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Fair enough. Anyway, as integrand pointed out these are trivial consequences of the pythagorean identity. I do find it silly that these aren't allowed to be stated though, given that they are natural steps in the derivation of the "standard t-formulae".

Cheers for the affirmation bro :). (Am 95% sure you understood that I was using painting as a noun.)

Cambridge is as good as anyone needs for HS.

I was fairly sure things like \sin(x)=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}} were in syllabus? I remember seeing the corresponding triangles in HS books, but I suppose I could be mistaken. Isn't that how the t-formula themselves would be derived anyway? Using the expressions for sin(x) and cos(x)...

\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}} \cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}=\frac{1}{\sqrt{1+\tan^2(x)}}

Because we are dealing with squares. I think \cos^2(x)=\frac{1}{1+t^2} is nicer to work with than \cos^2(x)=\left(\frac{1-t^2}{1+t^2}\right)^2. If there are any odd powers floating around though you will get square roots involved, which are why we normally use tan(x/2) I guess.

Was high as balls and more interested in the painting of a fractal on my wall than my then-gf. Second time was great though, didn't make that mistake again!

It is good (and indeed efficient) to have basic things like those trig ratios committed to memory, but imo you should NEVER use anything that you can't answer the question "why is this true?" about. It is bad practice, impedes understanding, and this hurts you more the higher level you are...

Yeah, sounds like you are just getting people to do your homework. If you can't do it yourself exams/assessments are going to be a rude shock lol.

Latex appears to be broken on this site right now, if you click the reply with quote button you will see the code for my simple notation (I just define D as the single variable differentiation operator). And yes it is meaningful to talk about irrational derivatives (equivalently irrational...

I can't think of any simple concept to directly relate it to as you can with the operator D:=\frac{d}{dx} and the concept of change. (You don't often get such a natural conceptualisation of the more advanced mathematical constructions.) But to motivate it, it is pretty natural to work with...

By the way, you should probably avoid the phrase "fractional calculus" in future. This phrase already has a very specific meaning (involving things like taking "half" of a derivative of a function).

Re: HSC 2015 4U Marathon - Advanced Level Do you insist that the domain contains 0? Because that forces \alpha \geq 1 otherwise we blow up at 0. And for \alpha >1 then we tend to 0 from above at both 0 and infinity, which makes injectivity impossible. For \alpha=1 this is of course just a...

Re: HSC 2015 4U Marathon - Advanced Level Your logic is only 1-directional when you introduce the arbitrary odd function O. Indeed any f must be of such a form (apart from the constant solution 1), but a generic odd O will not do. Eg O(x)=0 does not work.

Re: HSC 2015 4U Marathon - Advanced Level Find all functions f:\mathbb{Q}\rightarrow \mathbb{Q} such that: f(xy)=f(x)f(y)-f(x+y)+1.

Re: HSC 2015 4U Marathon - Advanced Level Note that composing a reflection about 0 and a reflection about 1 (in that order) gives us the translation x\mapsto x+2. This means that f(x)=f(x+2) for all real x. In particular, a_n=f(1/2n). Now, by a straightforward induction we have...

Re: HSC 2015 4U Marathon - Advanced Level x=0 tells you 0 is a root. x=k for positive integral k tells you k-1 is a root => k is a root. The only poly with infinitely many roots is the zero poly.

Re: HSC 2015 4U Marathon - Advanced Level By factorising the LHS we are forced to have x-y=1 and x+y=p. So x=y+1 and p=2y+1. So there are only solutions for odd primes, and for each odd prime p we have the unique solution pair ((p+1)/2,(p-1)/2).

Re: HSC 2015 4U Marathon - Advanced Level It has no POSITIVE integer solutions because x^2 < x^2+x+1 < (x+1)^2.

Re: HSC 2015 4U Marathon - Advanced Level No, it is just a relabelling so we don't have to always pay attention to n's parity. This is different to squaring both sides of your equation. (Which is generally invalid as a solution method.)