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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread sin( \alpha + \beta) = sin \alpha cos \beta + sin \beta cos \alpha \\ \\ \therefore sin (sin^{-1} \frac{4}{5} + sin^{-1} \frac{12}{13}) = \\ \\ \\ sin (sin^{-1} \frac{4}{5} )cos(sin^{-1} \frac{12}{13}) +...

you gotta show the questions for the guys who don't have the textbooks, potentially allowing more responses.

This proof gets asked a lot in trials, haven't seen it much in HSC though. \frac{d}{dx} (\frac{1}{2} v^2) = a

Re: HSC 2016 3U Marathon using the angle between 2 line formula tan \theta = \frac{m_1- m_2}{1+m_1m_2} since the lines are perpendicular, then the angle between the lines make a right angle or 90 degrees \theta = 90 that makes LHS undefined (tan90), so that would mean RHS is undefined...

Re: HSC 2016 3U Marathon Using the identity \\ \arccos { x } + \arcsin { x } =\frac { \pi }{ 2 } $ for$ -1 \leq x \leq 1 We add those 2 expressions together to get 2\arccos { x } = \frac { 5 \pi }{ 6 } \Rightarrow \\ \\ \\ \ x = cos(\frac { 5 \pi }{ 12 }) Then you can use the...

I would check out HSC math papers (the only subject I liked back then ahha) to see what kind of questions i could still do haha. Yeah that's not weird lol.

well for me, doing 4U math enhanced my problem solving abilities (I reckon the harder 3U topics had the most impact on that), I was able to do most of the last questions of the 3U paper that i couldn't do before

Re: HSC 2016 3U Marathon The theorem doesn't work as they are not subtended from a chord (or arc) in the circle

get a driver's license hahahahaha.

Re: Carrotsticks' Solutions 2015 Extension 2 HSC1 I got 64/70 in 2013 and i got 97

yeah just a rough sketch free hand is good enough, haven't lost a mark just because it's not to scale.

this is the hardest 3U bos trials by far.......

ahhhh damn you're right,I always screw up little things like that, luckily I won't get the wrong answer.

The coefficient of x^2 in \binom{9}{0}(1+x)^9 \ \ is \ \ \binom{9}{0}\binom{9}{2} and the x^2 in \binom{9}{1} (1+x)^8kx^2 is \binom{9}{1}\binom{9}{0}k then you add those 2 terms together.

inb4 Q14 is perms and combs :P but yeah I wonder what binomials they could come up with that's harder than my one in 2013 :P

((1+x)+kx^2)^9 = \binom{9}{0}(1+x)^9 +\binom{9}{1}(1+x)^8kx^2+\binom{9}{3}(1+x)^7(kx^2)^2+... \\ \\ \\ $Now we only focuss on the first two terms of the expansion as the other terms will have powers greater than x^2$ \\ \\ \\ $so the coefficient of$ \ \ x^2 \ \ $is only in$ \\ \\...

For the first question, construct a right angle triangle according sinx = k now tan(\frac{\pi}{2} + x ) = \frac{sin(\frac{\pi}{2} + x )}{cos(\frac{\pi}{2} + x )} = \\ \\ \\ \frac{-cosx}{sinx} = -\frac{-\sqrt{1-k^2}}{k} = \\ \\ \\ \frac{\sqrt{1-k^2}}{k}

yayyy great job buddy.

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Use complex analysis methods to find \int_{0}^{2 \pi} \frac{1+cos(3 \theta)}{5-4cos \theta} \ \ d \theta \int_{0}^{\infty} \frac{x^\frac{1}{4}}{x^2+4} \ \ \ dx \ \ \ \ $by taking the branch cut of$ \ \ \ x^{\frac{1}{4}} \ \ $ along the imaginary axis. Make sure to give details regarding...