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moodydog
02-02-12, 12:11
integrate ((4x-1) / (2x))

Hmm... the boundaries are between 8 and 2

If we split this up into (2x - 1/2x) and differentiate I get 2x - 1/2 ln(2x).

8
[2x -1/2 ln(2x)]
2

[16 - 1/2 ln(16)] - [4 - 1/2 ln(4)]

16 - 1/2 ln(16) - 4 + 1/2 ln(4)

12 + 1/2 ln(1/4)

The answer book gives 12 ln2. My answer is the negative of this. Also wolfram gives:

http://i40.************/2cervk8.gif

What has it done?


_____________________________________

sorry new question, the book I am using is actually swimming with errors int he answers.

If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

Thinking about it

Sin(x+x) = sinxcosx + sinxcosx hence 2sinxcosx
but then

Sin(x/2+x/2) should I imagine = 2sin(x/2)cos(x/2) and not 2sin(x/4)cos(x/4). Why is the two doubled? Same with anying else, sinx/6 = 2sin(x/12)cos(x/12)
Why is it doubled?


Secondly

Solving tan2x + tanx =0

I get X as 71.56, 251.56, 0 and 180

the way I did that:

2tanx/(1-tan^x) + tanx = 0

2tanx + tanx(1-tan^2x) = 0

2tanx + tanx - tan^3x = 0

tan^3x - 3tanx = 0

tan^2x - 3 = 0

tanx(tanx-3) = 0

tanx = 0 or tanx = 3

x = 0 or x= 71.56

Between 0 and 360

I get X as 71.56, 251.56, 0 and 180

I got two of those values right, but apparently tanx does not equal three? What is wrong with my method? Or like I said, the book is filled with errors.


Hey again, further down the page, there was this. :o

It uses the same principles as before, I have tried changing the cosec into (1 + cot) with ^2 theta, and tan into sec -1 and sin into 1 - cos, and then expanded everything out, but just came into a big mess.
and also tried the right hand side, and couldn't get further then sec -1...

Sorry this is a newer question I am stuck on;


The question wants me to rearrange the left hand side so it looks identical to the right, using trig identities.

hey, I am stuck on this question;

Solve to find x

(x-2)^2 = (root x) +2

I know one of the answers is 4, but I don't know how to get it. :(

Mystery-King
02-02-12, 12:18
So you know that the answer is 4, but you don't know why? Is that what you're saying? :)

moodydog
02-02-12, 12:51
So you know that the answer is 4, but you don't know why? Is that what you're saying? :)

Exactly. :)

I can expand the (x-2)^2 to get:

x^2 - 4x + 4 = (root x) + 2

But to solve that to find x on its on is really hard. I have no idea what I am supposed to do with the (root x)?

I looked at the answers, and it gives it as 4. However, there is no worked solutions in the answer.

I also know there must be two solutions, as it is x^2, however, the question states that x>2. So the other answer must be <2.

Can anyone please help? :)

jonathanrij
02-02-12, 15:57
I think this is it. I did it in as many steps as possible:

(x-2)^2 = (root x) + 2
(x-2)(x-2) = (root x) + 2
x^2 - 2x - 2x + 4 = (root x) + 2
x^2 - 4x + 4 = (root x) + 2
x^2 - 4x + 4 - 2 = (root x)
x^2 - 4x + 2 = (root x)
(x^2 - 4x + 2)^2 = x
x^4 - 8x^3 - 20x^2 - 16x + 4 = x

And follow Cochrane's steps below...

Cochrane
02-02-12, 16:09
Are you supposed to know that one of the answers is four? If yes, then take jonathanrij (which looks correct, but better check first), bring it in … = 0 form, and divide the polynomial by (x-4). This gives you a standard quadratic formula of the x^2 + px + q = 0 form, and solving that is trivial.

jonathanrij
02-02-12, 17:08
I'm so sorry, my answer was insufficient :o. For some reason I stopped and didn't think it through. So it's only part of the whole answer. I wanted to edit it after I noticed it, but I had no computer near me and my cell phone wouldn't let me log in.

