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Yeah but there's no expansion involved as you previously said :)

Well, if I sub z=2itanpi/10, then I can just use the result from part (i), which yields zero. Nothing too hard about that lol.

Hey, can someone please do this for me? Thanks

Hmm okay, how do you know that each term except the first and last cancels?

Could someone please help me out? Thanks

Hi again, Just 1 last question. For part (ii) couldn't I just sub z=2itan(pi/10) to show that it satisfies the equation, which proves that it's a root? Thanks

Thanks alot Jetblack. I really understand it now =D

Thanks jetblack, it makes alot of sense to me now. But 1 question though... Where did you get the tan(pi/10) < tan(pi/4) ? Thanks

lol, I'm thinking of setting ppl's calculators to round to the nearest integer, and setting the mode to radians for 2U or Physics candidatures before their trial exam. "Hey mate, can I borrow your calculator just to check something...?" xD wonder what would happen.

Hi, Could someone please help me out here? Thanks

Yeah thanks, I know what I'd done wrong. I've forgotten Year 9/10 factorisation :P

I think the answer solves it over C.

Prelim is too easy :P Just make sure you know how to do everything (ie. you don't need to come 1st in every exam, just make sure you understand everything) and it should be a strong foundation for you in year 12.

Hi, I have a polynomials question. I think the answers are wrong :( If\ P(x)=3x^4-11x^3+14x^2-11x+3,\ Show\ that\\P(x)=x^2[3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8]\ and\ hence\ solve\ P(x)=0\ and\ factorise\ P(x)\ over\ R

Two of the roots of the equation x3-13kx2+13kx-1=0 are k and 1/k where k>0 (i) Find the 3rd root <--- I can do this, it's x=1 (ii) Find both values of k The thing I don't get about (ii) is that I get k=1/3, but k=3 is also in the answers. Why is that?

Sorry man, this wasn't supposed outburst in such an argument. Good stuff, you really do understand Physics =D I'll recommend you to anyone that needs tutoring :)

This stuff is easy, it's all essentially the same. All you need to remember for a limit sum is that |r|<1 and a/(1-r) ... Not really that hard IMO.

Namu, the purpose of the question was to see if he actually did learn physics without rote learning. I really hate it when people fancily advertise just to get students. I could go on, but it's unnecessary, I've already made my conclusions :)

He doesn't know how to answer the very simple question :O

Just use your calculator.