VCE Specialist 3/4 Question Thread!

[Bad Student]

My school has done this type of analysis task in the past. Luckily, we had a normal SAC this year.

[hongkyho]

but that says one of the roots is pi/8, when in the question its 7pi/8 ?

You're absolutely right, mate.The worked solutions for the Maths Quest Specialist book is wrong as it assumes that the first root is pi/8. If you used B and expanded -cis(7pi/8), then you would get the point exactly pi distance from the inital point, which is correct.

Sorry about irrelephant post

[Lasercookie]

You're absolutely right, mate. Can you please post your 50 in accounting in your signature, please?

Stay on topic please. If you want to have other discussions, do that via PM.

[Planck's constant]

Guys, I have a Spesh SAC tomorrow. It's an Analysis task over 2 days, and its focuse on Partial Fractions, Trig, Co-ord Geom and Rational Functions. The thing is apparently the SAC is like a reasoning task, where you'll be given a series of multiple choice questions and you'll not only have to answer the question, but also explain why you picked that or why the others are wrong  For instance they could give you a rational function and tell you to pick how many asymptotes the graph had and then the following question could be why can't it have "3" asymptotes ..? Has anyone else had this type of SAC? Or have any tips as to how I could prepare for this? Confused!

Sounds like a good quality SAC.

[Holmes]

I'm having trouble understanding this concept.
Is sin (−θ ) = −sin (θ )
and cos (−θ ) = cos (θ )     only if (θ ) is between 0 and pi/2 degrees? It seems like a general property, but I don't really understand why. I get why it applies to angles if θ  is less than 90 degrees, because then it would be in quadrant 4 and so cos is positive and sin is negative, but why is this true in a general sense of the law?

[Ancora_Imparo]

These identities work for all values of θ. To understand why, you can look at all the different possible cases. Ie: Try a value of θ in each quadrant.

If θ is in the 1st quadrant, sin and cos are both +ve.
Thus -θ will be in the 4th quadrant. In the 4th quadrant, sin is -ve but cos is +ve.
Since sin(θ) changes signs, but cos(θ) is still the same sign, the identities hold true.

If θ is in the 2nd quadrant, sin is +ve but cos is -ve.
Thus -θ will be in the 3rd quadrant. In the 3rd quadrant, sin and cos are both -ve.
Since sin(θ) changes signs, but cos(θ) is still the same sign, the identities hold true.

If θ is in the 3rd quadrant, sin and cos are both -ve.
Thus -θ will be in the 2nd quadrant. In the 2nd quadrant, sin is +ve but cos is -ve.
Since sin(θ) changes signs, but cos(θ) is still the same sign, the identities hold true.

If θ is in the 4th quadrant, sin is -ve but cos is +ve.
Thus -θ will be in the 1st quadrant. In the 1st quadrant, sin and cos are both +ve.
Since sin(θ) changes signs, but cos(θ) is still the same sign, the identities hold true.

Therefore, these identities hold true for all values of θ.

[BubbleWrapMan]

It's generally true; sketch the graph of y = sin(x) and apply the transformation to bring it to y = sin(pi/2 - x). Then do the same thing with y = cos(x).

Another way to think about it: if you draw the angles θ and -θ on the unit circle, you'll notice that their y-coordinates are negatives of each other, while the x-coordinates are the same. This is exactly why sin (-θ) = −sin (θ ) and cos (-θ) = cos (θ ), since sine and cosine are the y- and x-coordinates of the unit circle, respectively.

[Henreezy]

Can somebody help me anti-differentiate 1/16+x^2 ??

I'm unsure about how to make 1 equal to a, intuitively I assumed that you would take out 1/4 from the fraction to get 4, but I'm unsure if this is the correct method. Could somebody show me how to do it? Thanks in advanced! :^)

[b^3]

Assuming this is .

We want to try and get out integral into the form , that is we need to make the numerator into . Now if we don't want to change our expression, so to keep everything in check, we need to multiply both the top and bottom by the same, that is we will multiply by , i.e. . Then we can take the factor of outside the integral, since it is a constant.


Hope that helps.

[Henreezy]

Ahh, thanks a bunch b^3!
In retrospect, I would have had 4/4+x^2 which is definitely wrong except in the case of 1

[Homer]

how would i find the range and domain of arccos(2x-1)+(pi/2)?

[Phy124]

how would i find the range and domain of arccos(2x-1)+(pi/2)?

Domain:



Therefore:



Range:

The usual range is:



It's be translated units up meaning the new range is:

[Jaswinder]

could someone help me with b) please the answer is pi 

[sin0001]

Could be a typo, maybe the inner function is supposed to be sin?

[jono88]

Does anyone know what mutually perpendicular vectors are? Say you had 3 vectors: a=4i+3j-k, b=2i-j+5k and c=yi+zj-2k
Find y and z, such that a,b,c are mutually perpendicular.
All i know is that if you scalar product 2 vectors, it should equal 0 for it to be perpendicular.