So go ahead with what Cochrane said :p.

tomekkobialka
02-02-12, 17:09
I think this is it. I did it in as many steps as possible:

(x-2)^2 = (root x) + 2
(x-2)(x-2) = (root x) + 2
x^2 - 2x - 2x + 4 = (root x) + 2
x^2 - 4x + 4 = (root x) + 2
x^2 - 4x + 4 - 2 = (root x)
x^2 - 4x + 2 = (root x)
(x^2)^2 - (4x)^2 + 2^2 = x
x^4 - 16x^2 + 4 = x

Shouldn't this be (x^2 - 4x + 2)^2 = x ? Which would be x^4 - 8x^3 - 20x^2 - 16x + 4 = x ? (I think :pi: )

jonathanrij
02-02-12, 17:25
^ Damn, you're right. I guess I just wasn't thinking clearly. In my defence, I was in a hurry... xD

tomekkobialka
02-02-12, 18:30
OK, this solution sets out to find all solutions of x using the rational roots theorem...hopefully you should be able to work out how to work out the specific answer to your question. :)

http://i.imgur.com/I2062.jpg


http://i.imgur.com/SPkaR.jpg


Sorry about the messy handwriting, it's really hard to write equations on plain paper. :p

moodydog
02-02-12, 22:26
Thank you all for your help. :)

@ tomekkobialka - Thank you for taking time to solve it, :hug: however, being only an A-level question, Rational Roots Theorem is not learnt, or expected to be used. And also, I don't think the question is looking for that much working out from the nature of the question.

@Cochrane - No the question does not give the answer, I just looked at the mark scheme. But thanks anyway.

I should explain the question a bit better.

f(x) = (X-2)^2

1) Find f-1(x):

f-1(x) = (root) x + 2

2) Solve f(x) = f-1(x)

:)

tomekkobialka
02-02-12, 22:33
@ tomekkobialka - Thank you for taking time to solve it, :hug: however, being only an A-level question, Rational Roots Theorem is not learnt, or expected to be used. And also, I don't think the question is looking for that much working out from the nature of the question.


I only wanted to provide a slight hint at what the final working out would be...I had a feeling from the question that RRT wasn't going to be needed. I don't want to give away the actual solution! :D



f(x) = (X-2)^2

1) Find f-1(x):

f-1(x) = (root) x + 2

2) Solve f(x) = f-1(x)

:)

So you have to find the solution for when x > 2?

I think that at A-Level standard, they wouldn't expect you to work out a value of x that isn't an integer. Therefore the answer should simply be worked out by re-arranging the formula to get the polynomial, then substituting the x value with different integers close to 0 until you get the correct answer(s). That's how I used to do it, anyway. :)

moodydog
02-02-12, 22:38
I only wanted to provide a slight hint at what the final working out would be...I had a feeling from the question that RRT wasn't going to be needed. I don't want to give away the actual solution! :D

ooh ok. Am I anywhere close?

Do I need to expand the brackets, move the two over to the other side to isolate the root X, and then square everything to get X4, X3, X2, X and C?

I completely forgot how to solve with power^4 :o

tomekkobialka
02-02-12, 22:45
Just had a look at the A-Level specification, it seems like you should be able to use inspection, which is the method I described in my previous post (trial and error). :)

moodydog
02-02-12, 23:56
Just had a look at the A-Level specification, it seems like you should be able to use inspection, which is the method I described in my previous post (trial and error). :)

integration by inspection is a topic in A-level, but isn't really related. TBH, I think some inspection work was in C2, which I haven't done in over two years haha. This question is a C3 topic, and i just got it from a book of progressive questions, so everything asked is relative to anything previous in the book (if that makes sense).

So its just a case of expanding the brackets, and substiting x=1, 2, 3 and 4... :) ok thanks ;)

moodydog
09-02-12, 20:38
Ok I am stuck on a later question

http://www4b.wolframalpha.com/Calculate/MSP/MSP12121a066996gbhg715d000010g9cd89af944e97?MSPSto reType=image/gif&s=63&w=149&h=42

where the equal sign is actually equivalent, therefore nothing can be crossed over, just re-arranged within its side of the = sign.

Cochrane
09-02-12, 20:47
That statement is true for all theta. So what is the question?

peeves
09-02-12, 20:50
Ok I am stuck on a later question

http://www4b.wolframalpha.com/Calculate/MSP/MSP12121a066996gbhg715d000010g9cd89af944e97?MSPSto reType=image/gif&s=63&w=149&h=42

where the equal sign is actually equivalent, therefore nothing can be crossed over, just re-arranged within its side of the = sign.

My guess would be sin squared (0/)

Soul
09-02-12, 20:51
I also don't get what you want to know but maybe this helps:

sinē(x)+cosē(x) = 1

and

(1+cos(x)) * (1- cos(x)) = (1-cosē(x)) = sinē(x)

Cochrane
09-02-12, 20:54
I also don't get what you want to know but maybe this helps:

sinē(x)+cosē(x) = 1

and

(1+cos(x)) * (1- cos(x)) = (1-cosē(x)) = sinē(x)

Ah, yes, of course. Use the second on the left side and both sides will be equal. Still not sure what the question is, though.

moodydog
09-02-12, 20:58
That statement is true for all theta. So what is the question?

The question wants me to show steps on how to transform the left hand side into the right hand side.

