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Not exactly, some signs change sometimes. This due to the following facts: $\noindent $\cosh z = \cos (iz)$ for all $z\in \mathbb{C}$ and $\sinh z= -i\sin (iz)$ for all $z\in \mathbb{C}$. Equivalently, $\cos z= \cosh (iz)$ and $\sin z = -i \sinh (iz)$ for all $z\in \mathbb{C}$.$ $\noindent The...

Those results are separate, it's just bad formatting (they should be on different lines and are not intended as steps in working). They are actually three results, namely: $(1)\quad $\log_a x^k = k\log_a x$$ $(2)\quad $\log _a a = 1$$ $(3)\quad $\log_a 1 = 0$.$

Haha it worked this time. And you should probably post that question in the extracurricular integration thread so that we don't derail this thread. :)

Haha I think this was an SOS thread. And in your question, did you mean sinh and cosh? These have LaTeX codes too, just write like \sinh and \cosh.

Your third last line seems to be three separate log laws joined together. Each equation in that line is actually a separate law.

$\noindent If we rationalise the numerator and apply difference of two squares, it becomes $\frac{-253}{\sqrt{n}-\sqrt{n+253}}=\frac{253}{\sqrt{n+253}-\sqrt{n}}$. Note that the denominator is monotone decreasing with limit 0. So for this to be a natural number, we need the denominator here to be...

$\noindent Using the change of base law,$ $$\begin{align*}\log _{a^x}b &=\frac{\log _a b}{\log _a \left(a^x\right)} \\ &= \frac{\log_a b}{x} \quad (\text{as }\log_a \left(a^x\right) = x).\end{align*}$$

Re: HSC 2016 3U Marathon Correct! $\noindent By the way, for any curious HSC students, here's an easier way that also explains why the answer is so simple in the end (and is probably what your intuition told you it should be).$ $\noindent Visualise $n$ blank spaces: $\_$ $\_$ $\_$ $\ldots$...

$\noindent For the circle one, let the radius of the circle be $r$ (in metres), and let the circle's centre be $O$. Then $OD = r-0.1$, $OC=r$, and $CD = 1$ (all units in metres). We can then use Pythagoras in $\triangle OCD$ to obtain a quadratic to solve for $r$.$

Re: HSC 2016 3U Marathon $\noindent The green marbles are not like indistinguishable particles (think of each green marble having a label with a number from 1 to $n-1$ if you want; but all green marbles are the same basically. Like there's $n-1$ of them, and we can't `see' any difference in...

Re: HSC 2016 3U Marathon A proof of Snell's Law is on this page: https://proofwiki.org/wiki/Snell's_Law

Well you don't know it won't have no practical applications – a lot of areas of pure maths from the past were considered to be forever 'useless', but are now used heavily in our lives, e.g. prime numbers for encryption etc. Do you actually want the things you do to have no practical application...

HSC Biology isn't needed or useful much. You should probably only pick it if you reckon you can get a good ATAR due to doing it.

Not true, at least for New South Wales.

Yeah, HSC Business Studies has a lot of rote learning (not sure whether more than HSC Biology, since that also has heaps of rote learning, in fact most of it is rote), and also has essay-writing involved, which you said you wanted to avoid. Do you think you'd be happy to do Business Studies...

Re: HSC 2016 4U Marathon No

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Observe that as $x$ varies in its domain $\Big{(}$which is $-1< x\leq 1$, noting that $x\neq -1$, since if $x=-1$, $\frac{1}{2}\cos ^{-1} x = \frac{\pi}{2}$, so $g(x)$ is undefined here$\Big{)}$, $\frac{1}{2}\cos...

It's easy to mess up if doing it by hand like that, so I'd prefer to do it using the component-by-component method.

$\noindent For adding two complex numbers using vectors, you can use the head to tail method I guess (or the parallelogram rule, whichever you prefer). Alternatively, you can add the vectors component-wise. So to get $z_1+z_2$, start at $z_1$ say, and then move in the $x$-direction...

This part is not the case (I think I know what you mean though). You meant that the complex numbers can be "flipped" from each other. But this doesn't mean the arg's are negatives of each other (this is the case for conjugates), but rather that the arguments differ by π.