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I believe it's a "he". Do some inspection!! inspection ~ investigation, where ~ is approximately equal to. Oh I like Arial coz its quicker to write If I dont put anything it looks like an "n". apparently you can just write "Times" (with no quotes its okay)

ooh... I did "(n-1)Int{ (cosx)^2(sinx)^n-2 dx" using integ by parts like in McLake solution (so I did integrations by parts 2 times, which return me to I_n) Part (iii) isn't tough with your hint ^^

hmm.. apparently maple syrup isnt that useful... oh thanks for the π symbol.

ngai did that while he was eating pancake with maple syrup, which is the secret to success in the HSC :D general solution... solution which is general enough to be understood by the ordinary people of our society, i.e. doesn't contain mathematical jargons

simbaaa.. your friend is calling you.... oh yes, I have exams.. how can I forget.. gotta hurry now... haven't studied at all :P so can we assume this identity or not?

I tried it but it gave me I_n (as in.. I_n = I_n, which is true but useless...)

Can you give a hint for part (b) (iii)? Thank you.

what formula? I assume that this is a serious question :P if not then... I'll go invisible Anyway, 1. It can be shown that the coefficients in the expanded form of (A+B)^n are increasing and then decreasing (and never increase again). Hence there's a peak where the coefficient is maximum...

That looks sOO hard Are you sure we're expected to be able to do this in 3U? Thanks.

hmm I think I support you... probably.. I havent done many exam questions but I think it's true. If you wanna save don't buy extra books at all.. just buy the compulsory one. Is it true? Having Cambridge myself I can tell it has such a long explanation at the start of each section....

oh, thanks McLake ^_^

I am not simba, neither am I a li0n

or you can just use Windows calculator to convert it :P EDIT: oh am I quoting an empty post?

Let I_n = integral of sin^n (x) dx from x=0 to x=pi/2, where n is an integer, n >= 0. Using integration by parts, show that for n >= 2, I_n = (n-1)/n * I_(n-2) Thanks.

so since there are... hmm... 25 letter-parts in this paper, it will take 25 weeks for the complete solutions to become available in this forum and that's the letter-part... the (a), (b) etc there are some parts that contain numerical (or rather semi-letter) parts like (i), (ii), (iii) and...

See the attachment. I did it by inspection :) cos4x + cos2x = sqrt2 . cos^2(x) + 1/sqrt2 . sin2x 2cos^2 2x - 1 + cos 2x = sqrt(2) (1/2 + 1/2 cos 2x) + 1/sqrt(2) sin 2x 2cos^2 2x - 1 = sqrt(2) (1/2 + 1/2 cos 2x) + [1/sqrt(2) sin 2x - cos 2x] transform the [...] into one cos function...

oh, this one is truely invisible so invisible that it is actually empty :p

If you don't mind, yes. Thanks again.

Oh I think it's been a few hours Have you finished?? Hurry...

What do you mean?? I suppose so.. LOL... actually I use that word a couple of times in my answers...