For example, the first move I made was to make the sin^2 theta into 1 - cos^2 theta.

But not sure what to do from there. The question wants me to rearrange the left hand side so it looks identical to the right. :)

EDIT I can't move anything across the equal sign.

Cochrane
09-02-12, 21:00
The question wants me to show steps on how to transform the left hand side into the right hand side.

For example, the first move I made was to make the sin^2 theta into 1 - cos^2 theta.

But not sure what to do from there. The question wants me to rearrange the left hand side so it looks identical to the right. :)

Ah, OK. Soul has the solution. Use the second formula, and you should have the result almost immediately. Or if you want to figure it out for yourself: Observe that 1 = 1^2 and see what you can do with that.

Soul
09-02-12, 21:00
That's a good start. Now you only have to get the (1-cos(theta)) to the other side and multiply it.

-> 1-cosē(theta) = (1-cos(theta))*(1+cos(theta))

moodydog
09-02-12, 21:05
That's a good start. Now you only have to get the (1-cos(theta)) to the other side and multiply it.

-> 1-cosē(theta) = (1-cos(theta))*(1+cos(theta))

cannot do that unfortunately. You cannot cross over the = sign. Its not actually an = sign, its an equivalent sign. So both halves need to be worked on independently.

OOh I see is

http://www2.wolframalpha.com/Calculate/MSP/MSP32501a066ace58gg016b000018762a73b3g3434c?MSPSto reType=image/gif&s=26&w=147&h=18

an identity I can sub in for

http://www2.wolframalpha.com/Calculate/MSP/MSP18711a066bhca032edb2000065hc657i18ab745h?MSPSto reType=image/gif&s=4&w=43&h=18 ?

Cochrane
09-02-12, 21:11
OOh I see is

http://www2.wolframalpha.com/Calculate/MSP/MSP32501a066ace58gg016b000018762a73b3g3434c?MSPSto reType=image/gif&s=26&w=147&h=18

an identity I can sub in for

http://www2.wolframalpha.com/Calculate/MSP/MSP18711a066bhca032edb2000065hc657i18ab745h?MSPSto reType=image/gif&s=4&w=43&h=18 ?

Of course.

moodydog
09-02-12, 21:17
Of course.

I see. :jmp: Thanks both of you. I don't know how I missed it. :o

moodydog
10-02-12, 13:03
Hey again, further down the page, there was this. :o

http://i43.************/2gwb0ut.gif

It uses the same principles as before, I have tried changing the cosec into (1 + cot) with ^2 theta, and tan into sec -1 and sin into 1 - cos, and then expanded everything out, but just came into a big mess.
and also tried the right hand side, and couldn't get further then sec -1...

EDIt

Oooh I think I have just worked it out: Multiply everything out in the brackets first, cosec tan - cosec sin

tan/sin - sin/sin = tan/sin -1

tan/sin = sec -1

sec = (1 + tan) = 1 + tan - 1 = tan

every tan sin and sec and cosec has ^2 theta, but I am not putting it in on here so its easier to read.

moodydog
24-02-12, 14:18
sorry new question, the book I am using is actually swimming with errors int he answers.

If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

Thinking about it

Sin(x+x) = sinxcosx + sinxcosx hence 2sinxcosx
but then

Sin(x/2+x/2) should I imagine = 2sin(x/2)cos(x/2) and not 2sin(x/4)cos(x/4). Why is the two doubled? Same with anying else, sinx/6 = 2sin(x/12)cos(x/12)
Why is it doubled?


Secondly

Solving tan2x + tanx =0

I get X as 71.56, 251.56, 0 and 180

the way I did that:

2tanx/(1-tan^x) + tanx = 0

2tanx + tanx(1-tan^2x) = 0

2tanx + tanx - tan^3x = 0

tan^3x - 3tanx = 0

tan^2x - 3 = 0

tanx(tanx-3) = 0

tanx = 0 or tanx = 3

x = 0 or x= 71.56

Between 0 and 360

I get X as 71.56, 251.56, 0 and 180

I got two of those values right, but apparently tanx does not equal three? What is wrong with my method? Or like I said, the book is filled with errors.

mind
24-02-12, 14:24
http://memeorama.com/wp-content/uploads/2012/01/nothing-to-do-here.gif

moodydog
24-02-12, 14:51
http://static.fjcdn.com/gifs/Nothing_2479d8_2822723.gif

what?

Spong
24-02-12, 14:55
GIF link fail :vlol:
Unless the GIF is meant to be nothing (judging by its name) :pi::p

peeves
24-02-12, 15:07
I didn't think that gif had no point whatsoever.

Legend of Lara
24-02-12, 15:11
I didn't think that gif had no point whatsoever.

That means you thought the gif had a point.

