Why $\frac{\pi}{12}$ equals to $\frac{\pi}{3} - \frac{\pi}{4}$ 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
 A: 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.
$$\frac{\pi}{3}-\frac{\pi}{4} = \pi\left(\frac{1}{3}-\frac{1}{4}\right)=\pi\left(\frac{1}{12}\right)=\frac\pi{12}$$
A: 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.
A: $$\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi}{4\cdot3}-\frac{3\pi}{3\cdot4}=\frac{4\pi-3\pi}{12}=\frac{\pi\left(4-3\right)}{12}=\frac{\pi\left(1\right)}{12}=\frac{\pi}{12}$$
