VCE Specialist 3/4 Question Thread!

[Kanye East]

Thanks so much!
I did this another way where I get , could you please tell me how I would eliminate

[brightsky]

neat trick:
we have y = cos(2x)-sqrt(3) sin(2x)
take the coefficients in front of the trig functions, i.e. 1 and sqrt(3). factor out the square root of the sum of the squares of the coefficients, i.e. sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2.
y = 2 (1/2*cos(2x) - sqrt(3)/2 sin(2x))
now we know that 1/2 = sin(pi/6) and sqrt(3)/2 = cos(pi/6)
y = 2(sin(pi/6)cos(2x) - cos(pi/6) sin(2x))
= 2sin(pi/6-2x)
yay

[Daenerys Targaryen]

neat trick:
we have y = cos(2x)-sqrt(3) sin(2x)
take the coefficients in front of the trig functions, i.e. 1 and sqrt(3). factor out the square root of the sum of the squares of the coefficients, i.e. sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2.
y = 2 (1/2*cos(2x) - sqrt(3)/2 sin(2x))
now we know that 1/2 = sin(pi/6) and sqrt(3)/2 = cos(pi/6)
y = 2(sin(pi/6)cos(2x) - cos(pi/6) sin(2x))
= 2sin(pi/6-2x)
yay

Does that work for all? Or only some in which can have special angles formed?

[Planck's constant]

neat trick:
we have y = cos(2x)-sqrt(3) sin(2x)
take the coefficients in front of the trig functions, i.e. 1 and sqrt(3). factor out the square root of the sum of the squares of the coefficients, i.e. sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2.
y = 2 (1/2*cos(2x) - sqrt(3)/2 sin(2x))
now we know that 1/2 = sin(pi/6) and sqrt(3)/2 = cos(pi/6)
y = 2(sin(pi/6)cos(2x) - cos(pi/6) sin(2x))
= 2sin(pi/6-2x)
yay

That's neat.
This is also neat.
I want the tan of my angle to be 1/sqrt(3)
What angle is this? pi/6
1 = 2sin(pi/6) and sqrt(3) = 2cos(pi/6)
Therefore, cos(2x) - sqrt(3) sin(2x) = 2sin(pi/6) cos(2x) - 2cos(pi/6) sin(2x) = 2sin(pi/6 - 2x)

[Stick]

neat trick:
we have y = cos(2x)-sqrt(3) sin(2x)
take the coefficients in front of the trig functions, i.e. 1 and sqrt(3). factor out the square root of the sum of the squares of the coefficients, i.e. sqrt(1^2+(sqrt(3))^2) = sqrt(1+3) = 2.
y = 2 (1/2*cos(2x) - sqrt(3)/2 sin(2x))
now we know that 1/2 = sin(pi/6) and sqrt(3)/2 = cos(pi/6)
y = 2(sin(pi/6)cos(2x) - cos(pi/6) sin(2x))
= 2sin(pi/6-2x)
yay

I thought this was the only method to do this problem. :| Look how much I know. XD

Also, sometimes they'll purposely throw a question where you'd have to use addition of ordinates (although this is probably more common in Methods).

[Phy124]

Thanks so much!
I did this another way where I get , could you please tell me how I would eliminate

You disregard the negative result as the amplitude is always a positive value.

So you'd have

And that (add pi due to the phase angle being in Q2)

So the equivalent function is which after apply the identity can be converted to the given answer


Simply, if you have something of the form then the equivalent single sine function is

Where (remember to take the positive value as the amplitude is positive)

and , if lies in Q1 or Q4,

, if lies in Q2 or

, if lies in Q3

We were taught in first year mathematics at uni (for those who hadn't studied anything higher than methods), to always disregard the negative A and have a positive amplitude, but considering you guys might be using other methods it might be allowed. Not to mention if you yield the correct answer I doubt they'd care anyway.

[zvezda]

Let b = i + 2j - k and a can be any vector such that a.b=2. The smallest possible value for mod a is?

needing help with this one, help is much appreciated. I feel like its really simple though

[lzxnl]

Return to the geometric definition of the dot product and see what happens.

[zvezda]

Return to the geometric definition of the dot product and see what happens.

Cheers for that. I think I'm a bit rusty on these

[zvezda]

Points A, B, C and D are defined by position vectors a, b, c and d respectively. If AB+CD=0:
A) express d in terms of a, b and c
B) show that AC  and BC bisect each other

I've done part a, Ive only put it up because it may or may not assist in the working out for part b.
Would I be able to just say that ABCD is a quadrilateral of some sort, or is this incorrect to do?

[Limista]

Points A, B, C and D are defined by position vectors a, b, c and d respectively. If AB+CD=0:
A) express d in terms of a, b and c
B) show that AC  and BC bisect each other

I've done part a, Ive only put it up because it may or may not assist in the working out for part b.
Would I be able to just say that ABCD is a quadrilateral of some sort, or is this incorrect to do?

Not sure if my answer is correct, but this is what I would do:

AB + CD = 0, therefore AB is opposite direction to CD and both are parallel.

You can draw any shape as long as you follow the two rules above. For simplicity, I drew a square (quadrilateral like you said). Then I showed the bisection on my diagram, by drawing a line from A to C, and drawing a line from B to C.

^ I'm not sure whether this is the appropriate method to prove it although. 

[Daenerys Targaryen]

Points A, B, C and D are defined by position vectors a, b, c and d respectively. If AB+CD=0:
A) express d in terms of a, b and c
B) show that AC  and BC bisect each other

I've done part a, Ive only put it up because it may or may not assist in the working out for part b.
Would I be able to just say that ABCD is a quadrilateral of some sort, or is this incorrect to do?

Im thinking:

AB=b-a and CD=d-c
If AB+CD=0 then;
b-a+d-c=0
d=a-b+c

[zvezda]

Yeah I had that for part a as well.

Also, in ex6I q11 part d in the essentials textbook, I'm sort of stuck.
Help is much  appreciated

[b^3]

Yeah I had that for part a as well.

Also, in ex6I q11 part d in the essentials textbook, I'm sort of stuck.
Help is much  appreciated

You have shown that for there to be a line that is parallel to the y axis (this will include ) that . Now we also have a relationship between and , which is . Substituting in will give a second equation with just and . You know have two equations with and only, and those two unknowns, so you should be able to solve for .

Hope that helps.

EDIT:

Spoiler

 

EDIT2: Fixed the wrong power.

[zvezda]

Ah yeah right. Thanks for the help