Ward Dragon
24-02-12, 15:17
If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

x/2=x/4+x/4=2(x/4)

So sin(x/2)=sin(2*x/4)=2sin(x/4)cos(x/4) :)


tan^3x - 3tanx = 0


After that line:

tan(x)*(tan^2(x)-3)=0

Then either tan(x)=0 or tan^2(x)=3 (which means tan(x) is positive or negative square root of 3).

For the tan(x)=0 case you get that x=0 or x=180 degrees which you already got yourself.

For the other case:
arctan(sqrt(3))=60 degrees
arctan(-sqrt(3))=-60=300 degrees

moodydog
24-02-12, 15:50
x/2=x/4+x/4=2(x/4)

So sin(x/2)=sin(2*x/4)=2sin(x/4)cos(x/4) :)



After that line:

tan(x)*(tan^2(x)-3)=0

Then either tan(x)=0 or tan^2(x)=3 (which means tan(x) is positive or negative square root of 3).

For the tan(x)=0 case you get that x=0 or x=180 degrees which you already got yourself.

For the other case:
arctan(sqrt(3))=60 degrees
arctan(-sqrt(3))=-60=300 degrees

in other words, you cant devide through by tanx to simplify the powers. :o

cheers that really explained a lot :')

Ward Dragon
24-02-12, 17:12
in other words, you cant devide through by tanx to simplify the powers. :o

cheers that really explained a lot :')

You're welcome :)

peeves
02-03-12, 13:28
That means you thought the gif had a point.

No that means the gif didn't have a point.

tomekkobialka
02-03-12, 17:03
I did not think that gif had no point whatsoever.

Double negative. :( A bit like how -1 x -1 = 1

Doesn't matter anyway. :p

Soul
08-03-12, 19:09
I have to find out the real part and the imaginery part of "cos(2i*pi)" and I can't find out how to do it.

Any help would be appreciated, as wolframalpha only tells me that iot's the same as cosh(2pi) but no idea how to get this result.

EDIT: would this be correct?:
cos(2*i*pi) = cos(i*pi + i*pi) = cosē(i*pi) + sinē(i*pi) = 1

EDIT2: Nope...the formula is cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

Cochrane
08-03-12, 19:25
I have to find out the real part and the imaginery part of "cos(2i*pi)" and I can't find out how to do it.

Any help would be appreciated, as wolframalpha only tells me that iot's the same as cosh(2pi) but no idea how to get this result.

EDIT: would this be correct?:
cos(2*i*pi) = cos(i*pi + i*pi) = cosē(i*pi) + sinē(i*pi) = 1

Both my calculator and Wolfram Alpha give 267.746… as a decimal approximation of the number. Specifically, cos(x + iy) = cos(x) cosh(y) - i sin(x) sinh(y) with x = 0 and y = 2π gives 1*cosh(2π) - i*0 = cosh(2π) = (e^(2*2π) + 1)/(2e^(2π))

Soul
09-03-12, 14:12
Specifically, cos(x + iy) = cos(x) cosh(y) - i sin(x) sinh(y) That's exactly what I needed. Thanks!

moodydog
09-03-12, 14:38
integrate ((4x-1) / (2x))

Hmm... the boundaries are between 8 and 2

If we split this up into (2x - 1/2x) and differentiate I get 2x - 1/2 ln(2x).

8
[2x -1/2 ln(2x)]
2

[16 - 1/2 ln(16)] - [4 - 1/2 ln(4)]

16 - 1/2 ln(16) - 4 + 1/2 ln(4)

12 + 1/2 ln(1/4)

The answer book gives 12 ln2. My answer is the negative of this. Also wolfram gives:

http://i40.************/2cervk8.gif

What has it done?

Ward Dragon
09-03-12, 18:30
integrate ((4x-1) / (2x))

Hmm... the boundaries are between 8 and 2

If we split this up into (2 - 1/2x) and differentiate I get 2x - 1/2 ln(2x).



The bold part has your mistake in it. 2-1/(2x)=2-(1/2)(1/x) so then when you integrate it you should get
2x-(1/2)(ln(x))

Also, the Wolfram site uses "log" to mean "ln" in case that's what was confusing.

moodydog
09-03-12, 19:26
The bold part has your mistake in it. 2-1/(2x)=2-(1/2)(1/x) so then when you integrate it you should get
2x-(1/2)(ln(x))

Also, the Wolfram site uses "log" to mean "ln" in case that's what was confusing.

ooh makes sense. Cheers. :)

Ward Dragon
09-03-12, 19:41
ooh makes sense. Cheers. :)

You're welcome :)

moodydog
15-03-12, 11:05
:jmp: I am actually at the point now where I am getting A's on my C3 Maths paper. :D
Now time to start C4 as well.