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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}\cos{x}&=\cos{3x} \\\cos{x}&=4\cos^3{x}-3\cos{x} \, (*)\\4\cos^3{x}-4\cos{x}&=0 \\\cos{x} \left (\cos^2{x}-1 \right )&=0 \\\cos{x}&= 0 \to x=2n\pi \pm \cos^{-1}{0}=2n\pi \pm \frac{\pi}{2} \\\cos^2{x}&=1...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}t^2-2t-1&=0 \\\textup{Completing the square:}& \\t^2-2t+1&=2 \\(t-1)^2&=2 \\t-1&=\pm \sqrt{2} \\t&=1 \pm \sqrt{2} \\\end{align*} If you solve this using your calculator, you get...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}\sec{x}+\sin{x}&=\cos{x} \\\frac{1+t^2}{1-t^2}+\frac{2t}{1+t^2}&=\frac{1-t^2}{1+t^2} \\\textup{Multiply through by }(1-t^2)&(1+t^2) \\(1+t^2)^2+2t(1-t^2)&=(1-t^2)^2 \\(1+t^2)^2-(1-t^2)^2+2t(1-t^2)&=0 \\\textup{Since...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread cos(3x) = cos(2x+x) = cos(2x)cos(x) - sin(2x)sin(x) = [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x) = 2cos³(x) - cos(x) - 2sin²(x)cos(x) = 2cos³(x) - cos(x) - 2cos(x)[1 - cos²(x)] = 2cos³(x) - cos(x) - 2cos(x) + 2cos³(x) =...

The (n-k)! is for arranging the first group, and the k! arranges the second group.

You have to consider different cases. So one case could be 3 unique letters (e.g. CHA), and another a pair of identical letters and one unique letter (e.g. CLL).

I think it is this: http://math.mit.edu/~jorloff/suppnotes/suppnotes03/h.pdf

The bottom of the motion is 15cm from the ground, and the top is 55cm from the ground. The weight is travelling 40cm from top to bottom. At halfway, it will have travelled 20cm. 15+20=35cm from the ground.

Any more estimates?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread This would mean there are no solutions. Here is another way of reaching the same answer: \begin{align*}2\cos^2{x} - 3\sin{x}\cos{x} + 2\sin^2{x}&=0 \\2(\sin^2{x}+\cos^2{x})-3\sin{x}\cos{x}&=0 \\2-3\sin{x}\cos{x}&=0...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread \begin{align*}\int \frac{4x^5 + 32x^3 + 66x}{x^4 + 8x^2 + 16} \, \mathrm{d}x &= \int 4x+\frac{2x}{(x^2 + 4)^2} \, \mathrm{d}x \\&= \int \left (4x+\frac{2x}{\underset{u}{(\underbrace{x^2 + 4}})^2} \right ) \, \frac{1}{2x}\times...

Why has it dropped since last time, when my ranks were a bit lower? (here)

School rank: 24 English Advanced: 92/131 Biology: 27/75 Chemistry: 36/106 Physics: 28/74 2U Maths: 3/104 3U Maths: 46/120

Re: HSC 2015 3U Marathon Didn't really consider it as I copied the question. I changed the question so that the triangle does intersect at a maximum.

Re: HSC 2015 3U Marathon Here is a projectile motion question I saw in another forum: $A particle is thrown over a triangle from one end of the horizontal base to the other end. At the top of it's motion it touches the top vertex of the triangle. If $\alpha$ and $\beta$ are base angles that...

Yea something wrong with the codecogs website. When I go to view that image I get 500 - Internal Server Error.

For both of these, you can combine the terms in displacement and velocity involving acos(nt) and bsin(nt) using the auxiliary angle method, to get the functions in a form such as Rsin(nt+c), where R=sqrt(a^2+b^2). The R term is the only one influencing the amplitude in this case. So the...

I think he means a scenario like this, where Result 5 of Trial 2 is an outlier which results in Trial 2 having a different average to the other trials:

Make sure the equal angles are in the same place. It might help if you draw them separately like so:

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Construct a line through Q, perpendicular to QP. Note that this line also passes through S. Let angle QPR = 2alpha. Since PS bisects angle QPR, angle QPS=angle SPR=alpha. Now, find angle RQP and angle PRQ, using the fact that...