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I'm going back to basic trigo for the sake of being able to help my kids and also being bad younger at it, I want to be able to overcome that lack of understanding and honestly, I hate unfinish business.

So please bear with me if you feel my question is really basic or stupid (and if you feel I should close it, please leave a comment)

I was going through this book and at the 17th page, it said something like that:

$$\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$$

Looking on the net, it says something about angles in the triangle but the lack of precision is appaling and I could not get it.

Your insights are more than welcomed

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    $\begingroup$ Can you subtract fractions? What is $1/3 -1/4$? $\endgroup$ – Ant May 12 '16 at 8:19
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    $\begingroup$ Well:$$\frac{\pi}{3} - \frac{\pi}{4}=\frac{4\pi}{12} - \frac{3\pi}{12}=\frac{4\pi-3\pi}{12} =\frac{\pi}{12}$$But this may not be what you're looking for? $\endgroup$ – StackTD May 12 '16 at 8:19
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    $\begingroup$ $1/3 - 1/4 = (4 - 3)/12$ is the simplest, but not sure if this is the approach you need or whether some property of the geometric object (by the way, the link you give doesn't go to any particular page, just to details of the book itself). $\endgroup$ – jim May 12 '16 at 8:19
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    $\begingroup$ @AndyK Since you say you are doing this for your kids, this is a good opportunity for you to see that without basic arithmetic skills (multiplication tables, fractions, some exponents, etc.) nobody has much chance to understand trigonometry or any other a little more advanced material. $\endgroup$ – DonAntonio May 12 '16 at 8:30
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    $\begingroup$ Hi @Joanpemo, I felt very very bad for not seeing something so simple... But the lesson has been learnt. :) And definitely, I'll even tell them that I missed that. There is no age to learn $\endgroup$ – Andy K May 12 '16 at 8:33
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$$\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi}{4\cdot3}-\frac{3\pi}{3\cdot4}=\frac{4\pi-3\pi}{12}=\frac{\pi\left(4-3\right)}{12}=\frac{\pi\left(1\right)}{12}=\frac{\pi}{12}$$

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  • $\begingroup$ Hi @jan-eerland, thanks for your answer. it seems to be downright basic and I'm blushing of shame looking at it. Are there no special rules here? I thought there was something more in it ... $\endgroup$ – Andy K May 12 '16 at 8:22
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    $\begingroup$ @AndyK If you have been away from mathematics for some years than it may not be obvious. Don't feel bad. $\endgroup$ – MathMajor May 12 '16 at 8:27
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    $\begingroup$ Hi @MathMajor, thanks for the uplifting comment. I was always bad at geometry. As I said in the comment's answer to Yves's comment, the teachers have some responsibilities about it. But anyway, it is never too late and Jan's answer made me understand something that eluded me. $\endgroup$ – Andy K May 12 '16 at 8:31
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    $\begingroup$ @AndyK You're welcome, I'm glad that I good help $\endgroup$ – Jan May 12 '16 at 8:45
  • $\begingroup$ Hi @JanEerland, I was going a bit mentally about my question yesterday and well a bit like the (in)famous XY problem, I've figured that I wanted to know how our 2 chaps were able to break it that way $\frac{\pi}{3} -\frac{\pi}{4}$ and not like that $\frac{\pi}{6}-\frac{\pi}{2}$? Any hints are more than welcomed ... $\endgroup$ – Andy K May 13 '16 at 12:16
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The hard way:

The Tchebytcheff polynomial of order $12$, such that $T_{12}(\cos(x))=\cos(12x)$, is

$$2048x^{12}-6144x^{10}+6912x^8-3584x^6+840x^4-72x^2+1.$$ We equate it to $-1=\cos(\pi)$ and we set $x^2=t$, giving

$$2048t^{6}-6144t^{5}+6912t^4-3584t^3+840t^2-72t+2=0,$$

which is a perfect square by symmetry, and

$$2(32t^3-48t^2+18t-1)^2=0.$$

We know that $\cos^2(\frac\pi4)=\frac12$ is a root, and by synthetic division we factor

$$2 (2 t-1)^2 (16 t^2-16 t+1)^2=0.$$

Then taking the appropriate root $$\cos^2\left(\frac\pi{12}\right)=\frac{\sqrt3+2}4$$ and $$\cos\left(\frac\pi{12}\right)=\sqrt{\frac{6+2\sqrt6\sqrt2+2}{16}}=\frac{\sqrt6+\sqrt2}4.$$

On another hand,

$$\cos\left(\frac\pi3-\frac\pi4\right)=\frac12\frac1{\sqrt2}+\frac{\sqrt3}{2}\frac1{\sqrt2}=\frac{\sqrt6+\sqrt2}4,$$ which seems to substantiate the claim.

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    $\begingroup$ I love the hard way. Thanks for your uplifting answer; You are a Teacher, a real one. Only minor issue is I will have to provide work, Ah well, definitely can do that :) $\endgroup$ – Andy K May 12 '16 at 9:19
  • $\begingroup$ +1 fun way! But of course one needs to be careful at the end as $\cos x = \cos y$ "not implies" $x = y$ (PS do you know how to make a "not imply" symbol? $\endgroup$ – Ant May 12 '16 at 13:38
  • $\begingroup$ @ant: precisely why I said "seems to substantiate". $\endgroup$ – Yves Daoust May 12 '16 at 13:38
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    $\begingroup$ This answer illustrates the potential to learn something interesting by clicking on any MSE question, no matter how modest it looks. :-) $\endgroup$ – Bungo May 12 '16 at 16:25
  • $\begingroup$ nice way to look at a simple problem. But i must say it is not just hard but rather too hard for the problem at hand. Anyway +1 to add the "fun quotient" on MSE. $\endgroup$ – Paramanand Singh May 13 '16 at 9:15
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Perhaps the discomfort here stemmed from the use of an abstract symbol. We're using a symbol $\pi$ to represent a number, $3.14159\dots$, which we otherwise cannot possibly write down in its entirety (the digits go on forever!). Remember that $\pi$ is just a number. It's somewhere between $3$ and $4$, closer to $3$. It acts like any other number during mathematical operations.

If you see something like $$\frac{\pi}{3}-\frac{\pi}{4}$$ you can try reading it verbally to clarify what is happening, "one third of pi minus one fourth of pi". If we disregard the $\pi$, then this is just a third of something minus a fourth of the same thing. What's a third minus a fourth of something? A twelfth of something.

$$\frac{1}{3}-\frac{1}{4} = \frac{4}{12}-\frac{3}{12} = \frac{1}{12}$$

Then we must have one twelfth of pi.

$$\frac{\pi}{3}-\frac{\pi}{4} = \pi\left(\frac{1}{3}-\frac{1}{4}\right)=\pi\left(\frac{1}{12}\right)=\frac\pi{12}$$

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  • $\begingroup$ Hi @zahbaz, I've tried that $$\frac{\pi}{6} - \frac{\pi}{2}$$ to see the result, that is way different. So I've got my answers. Overcoming my fear of formalization. Thank you again $\endgroup$ – Andy K May 12 '16 at 8:46

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