Help with VCAA 2007 Trig Question

[DisaFear]

Hi,
Trig is my weakness. Help me combat it!



Please go through it step by step, I really suck at trig

Thanks! 

[Lukey]

have you tried using itute solutions? there more indepth

http://www.itute.com/wp-content/uploads/2007_vcaa_mm_exam1_solvs2.pdf

[DisaFear]

I need to know how to get there (Yes, I'm that bad at trig)

[b^3]

(Image removed from quote.)
I need to know how to get there (Yes, I'm that bad at trig)

From your exact value table you know that sin^-1(root(3)/2)=pi/3
Now we want this but when sin is negative, i.e. int he 3rd and 4th quadrants (to get this use CAST or think sin values are the y co-ordinate so when which quadrants is y negative?)
So we want the corresponding agle in the 3rd and 4th quads.
So that will be pi+pi/3 and 2pi-pi/3
which is 4pi/3 and 5pi/3
Hope that helps.

[DisaFear]

Why can't the answer be 2pi/33pi/2 - the basic angle, and + the basic angle for the other side?
That's what got me

[b^3]

(Image removed from quote.)

Why can't the answer be 2pi/3 - the basic angle, and + the basic angle for the other side?
That's what got me

Because sin(2pi/3) is in the 2nd quadrant (i.e. the top left). In that quadrant sin is positive since it is above the x-axis but we are looking for the negative values. So thats why it cannot be 2pi/3.

[DisaFear]

Sorry, I meant 3pi/2, 270degrees woops
Why can't it be 270degrees plus/minus the basic angle

[b^3]

Sorry, I meant 3pi/2, 270degrees woops
Why can't it be 270degrees plus/minus the basic angle

Because then we won't end up with a sin agle that will result in plus or minus root(3)/2, it will result in plue or minus 1/2 (we will have a cos angle that will result in root(3)/2 though)
It's just like we don't work from pi/2, we work from 0, pi and 2pi/

[b^3]

EDIT: I don't know why it double posted.

[DisaFear]

So we always work from 0, pi and 2pi?
That'll help a lot, thanks

[b^3]

So we always work from 0, pi and 2pi?
That'll help a lot, thanks

Well yes (but I wouldn't say 100% yes). We work from pi/2 and 3pi/2 when we try to go from sin to cos because we have our rules like sin(x-pi/2)=cos(x)

[DisaFear]

Right.
While we are discussing trig, how do I get the negative pi answer for this question? And why is that the case

[b^3]

so you need to know your exact values for tan





Now tan is +ve in the 1st and 3rd quadrants
so that gives the first solution of
Now for the third quadrant we need to do (or we could have looked at it as )
But this is outside of the domain, so it is equivalent to

[DisaFear]

Makes sense, thanks
Whenever we need to fill a domain, or fit it, just plus/minus 2pi, yes?

[b^3]

Makes sense, thanks
Whenever we need to fill a domain, or fit it, just plus/minus 2pi, yes?

Well you can do it that way (it works) but it's better to come up with the right solutions in the first place (i.e. in this case using -pi instead or pi)