfrac{1}{sin(u)} + \dfrac{1}{sin(v)} \right ) = sin(u) + sin(v) \implies\\ \dfrac{p\{(sin(u) + sin(v)\}}{sin(u)sin(v)} = sin(u) + sin(v) \implies \\ sin(u)sin(v)\{sin(u) + \sin(v)\} = p\{sin(u) + sin(v)\}.[/MATH]Now there are several ways to go. One way is to note that if the sum is not equal to zero

[MATH]p = sin(u)sin(v).[/MATH]
Now set up q to equal the other expression and try to reduce it to the same simplicity.
Click to expand...
I looked at this again this morning and then later I took up my pen and started to work it out. I did not have access to the computer while I was working on this so it took a while and actually a long while and several false starts before I came to the sinu sinv expression on the LS. Then I went at the right and finally again after some goofing around managed to get the same on the RS. What follows is a record of my efforts:
sinsin2.PNG
sinsin1.PNG
Is there more to do here?
 
allegansveritatem said:
I looked at this again this morning and then later I took up my pen and started to work it out. I did not have access to the computer while I was working on this so it took a while and actually a long while and several false starts before I came to the sinu sinv expression on the LS. Then I went at the right and finally again after some goofing around managed to get the same on the RS. What follows is a record of my efforts:
View attachment 19513
View attachment 19514
Is there more to do here?
Click to expand...
I am a bit worried about the cancellation of sin(u) + sin(v) because of the 0/0 problem, but perhaps it is implicit that sin(u) cannot equal -sin(v).

I'll let someone who is better versed in proof theory (which means almost anyone except my dog) deal with that. I am beginning to think that the problem should have said

[MATH]\text {Given: } sin(u) \ne - sin(v) \text { and } sin(u) \ne 0 \ne sin(v) \ ...[/MATH]
Now Dr. Peterson and I both tried to warn you that a proof that shows

[MATH]y = f(x) \implies y = g(x) \text { and } z = h(x) \implies z = g(x)[/MATH]
does not necessarily prove that [MATH]y = z \text { for all } x.[/MATH]
You also have to show that

[MATH]z = g(x) \implies z = h(x) \text { and } y = g(x) \implies y = f(x).[/MATH]
In other words, you have to show that the whole process is reversible. Now I have no doubt that it is in this case, but my understanding is that you have to show that.
 
We generally don't pay close attention to domain issues (that is, whether everything is defined) in teaching identities, as there is enough difficulty in just finding a way through.

But here, it is clear that the LHS is defined only when sin(u), sin(v), and sin(u) + sin(v) are non-zero (the first two because we have the cosecants, and the third because the denominator would be zero in that case).

The RHS is defined only when sin(u) and sin(v) are non-zero and sin(u)sin(v) is not 1; the latter is true except when both sines are 1 or both are -1. So the domains are different; but we can say that the identity is true whenever both sides are defined, which is true of many other identities.

And that is the condition under which the transformations are reversible.

@allegansveritatem, I'll point out that you omitted some required parentheses in your work on each side; but we can tell what you meant.
 
JeffM said:
I am a bit worried about the cancellation of sin(u) + sin(v) because of the 0/0 problem, but perhaps it is implicit that sin(u) cannot equal -sin(v).

I'll let someone who is better versed in proof theory (which means almost anyone except my dog) deal with that. I am beginning to think that the problem should have said

[MATH]\text {Given: } sin(u) \ne - sin(v) \text { and } sin(u) \ne 0 \ne sin(v) \ ...[/MATH]
Now Dr. Peterson and I both tried to warn you that a proof that shows

[MATH]y = f(x) \implies y = g(x) \text { and } z = h(x) \implies z = g(x)[/MATH]
does not necessarily prove that [MATH]y = z \text { for all } x.[/MATH]
You also have to show that

[MATH]z = g(x) \implies z = h(x) \text { and } y = g(x) \implies y = f(x).[/MATH]
In other words, you have to show that the whole process is reversible. Now I have no doubt that it is in this case, but my understanding is that you have to show that.
Click to expand...
I think I can do that. I will do it tomorrow...I don't see how I can miss. All I need to do is follow the thread back on the LS.
 
Here it is. It really wasn't that hard.

When I was in high school my trig teacher always wondered why in many of my identity proof on exams I had a big gap in the middle.

IMG_0407.jpgIMG_0408.jpg
 
Dr.Peterson said:
We generally don't pay close attention to domain issues (that is, whether everything is defined) in teaching identities, as there is enough difficulty in just finding a way through.

But here, it is clear that the LHS is defined only when sin(u), sin(v), and sin(u) + sin(v) are non-zero (the first two because we have the cosecants, and the third because the denominator would be zero in that case).

The RHS is defined only when sin(u) and sin(v) are non-zero and sin(u)sin(v) is not 1; the latter is true except when both sines are 1 or both are -1. So the domains are different; but we can say that the identity is true whenever both sides are defined, which is true of many other identities.

And that is the condition under which the transformations are reversible.

@allegansveritatem, I'll point out that you omitted some required parentheses in your work on each side; but we can tell what you meant.
Click to expand...
I'm aware that the sine and cosine functions can be zero...and tangent can be zero and can also be undefined, (can sine and cosine be undefiined too?)all of which conditions could throw a wrench into the works...but, as you say, it is a little much for a beginner to worry overmuch about domains when in the throes of verifying identities. It is a bad enough business getting the SOB on the LS to look like the SOB on the RS without setting up a bunch of restrictions. My author doesn't pay too much attention to the matter (yet) of domain and definition as I note from the fact that he routinely throws in problems where the solution entails tan+cot and tan times cot should equal one.
 
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