Harvey's Question Thread

[paulsterio]

You have to remember that the range of the inverse tan function is from -pi/2 to pi/2 (not inclusive of course).

This is the first and the fourth quadrants, which is represented by B

D - is talking about the 1st and 3rd quadrants, you cannot get a result out of an inverse tan which lies in the 3rd quadrant, eliminating D

[HarveyD]

ah k, thanks!

just another quick question
with this question (attached)
for part c)

When it says "Hence", would it be sufficient to equate sin(pi/4 - pi/6) = sin(pi/12)
or do you have to use cartesian and polar and then equate

[paulsterio]

You would have to use Cartesian/Polar

If it said hence or otherwise, you could use pi/4 - pi/6 or even 1/2 of pi/6

[HarveyD]

how would I do this:

[Mao]

3 rules in maths that always applies (to real numbers):

1. cannot divide by 0
2. cannot square root a negative
3. cannot log 0 or negative

So, work out when your expression might break these rules, and that will give you the maximal domain:

e.g.
1. cannot divide by zero, so , therefore
2. cannot square root a negative, so . If you plot this, you will find that this is true for
3. not applicable

The maximal domain is therefore the intersection of the above, i.e.

[paulsterio]

If you plot this, you will find that this is true for

I think this might be on an Exam 1, so you can't exactly plot it, which means that you'll have to consider x^3 and x^2 - 1 seperately and find where they're both negative or both positive, which will give the same intervals

[HarveyD]

ahh okay, makes sense
thanks for that

one more question
how would I do aii) for this one:

[Lord of Tzeentch]

The equation would be abs(Z-conj(Z))=abs(Z+conj(Z)) (sorry only new on these forums, not sure how to write mathematical expressions nicely)
The definition for it is that the line would be made up of the points which are equidistant from conj(Z) and -conj(Z).