Recent content by TheMathStudent

I was actually referring to the entire chapter actually. But I suppose I could take the 100 integrals as well

Hi all, Is there a 4 unit integration chapter from coroneos floating around somewhere? Cheers

Hi all, I am currently a PhD student at University of Sydney (in Artificial Intelligence) and been tutoring both high school and university for the past 9 years (for Maths, Stats and Programming). I also tutor 1st year Business Stats for Sydney Uni (as in employed by Usyd, not privately)...

That's mostly because you can use tan as follows : \lim_{\theta \to 0} \frac{\tan \theta}{\theta} =\lim_{\theta \to 0} \frac{\sin \theta}{\theta \cos{\theta}}=\lim_{\theta \to 0} \frac{\sin \theta}{\theta}\times \frac{1}{\cos \theta}= 1 \times 1 =1

Hey people, I am running a few holiday courses for the Trials and HSC for 2 Unit, 3 Unit and 4 Unit. Will mostly be going through past papers, and each session is 3 hours. http://www.themathstudent.com.au/holiday-courses.html The free videos...

part b) so the series goes 6+12+18.... use sum formula as above n=k, a=6, d=6 part c) expand the sum formula so that you get k^2 (or n^2 if dealing with just end) and then solve the quadratic. Remember to chuck the negative solution out as n>0

^Think you meant to say x^2-26x+25 not 5x^2

1. \begin{align*}4cos^4(x)&=1+2cos(2x)+4cos^2(2x)\\&=1+2cos(2x)+2(2cos^2(2x))\\&=1+2cos(2x)+2(1+cos(4x)) \end{align*} expand and simplify 2. All you need to note is that cos^2(x)=\frac{1+cos(2x)}{2} now it should be easier to sketch

*bump

If you were after the general solution, the answer would be: \frac{(-1)^k\frac{\pi}{6}+k\pi}{3} where k is an integer http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#General_solutions

first one -\frac{1/6}{6x+1} use u=6x +1 if you cant see this straight away. Second one seems like its too hard for a 2U exam. Are you sure you don't want to differentiate instead? Anyway the only thing I can think of use integration by parts by using dv=1/x^2, but this isn't part of 2U

Second question would be easier if I knew the angle was between 0 and 90. Anyway, will leave the last few steps to you. S is sine and C is cosine

Since I'm lazy I will just do Q3 part 1. The method of doing these is as exactly described by RishBonjour. Remember that the all sides are equal and that the angle is 60 deg. \begin{align*} A & =\frac{1}{2}xh\\ h & =x \sin(60^{\circ})=\frac{\sqrt{3}}{2} x\\ \therefore A & = \frac{\sqrt{3}}{4}...

Keep in mind that the general equation for Sum is: \begin{align*}S_n &=\frac{n}{2}(a+(n-1)d) \\ S_n & =\frac{na+n^2 d -nd}{2} \\ S_n &=n^2 \frac{d}{2}+ n\frac{(a-d)}{2}\equiv n^2 - 3n \\ \therefore & \frac{d}{2}=1, \qquad \frac{(a-d)}{2}=-3 \end{align*} I'll assume you can solve the...

yup, you just have to sub in n=1 regardless of the rearrangement to get the first term a