VCE Methods Question Thread!

[sailinginwater]

Yes

So does that mean that sin(2pi), sin(4pi), sin(6pi) etc all equal sin(0)?

[darkz]

So does that mean that sin(2pi), sin(4pi), sin(6pi) etc all equal sin(0)?

Yes, and same for tan

[sailinginwater]

Yes, and same for tan

Would cos(pi), cos(3pi), cos(5pi) etc all be the same?

[darkz]

Would cos(pi), cos(3pi), cos(5pi) etc all be the same?

Yes, this is because, for example, cos(3pi) = cos(pi + 2pi) => cos(pi)

[fun_jirachi]

Ok i screwed that up, because I pressed alt+s and not alt + p oops, time to try again
So I have this neat trick my mate taught me
First it's important to note you can subtract or add which is one revolution around the unit circle to get down to the bounds of . From there you can use the angles of any magnitude and your ASTC [all stations to (whatever station you have in victoria that starts with c)] to find the angle.

Ok here's the actual trick that's saved me so much time
With sine, you have the three main angles right,
and look at this
and that similarly


And you can rearrange the RHS to get the forms your more familiar with, but essentially all of them are identical, the roots of 1, 2, 3 on 2 for sin 30, 45, 60
ITS ACTUALLY MAGIC

The reverse is true for cos ie.
and that similarly



And for tan, it just goes from smallest to largest
and that similarly



Just remember you can use angles of any magnitude + ASTC to put them into forms you recognise and then convert from there, its doable in your head with enough practice

[Lear]

Would cos(pi), cos(3pi), cos(5pi) etc all be the same?

You can just put this on the CAS and check if it is the same, you know.

[sailinginwater]

You can just put this on the CAS and check if it is the same, you know.

Exam 1 is without cas

[Lear]

Exam 1 is without cas

I meant right now not mid exam

[Lear]

Ok i screwed that up, because I pressed alt+s and not alt + p oops, time to try again
So I have this neat trick my mate taught me
First it's important to note you can subtract or add which is one revolution around the unit circle to get down to the bounds of . From there you can use the angles of any magnitude and your ASTC [all stations to (whatever station you have in victoria that starts with c)] to find the angle.

Ok here's the actual trick that's saved me so much time
With sine, you have the three main angles right,
and look at this
and that similarly


And you can rearrange the RHS to get the forms your more familiar with, but essentially all of them are identical, the roots of 1, 2, 3 on 2 for sin 30, 45, 60
ITS ACTUALLY MAGIC

The reverse is true for cos ie.
and that similarly



And for tan, it just goes from smallest to largest
and that similarly



Just remember you can use angles of any magnitude + ASTC to put them into forms you recognise and then convert from there, its doable in your head with enough practice

This is an excellent way to memorise these values. Thanks for your contribution
Another thing that helps me is to know that Tan(x) is simply the numerator of sin(x) divided by the numerator of cos(x)
Example - Tan(pi/6)
Sin(pi/6) = 1/2
Cos(pi/6) = sqrt3/2
Therefore Tan(pi/6) is simply 1 (numerator of sin) divided by sqrt3(numerator of cos)
 = 1/sqrt3

Even for Tan(pi/2) it is simply 1 / 0 which is undefined

[Brittank88]

Working with statistics and having trouble understanding a correct and realistically doable (in tech-free conditions) method of finding a solution.

The question states that p̂ is a random variable representing the sample proportions of customers who bring their own shopping bag to a shopping mall. It also states that from a sample consisting of all customers on a particular day, an approximate 95% confidence interval for the proportion p of customers who bring their own shopping bags was determined to be (4853/50000,5147/50000).

Part a isn't too difficult. It asks for the value of p̂, therefore you find the value in the middle of the confidence interval and add it to the lower bound of the confidence interval.

Part b is where my trouble begins. It asks, considering that 1.96 = 49/25 (I recognise this is the k value for a 95% confidence interval), find the size of the sample from which the approximate 95% confidence interval was obtained.

How does one go about solving this in a manner that is realistically doable under test conditions on a tech active exam? My solution was to apply the formula used to calculate the confidence interval bounds in relation to p̂ (in this case 1.96*sqrt(p̂(1-p̂)/n) ) and rearrange, however with the values presented I just can't believe this is the easiest method.

Thanks in advance!

[minhalgill]

im so confused with part (a), can someone please help,
thanks in advance

[S200]

im so confused with part (a), can someone please help,
thanks in advance

Isn't it just \(0.1+0.45+0.35\)? So \(0.9\)?
Basically, if 40 or less people turn up, they can all have a seat. If >40 turn up, some miss out.

[minhalgill]

ok so, in this question, how would we find the negative x intercept closest to the y axis? i know its pretty simple in this question, and i just found the positive x intercept and put a negative sign infront of it because the graph is symmetrical about the y axis, but say it was a more complicated graph with a translation in the x or y axis. how would i do this? if i were to equate the graph to 0 and  solve for x, would find the values for x in the negative quadrants, or would i find it in the positive quadrant, and then just put a negative sign infront of it?
any help would be appreciated and i hope u get what im trying to say,
thanks in advane

[minhalgill]

Isn't it just \(0.1+0.45+0.35\)? So \(0.9\)?
Basically, if 40 or less people turn up, they can all have a seat. If >40 turn up, some miss out.

isnt that the probability of people who actually turn up?

[S200]

isnt that the probability of people who actually turn up?

Well yeah, but I'm pretty sure it's still correct.

OK so, in this question, how would we find the negative x intercept closest to the y axis? i know its pretty simple in this question, and i just found the positive x intercept and put a negative sign in front of it because the graph is symmetrical about the y axis, but say it was a more complicated graph with a translation in the x or y axis. how would i do this? if i were to equate the graph to 0 and  solve for x, would find the values for x in the negative quadrants, or would i find it in the positive quadrant, and then just put a negative sign in front of it?
any help would be appreciated and i hope u get what im trying to say,
thanks in advance

Solving for zero works. If the graph has an x-translation, just shove the minus sign on and then add/minus whatever the x-translation is, to move it to the correct point compared to the new axis of symmetry.