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allegansveritatem

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Here is the problem:tanx1.PNG

Here is a small selection of what I tried:tanx 2.PNG

I worked on this for most of two hours today without making any headway. I was able to transform the left side into various strange configurations but could not for the life of me get anything that looked like the right side. How to approach this matter?
 
I used the angle-sum formula for the tangent, rather than for sin and cos (though you can derive the former from the latter). After writing that out, I could see sin(h)/h and 1+tan^2(x) within the expression I got, and head in the right direction. Give that a try.
 
Dr.Peterson said:
I used the angle-sum formula for the tangent, rather than for sin and cos (though you can derive the former from the latter). After writing that out, I could see sin(h)/h and 1+tan^2(x) within the expression I got, and head in the right direction. Give that a try.
Click to expand...
you mean tan (a+b) =( tan a + tan b)/ 1- tan a tan b, right? I have tried that too...but I will try again. It seems that this problem may require the use of both the angle sum formula and some kind of sin cos formulation. Then, there is the business of the sec^2. Sec^2 - 1 is tan^2, I know...well, I will give it a go again tomorrow. Thanks for the tip.
 
Yes, that's the formula. I will add that after applying it, I combined fractions with a common denominator, found that a term cancels, factored out a tan and rewrote that as sin/cos.

By the way, what does the problem have to do with "maximum opacity"? Are you just saying it's hard?
 
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Dr.Peterson said:
Yes, that's the formula. I will add that after applying it, I combined fractions with a common denominator, found that a term cancels, factored out a tan and rewrote that as sin/cos.

By the way, what does the problem have to do with "maximum opacity"? Are you just saying it's hard?
Click to expand...
Yes, Maximum opacity is a reference to the feeling I get when trying to see a path through to a solution to this thing.

I went at it again today and, sad to report, still couldn't find an opening. Instead of cracking the nut I broke my teeth. Here is what I tried:
tan 3.PNG

I would try something and see it couldn't work and start again. This went on for over an hour and I had to move on to another problem. Too frustrating. I have tried approaching this from both right side and left. I thought I might reverse engineer a solution but that didn't prove fruitful either. I kept trying to find a way to get rid of that 1 in the denominator of the angle-sum fraction. I thought that if I could do that I could split the fraction and cancel it piecemeal, so to speak. No could do. There is probably some little thing I am not seeing...but what?
 
The one place I see you really trying to use the tangent of a sum is on the last line, where you made an algebraic error.

Here are the first couple steps of my work:

[MATH]\frac{\tan(x+h)-\tan(x)}{h} = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \tan(x)\right) = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \frac{\tan(x)(1-\tan(x)\tan(h))}{(1-\tan(x)\tan(h))}\right) = \dots[/MATH]​

See where you can go from there.
 
Dr.Peterson said:
The one place I see you really trying to use the tangent of a sum is on the last line, where you made an algebraic error.

Here are the first couple steps of my work:

[MATH]\frac{\tan(x+h)-\tan(x)}{h} = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \tan(x)\right) = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \frac{\tan(x)(1-\tan(x)\tan(h))}{(1-\tan(x)\tan(h))}\right) = \dots[/MATH]​

See where you can go from there.
Click to expand...
Thanks I will give it another go tomorrow. I see a tan x^2 coming from the last expression there. Well, I will see what can be done tomorrow. I think I am clutching on this problem, treating it like a quicksand trap rather than something to play with. I'll have to get creative.
 
Dr.Peterson said:
The one place I see you really trying to use the tangent of a sum is on the last line, where you made an algebraic error.

Here are the first couple steps of my work:

[MATH]\frac{\tan(x+h)-\tan(x)}{h} = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \tan(x)\right) = \frac{1}{h}\left(\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)} - \frac{\tan(x)(1-\tan(x)\tan(h))}{(1-\tan(x)\tan(h))}\right) = \dots[/MATH]​

See where you can go from there.
Click to expand...
I worked on this again today for more than an hour. I derived the sec^2 but still didn't verify the identity. I would publish some of my results but they are unsightly and wrong in every sense on top of that so I will hold back for now. I intend to go at it again tomorrow--I "propose to fight it out on this line if it takes all summer", as the feller said, and I will post more results as they show more promise. I will say that it is certainly educational to try every stunt you can come up with even if nothing seems to work.
 
