problem with double arctan

[allegansveritatem]

I will present the problem in this panel and then in the one immediately below I will show what I have done. I have too may images for one panel:
Problem:

 

[allegansveritatem]

Here is what I have done with a) b) and c)
a)
b)
c)

I want to ask a question about a first: Namely, this seems correct to me. Is it?
and a question about b) How do I justify going from two arctans to tan(a+b)?
I will save c) for later.

 

[Dr.Peterson]

Yes, everything is correct.

I myself would have labeled the upper and lower parts of theta, probably as alpha and beta, so that theta = alpha + beta; then tan(alpha) = 8/x and tan(beta) = 2/x. From that, for (a), alpha = arctan(8/x), and so on; and for (b), theta = alpha + beta and we use the addition formula.

For (c), use the form from (b), not the form from (a). Why do you think they had you write it??

 

[allegansveritatem]

Yes, everything is correct.

I myself would have labeled the upper and lower parts of theta, probably as alpha and beta, so that theta = alpha + beta; then tan(alpha) = 8/x and tan(beta) = 2/x. From that, for (a), alpha = arctan(8/x), and so on; and for (b), theta = alpha + beta and we use the addition formula.

For (c), use the form from (b), not the form from (a). Why do you think they had you write it??

well, here is my dilemma: I don't see why we go from two arctans to two tans. I mean...what is the reasoning there? What is the logic? I know that the arctan is the angle. Arctan= theta or alpha or whatever. In short: Why do we use the tan(a+b) form? I mean, we are looking for an angle and not a tangent. Of course, one might say, if you have a tangent you can easily find an angle....but feels a little backdoorish to me somehow.
As for c) I am trying to recall my problem with that...I don't have access to my notebook right now but I will trace my staggering footprints today and comeback with my question. I do recall that when I tried to work out the problem I did use the form in b) but seems to me that I ran into a snag along the way.

 

[Dr.Peterson]

well, here is my dilemma: I don't see why we go from two arctans to two tans. I mean...what is the reasoning there? What is the logic?

I know that the arctan is the angle. Arctan= theta or alpha or whatever. In short: Why do we use the tan(a+b) form?

I mean, we are looking for an angle and not a tangent. Of course, one might say, if you have a tangent you can easily find an angle....but feels a little backdoorish to me somehow.

But in what I suggested for part (b), you don't go from arctan to tan, but from tan to arctan! I'll show it to you completely, since you don't seem to have followed. It's the arctan that is a back door, and a very useful one.



[MATH]\tan(a) = \frac{8}{x}[/MATH], [MATH]\tan(b) = \frac{2}{x}[/MATH], [MATH]\theta = a + b[/MATH][MATH]\tan(\theta) = \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} = \frac{\frac{8}{x} + \frac{2}{x}}{1 - \frac{8}{x}\frac{2}{x}} = \dots[/MATH]
Then theta is the arctan of the expression we end up with. And for part (c), you'll use what we have before taking the arctan.

 

[allegansveritatem]

But in what I suggested for part (b), you don't go from arctan to tan, but from tan to arctan! I'll show it to you completely, since you don't seem to have followed. It's the arctan that is a back door, and a very useful one.

View attachment 19311

[MATH]\tan(a) = \frac{8}{x}[/MATH], [MATH]\tan(b) = \frac{2}{x}[/MATH], [MATH]\theta = a + b[/MATH][MATH]\tan(\theta) = \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} = \frac{\frac{8}{x} + \frac{2}{x}}{1 - \frac{8}{x}\frac{2}{x}} = \dots[/MATH]
Then theta is the arctan of the expression we end up with. And for part (c), you'll use what we have before taking the arctan.

well, that is pretty much what I did too in part b. I used the addition formula for tangents. What I mean by "going from actan to tan" being hard to follow is: In part a we get the equation: actan 8/x plus actan2/x equals theta. Then in part b the author of the problem says to use the addition forumla for tangents to prove theta = actan 10x/x^2-16....I think what I need to do is to think it through more thoroughly. 10x/x^2-16 is the tangent of theta. And theta is composed by the addition of two tangents, so why wouldn't we use the addition formula? Good. I'm getting there. Thanks for the pointers here. Here, by the way is what I did for part c:
I changed the tan 45 into a decimal...I was not sure how to do the quad formula otherwise.

 

[Dr.Peterson]

Think again: What is tan(45°)? I'm not sure what you did there.

It turns out that you could have solved by factoring, and don't need decimals (though there's nothing wrong with decimals in this sort of problem).

Always. Check. Everything.

 

[allegansveritatem]

Think again: What is tan(45°)? I'm not sure what you did there.

It turns out that you could have solved by factoring, and don't need decimals (though there's nothing wrong with decimals in this sort of problem).

Always. Check. Everything.

Truth to tell, I don't like factoring quadratics unless the factors jump out at me for the obvious reason that otherwise it takes a long time to ferret the factors out. I use the lovely quadratic formula instead (thank you Mr Pythagoras). When you say always check I take it that you refer to what I came up with in part c. I will go over it today. plug in my results and see what happens. I will come back to this with results.

 

[Dr.Peterson]

Truth to tell, I don't like factoring quadratics unless the factors jump out at me for the obvious reason that otherwise it takes a long time to ferret the factors out. I use the lovely quadratic formula instead (thank you Mr Pythagoras). When you say always check I take it that you refer to what I came up with in part c. I will go over it today. plug in my results and see what happens. I will come back to this with results.

Actually, I was telling you to check everything you write down. Do you really think that tan(45°) = 1.61978? You evidently had your calculator in radian mode. You should know the trig functions of special angles.

I agree on factoring only when I can see it quickly; I tell students that I look at a quadratic equation for about ten seconds to see if factoring will be easy, then use the formula. My point was merely that the equation will be much simpler when you use the right value for the tangent; the solutions are rational (which is true when an equation can be factored), and you don't have to worry about using decimals.

But Pythagoras had nothing to do with the quadratic formula ...

 

[allegansveritatem]

Actually, I was telling you to check everything you write down. Do you really think that tan(45°) = 1.61978? You evidently had your calculator in radian mode. You should know the trig functions of special angles.

I agree on factoring only when I can see it quickly; I tell students that I look at a quadratic equation for about ten seconds to see if factoring will be easy, then use the formula. My point was merely that the equation will be much simpler when you use the right value for the tangent; the solutions are rational (which is true when an equation can be factored), and you don't have to worry about using decimals.

But Pythagoras had nothing to do with the quadratic formula ...

Right, I had the calculator in radian mode and did not catch the fact that the tangent it spit out could not possible be right for 45 degrees since opp and adj are equal. I caught that mistake today and redid part c thus:

and I checked it thus:
As for Pythagoras...no, maybe not. I recall in one of my texts coming across a story about archaeologists finding a proof for the quadratic formula in a cuneiform tablet so...my thanks were misplaced. I think this Babylonian proof was very much like the proof from geometry of Euclid, and might be where Euclid got it.

 

[Dr.Peterson]

Thanks for not commenting on my wrong claim that the equation could be solved by factoring; I did that in my head and made a sign error ...

The story you recall is not about the quadratic formula, but the Pythagorean Theorem, a^2 + b^2 = c^2, which was proved by Babylonians (apparently) and by Pythagoras (reputedly). The quadratic formula was proved (in a primitive form, before variables and negative numbers made it easier to state) much later, around 600 AD.

 

[allegansveritatem]

yes, it was the quadratic equation that was in question with the Babylonians not the quadratic formula. At any rate there was something about completing the square.