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[Savy]

The theta comes from eulars formula

e^(ix) = cos(x) + isin(x)

Which isn't actually an angle in terms of the problem itself it's just a byproduct of solving some part of the equation linking it back to argand diagrams. So the theta comes from this and is why it seems to appear out of nowhere.

Hence where the right angled triangle comes from and hence the relationship I mentioned in my previous post.

I got this from his full solution here http://www.numberphile.com/pi/Pi_Balls_Ed.pdf

Which is in absolutely no way obviously related (or linked) in the video itself.

edit - FUN FACT

When I was doing further maths at a-level the link e(ix) = cos(x) + isin(x) made it so I didn't have to remember lots things I was meant to learn for rote in normal a-level maths. A few times I got comments form the teacher saying I have no idea what you use to solve these problems but you seem to get them right every time.

[MadMojoMonkey]

The theta comes from eulars formula

e^(ix) = cos(x) + isin(x)

Which isn't actually an angle in terms of the problem itself it's just a byproduct of solving some part of the equation linking it back to argand diagrams. So the theta comes from this and is why it seems to appear out of nowhere.

I see.

Since we know that (1-x)/(1+x) will always lie between 0 and 1 for x <1, and we define x = m/M <= 1/16, it fits the domain.
Since cos(anything) will always return a value from -1 to 1, the range 0 to 1 is contained therein. Therefore cos(anything) can yield the same answer as the fraction, provided the correct input.

In this case, he's not really concerned with any physical representation of theta, just that it can encode the same result if chosen wisely to correspond to x, which he sets out in definition by saying cos(theta) = (1-x)/(1+x). It's not an approximation, it's a straight correlation he creates out of convenience.

Then the magic is that the result utilizes this transformation from x to theta without needing to really be back transformed. He shows that the complete kinematic equations for any time are simplified down to a sine and cosine function describing the 2 bodies' motions, given only the input of how many collisions they've had, and their relative masses.

He back transforms the answer just to pin down values of x = m/M, but that final result is kinda moot. We know that there are some values of x which have this property, the exact values are not really special.