M208 – Introduction finished

Well that took a bit longer than expected. I3 seemed to take forever to work through all the exercises. Think I got most of it, though some of the final section on Equivalence Relationships took a couple of goes to get through. I think I was expecting there to be some nice mathematical way to write some of the answers, but they were looking for a more ‘written’ answer.

I did enjoy the complex number section, and being able to use De Moivre’s Theorem to prove:
 \mbox{RTP: }\sin^5\theta=\dfrac{1}{16}(\sin5\theta-5\sin3\theta+10\sin\theta)\\ \mbox{We know: } \sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}\\ \Rightarrow \sin^5\theta=(\dfrac{e^{i\theta}-e^{-i\theta}}{2i})^5\\ \Rightarrow \sin^5\theta=\dfrac{1}{2^5i}(e^{i\theta}-e^{-i\theta})^5\\ \mbox{Expand $(e^{i\theta}-e^{-i\theta})^5$ using the Binomial theorem}\\ (e^{i\theta}-e^{-i\theta})^5 =\binom{5}{0}e^{{i\theta}^5}-\binom{5}{1}e^{{i\theta}^4}e^{-i\theta}+\binom{5}{2}e^{{i\theta}^3}e^{{-i\theta}^2}-\binom{5}{3}e^{{i\theta}^2}e^{{-i\theta}^3}+\binom{5}{4}e^{i\theta}e^{{-i\theta}^5}-\binom{5}{5}e^{{-i\theta}^5}\\ =(e^{{i\theta}^5}-e^{{-i\theta}^5})-5(e^{{i\theta}^3}-e^{{-i\theta}^3})+10(e^{i\theta}-e^{-i\theta})\\ \mbox{Subsitute back in:}\\ \sin^5\theta=\dfrac{1}{2^5}(e^{{i\theta}^5}-e^{{-i\theta}^5})-5(e^{{i\theta}^3}-e^{{-i\theta}^3})+10(e^{i\theta}-e^{-i\theta})\\ =\dfrac{1}{2^4}\dfrac{(e^{{i\theta}^5}-e^{{-i\theta}^5})-5(e^{{i\theta}^3}-e^{{-i\theta}^3})+10(e^{i\theta}-e^{-i\theta})}{2}\\ =\dfrac{1}{16}(\sin5\theta-5\sin3\theta+10\sin\theta) \mbox{ As required}\\

Was quite fun, as I usually get stumped by trig formulas so it was nice to see one way of working them out that doesn’t involve triangles.

I’ll make a start on the TMA this weekend. Though I probably won’t get too far into it as we’ve got our first tutorial next wednesday, so maybe see if there’s any guidelines handed out there for formatting, etc. The MS221 tutor last year was very good for showing the amount of working we should be putting into the TMAs.

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