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you don't actually need to construct a triangle although that's probably the most obvious method. You can also note tan2theta = +-sqrt(sec^22theta-1). since tan 2theta is in this case negative since 2arcsin3/4 is in the 2nd quad, tan2theta = -sqrt(sec^2(2theta)-1) = -sqrt(1/cos^2(2theta) -1 )) =...

yep looks legit

change sin2x into (sinx+cosx)^2-1

dr du the goat

😊

ye i think i just set it at that since that atar should be enough to get me into basically any course (except like usyd med)

Yep thanks for the info. I wanted to sit the bos trials last yr but was in queensland at that time so I'll definitely try to sit them this year

ye but bos papers are harder typically

yes but there's still a lot of relevant questions

what?

Hi, were there any BOS papers made before 2012 for ext 1/2 maths and if so, where can I find them?

you can also do f = qvb for the sea of delocalised electrons in the copper wire lmao

is it C?

easiest way to do should be considering how f(x+1-1) + f(1-(x+1)) = f(x) + f(-x) = 2(x+1)^2+8

Ur allowed to use out of syllabus techniques in the hsc i think as long as you don't make any mistakes. However for internals, you'd have to do as your teacher says if you don't want to lose marks. That being said, imo implicit diff is simply an application of the chain rule so I don't agree...

Honestly rn seems to be a bit early for an accurate atar estimate but if u wanna maximise atar, it'll probably be good to focus on improving physics and english

use the two forms of CB.CD

uh this is a bit late since i just saw this, but my method is as follows for the inductive step. First off, we split the k+1 pieces of bread falling into k pieces falling and then the last piece falling Then we take two cases: 1- after the first k bread have fallen, the number of bread slices...

otherwise 1/4 would be a square number cos it's (1/2)^2

However, a square number by definition is a square of another integer, in this case u have no way of guaranteeing m/n is an integer.