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the way i did bii) was proving that \cos\left(\frac{2^{n+2}\pi}{9}\right) - \cos\left(\frac{2^{n-1}\pi}{9}\right) = 0 using sum to product formula surely i get the full marks right

a=i b=j c=i is a counterexample im pre sure as the angle of a+b+c is smaller

for ii) i proved like cos(2^n+2) pi/9) = cos(2^(n-1) pi/9) and just like recursively found my way to -1/8 hopefully thats right lol

those equations will most likely not be examined as they are not accurate to "square" resistance

are you going to be streaming your reaction to the test again this year?

if u wanna do it just say above "let sinx = s and cosx = c" or equivalent just to be safe

prepare for the worst and hope for the best :)

nope i believe thats under induction. it probably falls under this one but idk im not a syllabus slave never looked at this thing before tbh

its a mix of proof an integration and im pretty sure its in cambridge

tbh it probably wont show up, but given that there was something similar in the math adv hsc (2022 q29) i would just say give the question a go its not too bad

ruse 2020

if it does appear then it will probably be some squeeze theorem with riemann squares. usually the questions arent too hard tho

wasnt q13 c last year an 8 marker projectile motion question? we probably will get one in q14 or 15 i reckon tho

yea tru we more likely to get some complex bashing geometry question but if the exam markers are lazy they could totally just steal an inequality off aops lol

we bouta get an imo inequality in q16

we dont, we let k=4p+r

since k is an integer then it can be written in the form 4m+r, where r is the remainder. we choose k instead of n as n is specifically a square integer (k^2) and if u do substitute it as 4m+r u probably wont be able to easily find the contradiction as u cant really do any working out if u...

h is only equal to \sqrt{\frac{2kt}{\pi}} if dh/dt = k/hpi. since the rate of change to empty the container is not equal the rate to fill the container then you cannot substitute h as \sqrt{\frac{2kt}{\pi}}. also in ur final line u cube rooted both sides of the equation but the 2 is only square...

in the final line u proved that the area is less than a value, which means that value is the maximum. since equality occurs when x=y=z, then the maximum occurs when it is an equilateral triangle

equality is achieved when x=y=z from the 2nd line and that holds for the rest of the proof