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|z|=\sqrt{\cos^4\theta+\sin^4\theta}=\sqrt{1-\frac{1}{2}\sin^22\theta} 0\le\sin^22\theta\le1\therefore\frac{1}{2}\le1-\frac{1}{2}\sin^22\theta\le1 \therefore\frac{1}{\sqrt2}\le|z|\le1 B

alternatively, u know \tan\theta-\sec\theta=1-\frac{2}{\tan\frac{\theta}{2}+1} and so \int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}(\sec^2\theta-\sec\theta\tan\theta)\...

yeah u can actually did u know that \tan\theta-\sec\theta=\sin\left(\frac{\theta}{2}-\frac{\pi}{4}\right)\csc\left(\frac{\theta}{2}+\frac{\pi}{4}\right) whereupon we may do this one without the limits \int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\...

now we redo the original question. i don't think at this stage we need to give up on your method. it works! but just other way around. \int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}(\sec^2\theta-\sec\theta\tan\theta)\...

u know why this? i show u now \begin{aligned}\lim_{\theta\rightarrow\frac{\pi}{2}}(\tan\theta-\sec\theta) &=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\sin\theta-1}{\cos\theta}\\ &=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\frac{d}{d\theta}(\sin\theta-1)}{\frac{d}{d\theta}\cos\theta}\\...

technically in the context of this integral one only needs to do the left-sided limit \lim_{\theta\rightarrow\(\frac{\pi}{2})^-}(\tan\theta-\sec\theta)=0 but it works from both sides anyway look at graph of y=tan x-sec x u can see it approaches 0 as x->pi/2

u can do limit \lim_{\theta\rightarrow\frac{\pi}{2}}(\tan\theta-\sec\theta)=0

here another way \int_0^\frac{\pi}{2}\frac{1}{\sin\theta+1}\ \!d\theta=\int_0^\frac{\pi}{2}\frac{1}{2}\csc^2\left(\frac{\theta}{2}+\frac{\pi}{4}\right)\ \!d\theta=1

It's the other way around \frac{1}{\sin\theta+1}=\sec^2\theta-\tan\theta\sec\theta Answer is 1 and \int_0^\frac{\pi}{2}(\sec^2\theta-\tan\theta\sec\theta)\ \!d\theta=1 so method is ok, but just do it the other way around

as i said before it is out-of-print it didn't come with ebook either nevertheless a small part of it is available online - lumi's solutions to the famous "coroneos 100": https://4unitmaths.com/coroneos100-lumi-sol.pdf

That's great! Are you aware that this has been done before? - by Jose Lumi back in 2008 but I think it's out-of-print now i have them though

now i make this one for the 2024 intermediate: https://4unitmaths.com/amc-2024-int-sol.pdf

i make this one for the senior 2024: https://4unitmaths.com/amc-2024-sen-sol.pdf it good? did i make typo? i don't think amt website made one yet

dunno i also dunno what they said about it at the nesa presentation on hsc marking at uts last week hopefully they clarify this in the recordings of it, but that doesn't seem to have come out yet. it is possible a remark may not be necessary i think it strange though they give marks for...

well here is the question to which we find in nesa's marking guidelines the answer [-1,1] which is the correct answer to the wrong question. so if we assume the answer is correct then the question is wrong

yeah. each course will get their own one, which is better i think.

here is a resource for the new syllabus for standard, advanced and extension 1 i was given recently. this is the first resource for new syllabus outside nesa website before new textbooks come out apparently they will make one for the ext. 2 later too

any set of points in x-y plane has an inverse because you can reflect any set of points about the line y=x. the result doesn't have to be a function. so question 14b was wrong

let's put it this way if u aim for 80% u might get 60% if u aim for 100% u may get 95% if u aim for 110% u may get 100% therefore do the enrichment questions