VCE Specialist 3/4 Question Thread!

[Mr. Study]

Hey,

Find the cube roots of 1-.

My teacher hasn't covered this question before. >.>.

Thanks.

NVM. I had to use another text book to find out how. :O

[#1procrastinator]

1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)

and

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)

[b^3]

1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)

Remember that and
So we have

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)

Let


So we know that
Now we know that the angles are restricted to and
So


Now you should be able to do the other half of the question, give it a go remembering that

Hope that helps

[Bhootnike]

solve for

I can't get past a certain step, but i probs screwed it up before that:p















-

yeah. wtf?

[pi]

I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore), but you definitely can't divide both sides by cos(x) as you are ridding the possible solutions of cos(x)=0


(I think)

[#1procrastinator]

1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)

Remember that and
So we have

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)

Let


So we know that
Now we know that the angles are restricted to and
So


Now you should be able to do the other half of the question, give it a go remembering that

Hope that helps

Excellent, thanks! I have a little idea of what's going (like for the first one, think I tried to think of it as measuring the angle from the y-axis or something) but I have no idea how to show anything, guess part of the reason is I have a crappy foundation haha

[rife168]

solve for

has no real solutions



edit: oops, forgot about the restriction:

[Bhootnike]

solve for

has no real solutions



edit: oops, forgot about the restriction:

goddamit.
How do people manage to pick such things up!

p.s.

I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore), but you definitely can't divide both sides by cos(x) as you are ridding the possible solutions of cos(x)=0


(I think)

how so ?  like, i dont get how thats algeabriclly incorrect

[pi]

Solve and you'll see

Spoiler

If you divide both sides by , which is incorrect, you are assuming that as you can't divide by 0 Similarly, in your question, if you divide by cos(x), then you are assuming that cos(x) can't equal 0 either, which is mathematically incorrect

[Bhootnike]

Solve and you'll see

x^2 -x =0
x(x-1) = 0

so x=0 and x=1

wait.

[pi]

Yeah, nice Why didn't you divide both sides by x then at the start?

[Bhootnike]

Yeah, nice Why didn't you divide both sides by x then at the start?

Aw sheeeeet.

rookie mistake.

i get it ahaha.. :p
thanks vege! . 

[Hutchoo]

Herrow guaiz.

See attached!

[TrueTears]

Which question do you need answered?

I'll answer both - 'whydoihavenofriends', pretty obvious, you're too hot, the hotness that radiates from your dp scares away all the nubs.

As for the other question - tan(70*180/pi) is the gradient of the tangent.

now diff f(x), then find the x values for when the derivative of f(x) is equal to tan(70*180/pi) subject to the restriction of the domain. In other words, solve f'(x) = tan(70*180/pi)

[Fishyiscool]

i + 2j - 3k and -9i +4j -k

^ find the Angle between the pair of vectors
(already found that the dot product is 2 from other part of the question)

And another question:

if (u-v) dot (u+v) = |v|^2, then:
A) u=v
B) u must be equal to the zero vector, O.
C) u perpendicular to v
D) u is multiple of v
E) None are true.

(nb: all us and vs etc have little squiggly things under them hehehe.
'dot' is that dot that just stands in space. If you get what i mean. The dot in dot product, the one that is off the floor, unlike this  . Why am i explaining all this, you prob figured anyway)
Thanks