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I think the same Q got asked and answered here: http://community.boredofstudies.org/238/extracurricular-topics/349271/laters-maths-help-thread.html#post7133125 .

You could also have answered the first Q. in a similar way to the Wikipedia proof linked. Then the Wikipedia proof becomes a generalisation of the first proof.

It's not really a rule of thumb, more like a rule. Like if you see the image here, the angle between two vectors is the angle when the vectors have their tails lined up: https://upload.wikimedia.org/wikipedia/commons/0/05/Inner-product-angle.png . If you lined up one head with one tail, the...

$\noindent When finding the angle between two vectors, we line them up so that their tails are starting at the same place (or it is equivalent to have their heads start at the same place). So you should find the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$ (draw a diagram of the...

We have things like tan(x) and tan(y) now, but want the answer in terms of cot(x) and cot(y) instead. So we replace each tan with a 1/cot, and simplify the whole thing.

This is because switching them around is the same thing. For fractions, 1/(a/b) = b/a.

Yeah, cot(x+y) = 1/tan(x+y) = [1 - tan(x)tan(y)]/[tan(x) + tan(y)]. Now convert things to cot.

This is the compound angle formula for cot. It can be derived from the corresponding formula for tan.

Yeah. Then you can simplify that to get it down to what the RHS is.

Using t-formulas, the RHS is (1+t)/(1-t), provided t =/= 1. You can also express the LHS in terms of t using the t-formulas, and show it simplifies to (1+t)/(1-t).

Yeah, you can do it via half-angle formulas (convert sin(x) and stuff into trig. functions with x/2 instead of x).

You could try using t-formulas.

That last Q. is essentially just based on using the higher-order derivative test ( https://en.wikipedia.org/wiki/First_derivative_test#Higher-order_derivative_test ). So you look for places where the first and second derivatives are simultaneously 0 for a horizontal point of inflexion, and out...

Use the identities sin^2 (x/2) = (1/2) (1 - cos(x)) and sin(x) = 2 sin(x/2) cos(x/2). I.e. Half-angle formulas, as Paradoxica alluded to. (Surely they didn't forbid use of these?)

Re: First Year Uni Calculus Marathon Yeah you can move the limit inside for a continuous function. If lim as x -> a of g(x) is b and f is continuous at b, then lim as x -> a of f(g(x)) = f(b).

Re: HSC 2016 MX2 Combinatorics Marathon Is this past HSC Q. the one about waiting times (I think might have been the last Q. of a 2U paper)?

Re: HSC 2016 MX2 Combinatorics Marathon I'd interpreted it as that too, main reason being that braintic had referred to a "length" in part (a), so I thought he was talking about this length again in part (b).

In order for this to be true, we need g to be of fixed sign (and integrable of course), rather than simply non-zero. (If g is given to be continuous though, then g(x) =/= 0 suffices.) A proof of that result is given here...

Re: HSC 2016 MX2 Combinatorics Marathon Let x be in [0,1]. Consider the unit square in the a-b plane (the region where A,B's values are taken). Basically find the area on the unit square satisfying |a – b| ≤ x. To do this, sketch the lines b = a+x and b = a-x, and you'll see they intersect the...

Re: HSC 2016 MX2 Combinatorics Marathon Yeah :). So for clarification purposes, what was the Q. you and Paradoxica were doing again? Basically, this?: Let A, B be (independent and) uniformly distributed on [0, L] (L > 0, real). Suppose Pr(|A – B| ≤ 1/3) = 1/3. Find L. ??