Andiio's SM Questions Thread

[pi]

even if you don't know Euler's Formula you can still justify the key step in that solution.


this can be justified using what you know in specialist. In fact just replace with in every instance in that solution and you still should be able to follow it using just specialist knowledge.

Makes using the CAS easier too

[Andiio]

Hey guys, could anyone help me with this question please?

Find the exact value of sin(x) if tan(2x) = 2√6 and -pi/2 < x < 0. Show all the working details.

Thanks!

[xZero]

Using quadratic formula





Since x is within the 4th quadrant, tan(x) should be negative, hence





Now draw a triangle with the angle x, the opposite should be -3, adjacent should be and let hypotenuse = h







Hence



I hope I'm correct 

[david10d]

Yeah that's the right answer but you won't get full marks for that solution as '-pi/2 < x < 0' (angle can't be negative)

Alternatively, you can use cot^2(x)+1=cosec^2(x) to solve it.

Andiio are you doing Dr He's methods classes? If so which day/time? I'm going to hunt you down.

[Andiio]

Hey guys, for sketching trig reciprocal graphs, what do you guys find is the best method(s) for determining the asymptotes?

Just figuring out/solving for when the orig graph = 0, or 'applying' the "asymptotes occur at: pi/2 + npi, where n E Z" things?

Thanks!

[vea]

The first way is better IMO... it seems more conceptual as well.

[Andiio]

The first way is better IMO... it seems more conceptual as well.

True, quite tedious and annoying though
Do you just do that? i.e. y = 0, solve blablabla

[Liuy]

Ripping through the questions Andrew

[luken93]

The first way is better IMO... it seems more conceptual as well.

True, quite tedious and annoying though
Do you just do that? i.e. y = 0, solve blablabla

Yeah I prefer just graphing it, and you can pretty much see them emerging anyway

[vea]

The first way is better IMO... it seems more conceptual as well.

True, quite tedious and annoying though
Do you just do that? i.e. y = 0, solve blablabla

If you know where y=0 is for the standard graphs then you could just apply your transformations to get them. Otherwise, solving for y=0 is the safe way.

[TrueTears]

Hey guys, for sketching trig reciprocal graphs, what do you guys find is the best method(s) for determining the asymptotes?

Just figuring out/solving for when the orig graph = 0, or 'applying' the "asymptotes occur at: pi/2 + npi, where n E Z" things?

Thanks!

just apply the general formulas, but understand where they come from though, they're really simple and easier to apply, the former is more tedious and i wouldn't recommend using it in an exam or sac

[Andiio]

On an Argand diagram, O is the origin and P represents the point 3 + i. The point Q represents a + bi, where a and b are both positive. If triangle OPQ is equilateral, find a and b. (TECH FREE)

Thanks in advance!

[brightsky]

r = sqrt(9 + 1) = sqrt(10)
so we want sqrt(a^2 + b^2) = sqrt(10)
a^2 + b^2 = 10
a = sqrt(10 - b^2)

We also want sqrt((b-1)^2 + (3-a)^2) = sqrt(10)
(b-1)^2 + (3-a)^2 = 10
b^2 - 2b + 1 + 9 - 6a + a^2 = 10
10 - 2b + 10 - 6(sqrt(10-b^2)) = 10
6(sqrt(10-b^2)) = 10 - 2b
36(10-b^2) = 100 - 40b + 4b^2
360 - 36b^2 = 100 - 40b + 4b^2
40b^2 - 40b - 260 = 0
b = (1+3sqrt(3))/2 (since b >0)
so a = (3 - sqrt(3))/2

[VCE247]

how do you know what to restrict

[Andiio]

When sketching rays/lines/circles/ellipses on the complex region/plane,

for sketching a line, there would never be an open circle/filled in circle, right?

And when sketching a ray, if there are no translations then there'd be an open circle at the origin - conversely, however, if the ray IS translated vertically/horizontally etc etc then is the endpoint of the ray still an open circle?

Thanks!