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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We don't need triple angle formulae for this: 3x = x + 2kπ 2x = 2kπ x = kπ OR 3x = (2π - x) + 2kπ 4x = 2(k+1)π x = (k+1)π/2 which is identical to: x = kπ/2 This answer includes all answers from the first solution. So the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Why don't you combine those last 2 answers into just x = nπ

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread When you multiply through by (1-t²) (1+t²), you can't get a t^6 term.

Look your school up here to see how many E4s they have got in maths in recent years: http://topscores.info/select-nsw.php#pagehere

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread I meant multiply after applying the t-results.

As I said in an earlier post, the answer cannot possibly be as high as 72.

Re: MX2 2015 Integration Marathon When I try it on Wolfram, it just repeats the question as the answer, and draws the region. It is not telling me anything about convergence.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Once you multiply to remove fractions then simplify, it is actually very uncomplicated. The only trick is at the last step.

Re: MX2 2015 Integration Marathon So nobody is concerned that ln[sin(0)] does not exist?

Re: MX2 2015 Integration Marathon In an exam, you can't quote integrals that are not on the standard integrals sheet.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Yes - my mistake.

Re: MX2 2015 Integration Marathon Wolfram Integrator confirms 1/√6

A quadratic is a quadratic. It doesn't matter that is has resulted from the solution of a quartic - it is still solved the same way. You haven't had practice at solving quadratics which give complex solutions?

See my post 5 before this one.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread If sin2x = 1, then 2x = 90 OR 270 (0 < 2x < 360) x = 45, 135 (0 < x < 180)

Perhaps it should be called Astigmatic Maths.

His explanation is not right. There is only one way of arranging the two groups. But doing it the way you are describing, the number of ways of arranging (n-k+1) things in a circle is (n-k)! (You have learned this rule?)

As I said below, if there is a group of k, then the other n-k also form a group. There isn't any freedom to arrange two groups around a circle - there is only one way. The k! and (n-k)! come from arranging within the groups. And it looks like your textbook got all three answers wrong. Which...

I didn't know "denominate" was a word. I would have thought that would have a different meaning: "I nominated him for captain, but he pissed me off so I denominated him".

It is easier than that. If there is a group of k, then the other n-k also form a group.