Author Topic: Specialist 1/2 Question Thread! (Read 120684 times) Tweet Share

Elsa//768 · « **Reply #180 on:** November 27, 2017, 04:18:06 pm »

Hey I just started Specialist today and i found it quite challenging... But the thing is that I actually got 95% as overall mark for my year 10 maths exams and i was in an accelerated maths class... Since im also doing two unit 3/4 subjects right now, should i give up specialist and switch into legal studies instead, or should i try to stick to specialist and see if things will work out?? THANKS A LOT!!

Sine · « **Reply #181 on:** November 27, 2017, 04:41:38 pm »

Quote from: Elsa//768 on November 27, 2017, 04:18:06 pm

Hey I just started Specialist today and i found it quite challenging... But the thing is that I actually got 95% as overall mark for my year 10 maths exams and i was in an accelerated maths class... Since im also doing two unit 3/4 subjects right now, should i give up specialist and switch into legal studies instead, or should i try to stick to specialist and see if things will work out?? THANKS A LOT!!

Hey Elsa

I don't think anyone can say anything regarding the 95%/accelereated maths since it would be different at every school and generally years 7-10 and units 1/2 can't say anything about how well you will do for 3/4. I would stick with spec at least for 1 year and just see how it goes then if you don't like it / unhappy with it switch into legal. Which subject would you prefer spec or legal?

Elsa//768 · « **Reply #182 on:** November 27, 2017, 06:00:39 pm »

Quote from: Sine on November 27, 2017, 04:41:38 pm

Hey Elsa

I don't think anyone can say anything regarding the 95%/accelereated maths since it would be different at every school and generally years 7-10 and units 1/2 can't say anything about how well you will do for 3/4. I would stick with spec at least for 1 year and just see how it goes then if you don't like it / unhappy with it switch into legal. Which subject would you prefer spec or legal?

I see... I'm not sure yet but I think im going to stick with specialist xD Thank u xD

RuiAce · « **Reply #183 on:** December 04, 2017, 11:22:18 pm »

Quote from: Shadowxo on November 12, 2017, 10:59:53 pm

In this case, you can't take out t as it's inside the function (although they could have been clearer). If you could, however, you should know that cos(2π)=1 and sin(2π)=0
$2\cos (2\pi t)+2\sin (2\pi t)=\sqrt 2\\ \cos (2\pi t)+\sin (2\pi t)=\frac{\sqrt 2}{2}\\ \text{Since they're not squared and using double angle formulae wouldn't help solve it, we have to square both sides}\\ (\cos (2\pi t)+\sin (2\pi t))^2=\frac{2}{4}\\ \cos^2 (2\pi t)+2\cos (2\pi t)\sin (2\pi t)+\sin^2 (2\pi t)=\frac{1}{2}\\ (\cos^2 (2\pi t)+\sin^2 (2\pi t))+2\cos (2\pi t)\sin (2\pi t)=\frac{1}{2}\\ 1+2\cos (2\pi t)\sin (2\pi t)=\frac{1}{2}\\ 2\cos (2\pi t)\sin (2\pi t)=\frac{-1}{2}\\ \sin(4\pi t)=\frac{-1}{2}\\ 4\pi t=\frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}{6}, \frac{23\pi}{6}, \frac{31\pi}{6},\frac{35\pi}{6}, \frac{43\pi}{6}, \frac{47\pi}{6} \\ t=\frac{7}{24}, \frac{11}{24}, \frac{19}{24}, \frac{23}{24}, \frac{31}{24}, \frac{35}{24}, \frac{43}{24}, \frac{47}{24}\\ \text{Our squaring may have introduced false solutions, we only want when }\cos (2\pi t)+\sin (2\pi t)>0\\ t=\frac{7}{24}, \frac{23}{24}, \frac{31}{24}, \frac{47}{24} $
I think this is a harder question than you'd normally expect.

Just saw this... ew why don't they teach you guys the auxiliary angle transform?

Syndicate · « **Reply #184 on:** December 05, 2017, 12:14:29 am »

Quote from: RuiAce on December 04, 2017, 11:22:18 pm

Just saw this... ew why don't they teach you guys the auxiliary angle transform?

It's part of the course but nobody pays that much attention to it

Shadowxo · « **Reply #185 on:** December 05, 2017, 11:40:42 am »

Quote from: Syndicate on December 05, 2017, 12:14:29 am

It's part of the course but nobody pays that much attention to it

I don't recall ever being taught it, nor seeing a question like that $:-\$

Quote from: RuiAce on December 04, 2017, 11:22:18 pm

Just saw this... ew why don't they teach you guys the auxiliary angle transform?

They should

keltingmeith · « **Reply #186 on:** December 05, 2017, 01:17:11 pm »

Quote from: Syndicate on December 05, 2017, 12:14:29 am

It's part of the course but nobody pays that much attention to it

Mirroring Shadowxo, auxiliary angle is not in the course.

But also, fuq dat. We don't do enough trig to warrant learning it.

RuiAce · « **Reply #187 on:** December 05, 2017, 01:27:29 pm »

Quote from: keltingmeith on December 05, 2017, 01:17:11 pm

Mirroring Shadowxo, auxiliary angle is not in the course. But also, fuq dat. We don't do enough trig to warrant learning it.

keltingmeith · « **Reply #188 on:** December 05, 2017, 01:34:32 pm »

Quote from: RuiAce on December 05, 2017, 01:27:29 pm

<Solves asin(x)+bcos(x)=c equations>

In a TEXTBOOK. Where they ask stupidly fucked questions that have never been asked before in exams. This is one example of one of those types of questions.

undefined · « **Reply #189 on:** December 30, 2017, 08:07:50 pm »

hi all how would you go by answering this question?

