confused (1 Viewer)

[skool :(]

hey
i hav an exam on tuesday for 2 unit maths.. like about exponential and logarithmatic functions. im also doin 3 unit so like am i allowed to apply any of the 3 unit rules to the 2 unit questions. like for example. when im integrating (cos^2)x (as in cos squared x) can i apply the rule where its cos2x (cos two x) minus a half? cos like i was doin past hsc papers n they were using (cosx)^2 (as in cosx all squared). we dont really get the same answers so im just really confused :S
plzzzzzzz help

 

[ND]

(cosx)^2 = (1 + cos2x)/2, not cos2x - 1/2. That may be why you're getting it wrong.

 

[sukiyaki]

you cant apply 3uint stuff to 2uint =(
thats what i heard

 

[Harimau]

Originally posted by sukiyaki
you cant apply 3uint stuff to 2uint =(
thats what i heard

You can...

 

[sukiyaki]

Originally posted by Harimau


You can...

what you mean you can?
i got mark wrong in my last test 0_O

 

[babydoll_]

i think the rule at my school is you can use whatever methods you know... should be the same at other schools too

 

[professor]

for a 3u maths, u can't use 4u method i think...and for 2u, i think u can use 2u

but majority of time, question will state which method to use to solve the problem

 

[kini mini]

AFAIK in all maths courses, you are free to use whatever method you want to solve a question unless of course a method is specified. People often use 4U methods in 3U, and all kinds of things are done to solve problems in 4U that aren't necessarily in the course .

 

[wogboy]

I've always used 3U stuff in 2U and 4U stuff in 3U, and never been penalised for it when I did the HSC. Hell, I even remember in a 3U test, there was a question asking to find the value of cos5x in terms of cosx and/or sinx, and I used De Moivre's theorem and still got full marks, as opposed to using the method that cos(a +b) = cosacosb - sinasinb.

I think it is plain stupid for teachers to mark people down for using a 3U method to solve a 2U question and so on, unless the question explicitly asks you to solve it in a particular way.

 

[wogboy]

[Edit] Double post

 

[professor]

yea, i was thinking like that even though i dun do 4u...coz my friend did a practice paper for 3u and for an sub.integraton q's...he applied some 4u theory and dun need proving and when the teacher asked him why he didn't show it, he just answered 'i dun need to'...

 

[ND]

Originally posted by wogboy
Hell, I even remember in a 3U test, there was a question asking to find the value of cos5x in terms of cosx and/or sinx, and I used De Moivre's theorem and still got full marks, as opposed to using the method that cos(a +b) = cosacosb - sinasinb.

Sorry, just an unrelated question: How did you go about expanding (cosx + isinx)^5? For me it would be much easier (and quicker) to do it by the 3u method. Mind you i haven't done binomial theorm yet, does it involve that?

 

[wogboy]

I find it easier (and more interesting) to use complex no's to solve sinnx or cosnx when n is a bit large like say greater than 4. Yep, it does involve bionomial expansions:

let z = cos@ + isin@

therefore, z^5 = cos5@ + isin5@ (De Moivre)

For the expansion you must use the formula:

(a + b)^n = nC0*a^n*b^0 + nC1*a^(n-1)*b^1 + nC3*a^(n-3)*b^3 + ... + nCn*a^0*b^n

The C means the function "choose" (there should be a choose function on your calculator, written nCr or something. On the Casio fx-100s it's shift+division button). You'll learn more about the choose function and expansions when you do Bionomials, don't worry about it for now.

The expansion becomes:

z^5 =

cos^5(@) + 5i*cos^4(@)sin@ - 10*cos^3(@)*sin^2(@) - 10i*cos^2(@)*sin^3(@) + 5*cos@*sin^4(@) - i*sin^5(@)

therefore,

Re(z^5)
= Re(expansion)
= cos^5(@) - 10*cos^3(@)*sin^2(@) + 5*cos@*sin^4(@)

But Re(z^5) = cos5@, from De Moivre

so,

cos5@ = cos^5(@) - 10*cos^3(@)*sin^2(@) + 5*cos@*sin^4(@)

(You can also change those sin^2(@) and sin^4(@) to cos^2 if you want to get it completely in terms of cos@)

It might look hard now, but once you learn the bionomial theorem it's quite easy. You're almost guaranteed to get a Q like this in the HSC 4U paper, asking you to do it this way.

 

[ND]

Ah ok, i had a similar question (to find sin3@ in terms of sin@ or something, using de moivre's) in my first 4u assessment, but that was easy because i knew the expansion of (a+b)^3.

 

[Harimau]

Originally posted by kini mini
AFAIK in all maths courses, you are free to use whatever method you want to solve a question unless of course a method is specified. People often use 4U methods in 3U, and all kinds of things are done to solve problems in 4U that aren't necessarily in the course .

I used Matrix in a 4U exam and i got marked wrong...

 

[kini mini]

Originally posted by Harimau


I used Matrix in a 4U exam and i got marked wrong...

I don't think that is allowed under the BOS rules, so if it was an HSC assessment you could have appealed. My teachers were happy to see innovations.

 

[ensignkim113]

I thought that you are welcome to use techniques above that level test (i.e. 3u techniques in a 2u test) but I thought that they design the tests so that you can't use the higher techniques so that the 3u students don't have any advantage........