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How to simplify this?

It's [8/(1-x) - 4]^2 - 16 (I forgot the -16)
(edited 7 years ago)
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Original post by NeverLucky
You learning FP2 right now? It is a bit of work to get your head round everything but it'll be easier once you do. A fair bit of FP2 is just memorising things.


I'm doing AQA's FP3 as I've covered all the other FP's from that board. I'm not sure how much memorising there is at it's all just limits, polar coordinates and first-second order differentials. I like it and it's not as difficult as I first made it out to be; but that's to be decided once I get to second order differentials aha.

Original post by kiiten
Anyone know how to answer this using a CAST circle (i dont understand the graph method so i stick to CAST circles :smile:)


I would if I knew what those were. Honestly, put some time into getting to know the general solutions to sine, cosine and tan otherwise these questions won't always make sense and they're a bit awkward to explain from my experience.
Original post by kiiten
Am i doing this wrong?

Solve: 5+2sin(2x+1)=6 for 0<=x<=720

rearranged to sin(2x+1)= 1/2

let u = 2x+1
sinu=1/2
u=1/6 pi

change the bounds to 1 <=u<=4pi +1
using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

*sorry if this doesnt make sense - let me know and ill post the full working


Sorry its a bit messy 😬

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Original post by ManLike007
How does one simplify this? I've spent quite some time on it and have made little progress.


It doesn't go to that...

For simplicity; 4≡4(1−x)1−x4\equiv \frac{4(1-x)}{1-x} and now you can subtract the two fractions inside the bracket as they have the same denominator. Tidy it up and square both halves of the fraction. Take out any common terms. You won't get that answer.
(edited 7 years ago)
Original post by kiiten
Sorry its a bit messy 😬

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Okay your diagram is correct. You should get that angle to be π6\frac{\pi}{6} and you've deduced your range to be 1≤u≤4π+11\leq u \leq 4\pi +1 where u=2x+1u=2x+1 which makes sense.

Now follow your circle and the first angle should be u=0+π6u=0+\frac{\pi}{6} therefore u=0.524↦x=−0.238u=0.524 \mapsto x=-0.238 which is outside the range. Disregard this.

Your second should be u=0+π−π6u=0+\pi - \frac{\pi}{6} when you go half a circle and then back by pi/6 radians. This would give you 2x+1=2.62→x=0.809...=0.812x+1=2.62 \rightarrow x=0.809...=0.81 and this would be your very first answer within the range.

Your third one is after the first period of sine; therefore at u=2π+π6↦x=2.9u=2\pi + \frac{\pi}{6} \mapsto x=2.9 and so on.

General solutions is this circle simplified so I'd still recommend you to learn it rather than needlessly draw circles and alter the range via substitution as this can lead to increased likelihood for errors.
(edited 7 years ago)
Original post by RDKGames
Okay your diagram is correct. You should get that angle to be π6\frac{\pi}{6} and you've deduced your range to be 1≤u≤4π+11\leq u \leq 4\pi +1 where u=2x+1u=2x+1 which makes sense.

Now follow your circle and the first angle should be u=0+π6u=0+\frac{\pi}{6} therefore u=0.524↦x=−0.238u=0.524 \mapsto x=-0.238 which is outside the range. Disregard this.

Your second should be u=0+π−π6u=0+\pi - \frac{\pi}{6} when you go half a circle and then back by pi/6 radians. This would give you 2x+1=2.62→x=0.809...=0.812x+1=2.62 \rightarrow x=0.809...=0.81 and this would be your very first answer within the range.

Your third one is after the first period of sine; therefore at u=2π+π6↦x=2.9u=2\pi + \frac{\pi}{6} \mapsto x=2.9 and so on.

General solutions is this circle simplified so I'd still recommend you to learn it rather than needlessly draw circles and alter the range via substitution as this can lead to increased likelihood for errors.


Ahh thanks :smile: so is it because pi/6 is less than 1 so it isnt counted (bound starts from 1).

Is general solutions a topic that you learn in C3? - where can i learn about it?
Original post by kiiten
Ahh thanks :smile: so is it because pi/6 is less than 1 so it isnt counted (bound starts from 1).

Is general solutions a topic that you learn in C3? - where can i learn about it?


General solutions to trig equations are covered in FP1 if you want to take a look.
Original post by B_9710
General solutions to trig equations are covered in FP1 if you want to take a look.


Ohh im only doing pure maths. Ill take a look though :smile:
Original post by kiiten
Ohh im only doing pure maths. Ill take a look though :smile:


What do you mean?
Original post by kiiten
Ahh thanks :smile: so is it because pi/6 is less than 1 so it isnt counted (bound starts from 1).

Is general solutions a topic that you learn in C3? - where can i learn about it?


Exactly that. And the rest you would have to convert into x from u whichever ones are within the range.

General solutions are, as B_9710 said, in FP1. They are extremely simple to learn because they link to the graphs of sine, cosine and tan in order to explain where the formulae come from for the solutions (I am genuinely surprised they are not taught at C2). General solutions are much easier to work with than drawing circles and can give you every solution you would ever need for these types of questions by simply plugging in some integers. You can take a look inside FP1 for its chapter to learn about them; otherwise you can use this and I can walk you through it in PM's if you wish.
Original post by RDKGames
Exactly that. And the rest you would have to convert into x from u whichever ones are within the range.

General solutions are, as B_9710 said, in FP1. They are extremely simple to learn because they link to the graphs of sine, cosine and tan in order to explain where the formulae come from for the solutions (I am genuinely surprised they are not taught at C2). General solutions are much easier to work with than drawing circles and can give you every solution you would ever need for these types of questions by simply plugging in some integers. You can take a look inside FP1 for its chapter to learn about them; otherwise you can use this and I can walk you through it in PM's if you wish.


Thanks :smile: - with C2 the only thing i can think of is using a graph for these kind of questions. I preferred the CAST method so I just stick to that. But, now you mention it ill look at the general solutions too.

Original post by B_9710
What do you mean?


Pure maths as in C1-4 and 2 applied units e.g. M1, D2 etc.
Original post by kiiten
Pure maths as in C1-4 and 2 applied units e.g. M1, D2 etc.


FP is still pure maths - Further Pure. So... :tongue:
Original post by RDKGames
FP is still pure maths - Further Pure. So... :tongue:


I know that xD but in further pure you do more modules right? Like dont you do FP1 and C1?
Original post by kiiten
I know that xD but in further pure you do more modules right? Like dont you do FP1 and C1?


"in further pure"

No. In Further Maths we do more modules. Further Pure Core modules (FP1-4) AND Pure Core modules (C1-4). I think that's what you mean.
Original post by RDKGames
FP is still pure maths - Further Pure. So... :tongue:


Well there's no real pure maths at A level.
Original post by B_9710
Well there's no real pure maths at A level.


Can't tell if a joke about complex numbers or an insult on difficulty... :holmes:
Original post by RDKGames
Can't tell if a joke about complex numbers or an insult on difficulty... :holmes:


Neither. I mean that what gets labelled as pure maths isn't really pure maths at all as such.
y=4+3sin(2x-1)

is the amplitude of this sine wave 3 because the sin has a 3 in front of it?
Original post by kiiten
y=4+3sin(2x-1)

is the amplitude of this sine wave 3 because the sin has a 3 in front of it?


"because it has a 3 in front of it" is not quite a good explanation. You're right, but your explanation is bad. :smile:
Original post by RDKGames
"because it has a 3 in front of it" is not quite a good explanation. You're right, but your explanation is bad. :smile:


:tongue: is it because its a stretch in the y-direction, scale factor 3?

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