RuiAce · « **Reply #190 on:** December 30, 2017, 08:30:57 pm »

Quote from: undefined on December 30, 2017, 08:07:50 pm

hi all how would you go by answering this question?

$\text{The claim is that the terms}\\ a+3d, a+6d\text{ and }a+15d\\ \text{form a geometric sequence.}$
where $a$ is the first term and $d$ is the common difference.
$\text{In that case, a common ratio must exist between them.}\\ \text{Call it }r.\text{ Then,}\\ r = \frac{a+6d}{a+3d} = \frac{a+15d}{a+6d}$
$\text{Considering the second equality, we can solve for }d: \\\begin{align*}\frac{a+6d}{a+3d}&=\frac{a+15d}{a+6d}\\ (a+6d)^2 &= (a+15d)(a+3d)\\ a^2+12ad+36d^2&=a^2+18ad+45d^2\\ 6ad+9d^2&=0\\ 3d(2a+3d)&=0\end{align*}$
$\text{Hence }\boxed{d=0}\text{ or }\boxed{d=-\frac{2a}{3}}$
Remark: Technically, the common difference is allowed to be 0; this is just the constant sequence; every number in the sequence is all the same. But I will assume that this can't happen purely because I don't know if this is of interest in the VCE course. I will only consider the other case to be correct.
$\text{Since }d=-\frac{2a}{3},\text{ we can now sub into }r=\frac{a+6d}{a+3d}\\ \text{to obtain }r=\frac{a-4a}{a-2a} = 3$

pepper77 · « **Reply #191 on:** February 04, 2018, 11:58:57 pm »

Hello everyone.

I just started Specialist Maths last week (we're doing complex numbers) and it's... really easy. Lots of fun, but easier than Methods easy. (off topic but even though I got an A+ on the final maths test last year I forgot just about everything, RIP.) Does Specialist pick up later?

Also, the teacher ran through partial fractions for like 40 minutes then moved on and I don't think I 100% understand them. There's nothing in my textbook about them, so if anyone knows where to find practice questions I would really appreciate it.

Sine · « **Reply #192 on:** February 05, 2018, 12:09:58 am »

Quote from: pepper77 on February 04, 2018, 11:58:57 pm

Hello everyone.

I just started Specialist Maths last week (we're doing complex numbers) and it's... really easy. Lots of fun, but easier than Methods easy. (off topic but even though I got an A+ on the final maths test last year I forgot just about everything, RIP.) Does Specialist pick up later?

Also, the teacher ran through partial fractions for like 40 minutes then moved on and I don't think I 100% understand them. There's nothing in my textbook about them, so if anyone knows where to find practice questions I would really appreciate it.

none of the maths is meant to be hard it's all about where you fall amongst everyone in the state which will depend on how many careless errors you make.

You should be able to make your own questions for partial fractions if not they are in the 1/2 spec book i think. Also you could also just google "Partial fractions questions"

PolySquared · « **Reply #193 on:** February 24, 2018, 10:19:14 am »

Hi, can I get some help on this question? I have no idea on how to start this question.

Find the value of sec θ if tanθ = 0.4 and θ is not in the 1st quadrant.

RuiAce · « **Reply #194 on:** February 24, 2018, 10:24:28 am »

Quote from: PolySquared on February 24, 2018, 10:19:14 am

Hi, can I get some help on this question? I have no idea on how to start this question.

Find the value of sec θ if tanθ = 0.4 and θ is not in the 1st quadrant.

$\tan\theta\text{ is positive, so }\theta\text{ must be}\\ \text{in the 1st or 3rd quadrant.}$
$\text{But they say that }\theta\text{ is not in the 1st quadrant,}\\ \text{so therefore it must be in the third.}$
$\text{Since }\cos\theta\text{ is negative when }\theta\text{ is in the 3rd quadrant}\\ \text{so must }\sec\theta\text{. So we now know }\sec\theta\text{ is negative.}\\ \text{We then attempt to find the magnitude.}$
______________________________________________________________________________
$\text{We have }\tan\theta = \frac{2}{5}.\\ \text{As }\tan \theta = \frac{opp.}{adj.}\text{, we can sketch a right-angled triangle, with}\\ \begin{align*}opposite &= 2\\ adjacent &= 5\end{align*}$
$\text{Using Pythagoras' theorem, }hypotenuse = \sqrt{29}.\\ \text{So since }\sec\theta = \frac{hyp.}{adj.}\text{, we deduce}\\ \boxed{\sec\theta = -\frac{\sqrt{29}}5}$

Alternate method using trigonometric identities

\begin{align*}\sec^2\theta &= 1+\tan^2\theta\\ &= 1 + \left(\frac25\right)^2\\ &= \frac{29}{25}\end{align*}
$\text{So since we deduced }\sec\theta \text{ is negative from earlier}\\ \text{taking negative square roots gives }\boxed{\sec\theta = -\frac{\sqrt{29}}5}$

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Author Topic: Specialist 1/2 Question Thread! (Read 120684 times) Tweet Share

Elsa//768

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Sine

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Elsa//768

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Syndicate

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