Here's another hint, in case you want it: what happens if you split the tan(x) that is a factor of the numerator, and multiply the rest of the numerator by sin(x) and the denominator by cos(x)?
 
Dr.Peterson said:
Here's another hint, in case you want it: what happens if you split the tan(x) that is a factor of the numerator, and multiply the rest of the numerator by sin(x) and the denominator by cos(x)?

Thanks. I will give that a try. Once I do get this problem I am going to brand it into my brain so I don't get tripped up again, at least not in exactly the same way.
Click to expand...
 
Dr.Peterson said:
Here's another hint, in case you want it: what happens if you split the tan(x) that is a factor of the numerator, and multiply the rest of the numerator by sin(x) and the denominator by cos(x)?
Click to expand...
I tried this approach, but got into a terrific bind with it ( I was not sure how you meant it to be applied and probably applied it in a cockamamy way)so I went back to the drawing board and came up with this--it is not what I want and I don't even know if it is in any way correct because I haven't yet input it into a graphing facility. but at least it has the virtue of having a sec squared. This:
tan 6.PNG

I have been fooling with this for about an hour a day for a week. Oh boy! Well, it is not so bad because I have reviewed a lot of algebra and tested a lot of my ideas with little models to be sure that they can work.
 
I don't know how you got the last two lines. Look at the start of the third line from the bottom. Compare to the goal.

You don't want the [MATH]\sec^2(x)[/MATH] (which you wrote wrong, by the way) to be where it is; you want a [MATH]\sin(x)[/MATH] there. Pull the [MATH]\sec^2(x)[/MATH] to the left, and split the [MATH]\tan(x)[/MATH] in the numerator into [MATH]\frac{\sin(x)}{\cos(x)}[/MATH]. Then move the sine where your [MATH]\sec^2(x)[/MATH] is ...

A key to working with identities is to keep your eye on the goal. You aren't just making random changes; you are trying to make "this" look like "that". So never forget what "that" is!
 
Dr.Peterson said:
I don't know how you got the last two lines. Look at the start of the third line from the bottom. Compare to the goal.

You don't want the [MATH]\sec^2(x)[/MATH] (which you wrote wrong, by the way) to be where it is; you want a [MATH]\sin(x)[/MATH] there. Pull the [MATH]\sec^2(x)[/MATH] to the left, and split the [MATH]\tan(x)[/MATH] in the numerator into [MATH]\frac{\sin(x)}{\cos(x)}[/MATH]. Then move the sine where your [MATH]\sec^2(x)[/MATH] is ...

A key to working with identities is to keep your eye on the goal. You aren't just making random changes; you are trying to make "this" look like "that". So never forget what "that" is!
Click to expand...
Good. I will try this today. Keep eye on goal is a good tip...but the goal here is so various and, you might even say (considering its parent) so exotic, that it is hard to look at without flinching. Nevertheless, I will soldier on.
 
The goal is very specific! Look for ways to obtain each part of the RHS, as you already did for the sec^2. The long denominator will be the last thing to hope for, and should drop out automatically once you have the rest.
 
Dr.Peterson said:
The goal is very specific! Look for ways to obtain each part of the RHS, as you already did for the sec^2. The long denominator will be the last thing to hope for, and should drop out automatically once you have the rest.
Click to expand...
I went at it again today and at first I had a problem once more with trying to see how I could do anything with a tan x in as much as, of the three tan x's that were in my numerator, one went into making up the sec^2 x and the other two cancelled each each other. The more I pondered this dilemma I saw that there was a way to do something, if instead of splitting a tan x I split the remaining tan h. When I tried this, Lo! everything fell into place like magic! It was clear to me then that when you suggested I split tan x you meant tan h. Once I split the tan h in the numerator, I was able to get a sin h in the denominator by splitting the tan h there and cancel out the cos h's. Here is my work:tan 7.PNG

That QED was hard earned and if it were money I would have to split it with you, say 30/70?, in your favor.

It takes a long time to do all the exercises at the end of the sections in this book but I am studying this subject because I have time now and because it interests me--so what would be the point in glossing over whatever I could get away with like a kid with too much on his plate to do anything else?
 
Yes, I was probably seeing too many tans in your writing and got mixed up. But your final work here looks identical to mine.

I trust the extra work only made the final pleasure of conquest sweeter.
 
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