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Why don't you ask Drongoski.

You've done very well to do the integration with the suggested substitution. I tried x^2 = tan@ - fruitless.

I've checked. Seems to be correct. Congratulations. I would not expect you to have Michael Spivak's textbook.

trig

If you are serious contact me: message me 1st as I do not answer calls from unknown numbers. I've been tutoring Maths for 20 years now, all the way to Ext 2. Better if you are closer to Epping. Tel: 0490 - 780 347.

In Ext 1 Maths, you can assume: \lim _{\theta \to 0} \frac {sin \theta}{\theta} = 1

\lim _{\theta \to 0} \frac {cos\theta - 1}{\theta} = \lim_{\theta \to 0} (-\frac {1-cos\theta}{\theta} )= -\lim _{\theta \to 0}\frac {2sin^2(\frac {{\theta}}{2})}{\theta}\\ \\ =-\lim_{\theta \to 0} (sin \frac {\theta}{2}) \times \lim _{\theta \to 0} (\frac {sin \frac {\theta}{2}}{\frac...

Pardon my ignorance: I've always been curious, being from a much older generation, why/how young people develop bulimia? In my youth, we've never heard of this eating disorder.

\sqrt{5} \pm 1 $ and $ \2\sqrt{2} \pm 1 ???

One thing to ask yourself is: can I be salvaged? If so, how? Did you get 49% because that's a fair reflection of your ability? If you think you are capable of doing much better, then there is hope. Are you doing any tuition? If not, some additional assistance may help. If you're already going to...

OK! Thanks for the info. I've only betrayed my ignorance. I didn't know there is a NESA course "Mathematics Life Skills". My apologies to coolpankake for my post.

This reasoning is incorrect. You should point out that the quadrilateral so formed is a rhombus (all 4 sides equal, in this case = 1), whose diagonals bisect the angles of the rhombus. In general, diagonals of a parallelogram do not bisect their incident angles.

This depends on your current knowledge level and foundation. It is just possible that your grasp of the fundamentals of maths is not that solid: so you cannot understand the book's explanation - hence you'd think the book explains it badly. The authors may assume you already have prior knowledge...

No. Your method is the type I was looking for. It's way better than Drongoski's plodding method.

z and w lie on the same line. Let: z = a cis\theta $ and $ w = b cis \theta \\ \\ \therefore acis\theta + bcis\theta = |a|bcis\theta\\ \\ \therefore (a+b)cis\theta = |a|bcis\theta\\ \\ \therefore a+b = |a|b \\ \\ |a+b| =||a|b| = |a|\times |b| Case 1) a>0 & b>0 \therefore a + b = ab \implies...

Not as straightforward as I thought. Maybe there is a simple method.

These are well-established conventions. But I don't know if you'd be penalised for their use in your HSC exams. If I were a marker, I'd accept it. But it is possible some markers are themselves not familiar with this shorthand and be confused; but most "smart enough" markers should be able to...

(1 + 1)^2n has 2n+1 (odd number!) of terms. The expression 2nC0 + 2nC1 + . . . + 2nCn has n+1 terms. By adding an additional 2nCn to the remaining expression, I make it equal to the 1st half. I did not notice this at the beginning and that's why I thought the identity was incorrect at first.