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Why would you break out laughing at them though? Isn't it their choice if they choose to pursue something based on prestige or materialistic purposes?

Re: Flop math question thread $\noindent leehuan has put up a perfectly valid method above. For my method, we have$ $$\begin{align*}\cos x &=\sin \frac{x}{2} \\ \Rightarrow \cos x &= \sqrt{\frac{1}{2} -\frac{1}{2}\cos x}\\ \Rightarrow \cos^2 x &=\frac{1}{2}-\frac{1}{2}\cos x, \end{align*}$...

Re: Flop math question thread The one for the sine one that Drsoccerball wrote is more intuitive for most people, whereas the one you wrote is more elegant to write down (though some people find it harder to remember).

Re: Flop math question thread Maybe he prefers the other formula because it has more similarities to the ones for cos and tan?

Re: Flop math question thread That general solution formula is in the Year 12 3 Unit Pender textbook.

Re: Flop math question thread $\noindent Yeah, it's from that and replace $x$ with $\frac{x}{2}$. If you want to know how to obtain \textit{that} identity, it's in the HSC 3U Maths course and is a rearrangement of the double angle formula for cosine, which can be derived from a unit circle...

Re: Flop math question thread Maybe he doesn't want people to memorise those.

Re: Flop math question thread $\noindent By the way, we don't need to write the $\pm$, because integers can be negative as well, so saying that $k$ is any integer is already sufficient.$

Re: Flop math question thread $\noindent Note that $\sin \frac{x}{2} \geq 0$ as $x$ is acute. So we use this trig. identity (sub. this for the R.H.S. of the given equation): $\sin \frac{x}{2}= \sqrt{\frac{1}{2} -\frac{1}{2}\cos x}$ (taking non-negative square root as we know the sine is...

Re: Flop math question thread $\noindent Replace $\cos^2 x$ with $1-\sin^2 x$ to obtain a quadratic equation in $\sin x$, which you can solve similarly to the $\tan$ one. Was this one given a range of values that $x$ should be between, or are they asking for general solutions?$

Re: MX2 2016 Integration Marathon $\noindent It is to do with the substitution. The substitution isn't one-to-one in the domain of integration. To right this, do all integrals from 0 to $\infty$ (so that it is one-to-one) and double the answer at the end. As we can see, if we do it from $x=0$...

$\noindent Let $y =\sqrt{x}\log x$, then by the product rule,$ $$\begin{align*} y^\prime &= \frac{1}{2\sqrt{x}}\log x + \sqrt{x}\cdot \frac{1}{x} \\ &= \frac{1}{2\sqrt{x}}\log x + \frac{1}{\sqrt{x}} \\ &= \frac{\frac{1}{2}\log x +1}{\sqrt{x}}.\end{align*}$

They probably didn't read through the whole thread carefully.

They mistook it as a general thread for everyone to post their Q's in I think.

$\noindent We will assume $l\neq 0$ (if $l=0$, the product is clearly $m^2$). We have$ $$\begin{align*}(l\alpha + m)(l\beta + m) &= l^2 \alpha \beta + ml (\alpha + \beta) + m^2 \quad \text{(expanding)} \\ &= l^2 \cdot \frac{n}{l} + ml \cdot \left( -\frac{m}{l}\right) + m^2 \quad (\text{sum...

$\noindent Use primary school long division. Doing this, we see the answer is $2.8333...$.$

From Google, it means: ''noun a number denoting quantity (one, two, three, etc.), as opposed to an ordinal number (first, second, third, etc.).'' (Further reading: http://www.mathsisfun.com/definitions/cardinal-number.html)

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Yes, that's correct.

Re: HSC 2016 2U Marathon So were these terms ("burning lady log rule" and "broken pipette rule") basically made up by you and your friends? (I can't see the connection between that log law and those terms haha.)

Just to clarify, for NSW, "intermediate" maths as mentioned by the article would be HSC 2U Maths, right? And would the "advanced maths" that the article refers to be HSC Maths Extension 1 and 2?