How to approach this

[JeffM]

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.

Exactly. In most cases, it is a pure formailty. Indeed, if you use correctly the following notation, the formality is addressed as you go

[MATH]p = \dfrac{sin(u) + sin(b)}{csc(u) + csc(v)} \iff \\ p\{csc(u) + csc(v)\} = sin(u) + sin(v)[/MATH]In your presentation you explicitly affirm reversibility at each step. Of course if you can't show reversebility at each step, then your proof is flawed.

 

[allegansveritatem]

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.

I ask in the above if the sine and cosine can be undefined. It dawned on me shortly after waking this gray morning that of course they cannot in consequence of the simple fact that a hypotenuse can never be zero.

 

[allegansveritatem]

OK, cool. I apologize you posting it. I like you attitude!! Good for you. Please post back with your work so we can how well you have done.

Will do.

 

[allegansveritatem]

Exactly. In most cases, it is a pure formailty. Indeed, if you use correctly the following notation, the formality is addressed as you go

[MATH]p = \dfrac{sin(u) + sin(b)}{csc(u) + csc(v)} \iff \\ p\{csc(u) + csc(v)\} = sin(u) + sin(v)[/MATH]In your presentation you explicitly affirm reversibility at each step. Of course if you can't show reversebility at each step, then your proof is flawed.

If I can do it, I can undo it. We shall see, at any rate.

 

[JeffM]

I ask in the above if the sine and cosine can be undefined. It dawned on me shortly after waking this gray morning that of course they cannot in consequence of the simple fact that a hypotenuse can never be zero.

In fact, if you think in terms of a unit circle, the hypotenuse equals 1 by definition. Sleep does wonders for the mind.

 

[allegansveritatem]

In fact, if you think in terms of a unit circle, the hypotenuse equals 1 by definition. Sleep does wonders for the mind.

Absolutely

 

[allegansveritatem]

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.

View attachment 19516View attachment 19517

I did it this way: I went from the RS and got to sinu sinv and then reversed by going from sinu sinv to the LS expression where the whole thing started thus:

Is this acceptable. I know it is not the way you did it but...

 

[Steven G]

I did it this way: I went from the RS and got to sinu sinv and then reversed by going from sinu sinv to the LS expression where the whole thing started thus:
View attachment 19561
Is this acceptable. I know it is not the way you did it but...

Looks good to me. Good job!

 

[JeffM]

I did it this way: I went from the RS and got to sinu sinv and then reversed by going from sinu sinv to the LS expression where the whole thing started thus:
View attachment 19561
Is this acceptable. I know it is not the way you did it but...

tkhunny says that correct answers do not care how you find them.