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$\noindent For 2), recall the trig. identities $1-\cos 2\theta = 2\sin^2 \theta$ and $1+\cos 2\theta = 2 \cos^2 \theta$. So $\frac{1-\cos 2\theta}{1+\cos 2\theta}=\frac{2\sin^2 \theta}{2\cos^2 \theta}=\tan^2 \theta$, provided no denominators are 0.$ $\noindent For 4), it's just the $t$-formulas...

1) Let t = tan(22.5 deg), then 2t/(1 – t^2) = tan(45 deg) = 1, by the double angle formula. So 2t = 1-t^2 ==> t^2 +2t = 1. So (t+1)^2 = 2 i.e. t = 1 + √(2) (taking the positive solution; the negative solution is 1 – √(2), which we know isn't the right one, since t > 0 as 22.5 deg is acute).

1) Use the identity sin(t)cos(t) = (1/2)*sin(2t). 2) Essentially proved here: http://community.boredofstudies.org/14/mathematics-extension-2/350598/do-we-need-memorise-identity.html#post7159341 3) Double angle formula for cos 4) cos(theta)

First one: double-angle formula for tan, noting tan(45 deg) = 1. Second one: Complementary and Pythagorean trig. identities.

Both are clearly equal to 1 (can you see why?).

$\noindent a) Note that $\cos \theta <0$ as $\theta$ is in quadrant 2. So $\cos \theta = - \sqrt{1-\sin^2 \theta}=\sqrt{1-\frac{9}{16}}=-\frac{\sqrt{7}}{4}$. Hence $\sin 2 \theta = 2\sin \theta \cos \theta = 2\times \frac{3}{4}\times \left(-\frac{\sqrt{7}}{4}\right)=-\frac{3\sqrt{7}}{8}$.$ The...

1) Do sin(195°) = sin(135° + 60°), for example, since we know the values of trig. functions at 135° and 60°.

It is the same – rationalise the denominator of my answer to see this (I assume you've learnt how to do this).

I used a tan compound angle formula.

$\noindent Note that $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$. Thus$ $$\begin{align*}\tan 15^\circ &= \tan \left(45^\circ - 30^\circ\right)\\ &= \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \\ &= \frac{\sqrt{3}-1}{\sqrt{3}+1}.\end{align*}$$ (LaTeX may take a while to...

What were your attempts?

Re: HSC 2016 3U Marathon Equating the first method's x to 6 isn't quite right, it should be to -6. This is because by choosing epsilon = pi/2, you made it so the particle starts at the origin but moves towards the negative side first, so you should equate the x to -6. This should then yield...

No not quite, because the kj entry of of BC isn't bc (after all, what even is bc?). Note the kj entry of BC is itself a sum. leehuan essentially proved it on one of his threads I think. It relies on being able to switch summation orders as I said a bit earlier on this thread.

For the first one (it's a line, not a plane), let (x-1)/3 = (z+1)/6 = t, where t will be a real parameter. So x = 3t + 1 and z = 6t - 1. Also, y = 4. So on the curve, x = (x,y,z) = (3t+1,4,6t-1) = (1,4,-1) + t(3,0,6) (parametric equation of the line). For the second one (which is indeed a...

Those are both false as Drongoski has said (counter-examples may be found). A correction would be to use intersections rather than conditioning: Pr(A) = Pr(A∩B) + Pr(A∩Bc). Using the rule Pr(A∩B) = Pr(B)Pr(A|B), we can write this as Pr(A) = Pr(A|B)Pr(B) + Pr(A|Bc)Pr(Bc), which is basically...

I think the reason he asked for cis notation is that he's (at least for now) more used to that than exponential form since he remembers it from 4U (whereas exponential form isn't in 4U).

The notation "cis" is commonly used in HSC 4U complex numbers (short for cos + i.sin. So cis(theta) is used to mean cos(theta) + i.sin(theta).).

$\noindent Here it is with cis notation. Note $-64 = 64 \, \mathrm{cis} \left( \left(2k+1 \right) \pi \right)$, for $k$ integer. So the equation we have is $z^6 = 64\, \mathrm{cis}\left( \left(2k+1\right) \pi\right) \Rightarrow z = 2 \, \mathrm{cis}\left( \frac{\left( 2k+1 \right)...

They are in fact the same thing: cis(t) = e^(it).

$\noindent It is due to expanding $\left(z-2\omega\right)\left(z-2\overline{\omega}\right)$, where for convenience we are writing $\omega = e^{i\frac{\pi}{6}}$, so note that $e^{-i\frac{\pi}{6}}=\overline{\omega}$. So (noting $\omega \overline{\omega}= \left |\omega\right |^2 = 1$), we have$...