# Edexcel C3 (20/06) Revision ThreadWatch

This discussion is closed.
13 years ago
#101
(Original post by ntrik)
tan2x = -1
2x = -45
x = -90

if x = -90

then cos(-180) + sin(-180) = 0
-1 + 0 = 0

The answers are 67.5 and 157.5 apparently
Care with the simple steps (it is always where I mess up )

cos2x + sin2x = 0

=> cos2x = -sin2x

=> -tan2x = 1

=> tan2x = -1

=> 2x = -45, 135, 315

=> x = -22.5, 67.5, 157.5

I don't know what the interval is so probably reject the negative.
0
13 years ago
#102
Argh! I can't believe it

I think I might take a small break and then get back to the maths later.
0
#103
How to do 6b?
0
13 years ago
#104
(Original post by Seth)
Ahhh noo u shouldn't have given up...I think I have the answer...

sin3x=cosx
sin(2x+x)=cosx
sin2xcosx+cos2xsinx=cosx
(2sinxcosx)cosx+(1-2sin²x)sinx=cosx
2sinxcos2x+sinx-2sin³x=cosx

2sinx(1-2sin²x)+sinx-2sin³x=cosx
2sinx-4sin³x+sinx-2sin³x=cosx
(3sinx-6sin³x)²=(cosx)² <----(square both sides)
9sin²x-36sin^4 x+36sin^6 x=1-sin²x
36sin^6 x-36sin^4 x+10sin²x-1=0

let y=sin²x:
36y³-36y²+10y-1=0
gives 3 solutions...2 of which are complex, 1 which is real.

The real one is y=0.6282

so...

sin²x=0.6282
sinx= √0.6282
x = 52.4°

You'd be extremely unlucky to get a question like this,especially because complex numbers aren't in C3(at least I don't think so), took me a long time to solve and got lots of help from a further mathmo..
Shouldn't that be cos²x not cos2x lol
0
13 years ago
#105
Where did you get these papers from? =S
0
13 years ago
#106
(Original post by Featherflare)
Shouldn't that be cos²x not cos2x lol

Yeh small typo there.. sorry!.. I've considered it as cos²x later on so is not too much of a problem
0
13 years ago
#107
(Original post by TheWolf)
How to do 6b?
you know from your proof in 6a) that:

1-cos 2θ/ sin 2θ =tan θ

and 6b) is basically 6a) 'fiddled' around a bit until you get:

1 = 2sin 2θ
sin 2θ = 0.5
2θ = π/6, 5π/6,
θ = π/12, 5π/12,
0
13 years ago
#108
(Original post by TheWolf)
How to do 6b?
2(1 - cos2θ ) = tanθ
2(1 - (1 - 2sin²θ )) = tanθ
2(1 - 1 + 2sin²θ ) = tanθ
4sin²θ = tanθ
4sin²θ - tanθ = 0
4sin²θcosθ - sinθ = 0 (multiply by cosθ )
sinθ(4sinθcosθ -1) = 0
sinθ(2sin2θ - 1) = 0
sinθ = 0 [no solution for this one] or
sin2θ = 1/2

then do the rest
0
13 years ago
#109
(Original post by Featherflare)
sinθ = 0 [no solution for this one]
Why no solutions..

surely solutions for sin θ=0:

θ= 0, π
0
13 years ago
#110
(Original post by Seth)
Why no solutions..

surely solutions for sin θ=0:

θ= 0, π
0 < θ < π

so does not include 0 or π
0
13 years ago
#111
Meh, can someone give me the link to where i can get the specimen papers for C3? Perleeeeez. And also.. how many are there?
0
13 years ago
#112
(Original post by violet)
Meh, can someone give me the link to where i can get the specimen papers for C3? Perleeeeez. And also.. how many are there?
Try here.
0
13 years ago
#113
I am a bit confused about the quotient rule, I can use it but on a primative form ( the book is rather **** in comparison to the earliar chapters). Take the following example:

6x/(5x+3)^1/2

vdu/dx-udv/dx/v^2 - quotent rule

originally when answering this question I ignored using the chain rule on the denominator, but after solving it that way I got the wrong answer. Can someone solve this step by step so I can see how the quotient rule is correctly applied. Thanks
0
13 years ago
#114
(Original post by PoseidonX)
Try here.
Ahh thanks... but the link for paper A4 doesn't work..
Sorry to be so annoying.. I just need papers to do for.. tomorrow
0
13 years ago
#115
(Original post by DOJO)
I am a bit confused about the quotient rule, I can use it but on a primative form ( the book is rather **** in comparison to the earliar chapters). Take the following example:

6x/(5x+3)^1/2

vdu/dx-udv/dx/v^2 - quotent rule

originally when answering this question I ignored using the chain rule on the denominator, but after solving it that way I got the wrong answer. Can someone solve this step by step so I can see how the quotient rule is correctly applied. Thanks
u = 6x
v = (5x+3)1/2

du/dx = 6
dv/dx = 1/2(5x+3)-1/2 *5 = 5/2(5x+3)-1/2

so using rule,
6(5x+3)1/2 - 6x(5/2(5x+3)-1/2)/(5x+3)
=[6(5x+3)1/2 - 15x(5x+3)-1/2]/(5x+3)
0
13 years ago
#116
(Original post by Featherflare)
u = 6x
v = (5x+3)1/2

du/dx = 6
dv/dx = 1/2(5x+3)-1/2 *5 = 5/2(5x+3)-1/2

so using rule,
6(5x+3)1/2 - 6x(5/2(5x+3)-1/2)/(5x+3)
=[6(5x+3)1/2 - 15x(5x+3)-1/2]/(5x+3)
Thanks for your reply, I managed to get it into that form also after using the chain rule.

The answer in the back of the book is :

3(5x+6)/(5x+3)^3/2

I am aware that they got the numerator into that form through simple factorisation, however how did they get the denominator into the form (5x+3)^3/2?

Thanks
0
13 years ago
#117
[6(5x+3)1/2 - 15x(5x+3)-1/2]/(5x+3)

Multiply top and bottom by (5x+3)1/2:

= [6(5x+3) - 15x]/(5x+3)3/2

= (30x + 18 - 15x)/(5x+3)3/2

= (15x+18)/(5x+3)3/2

= 3(5x+6)/(5x+3)3/2
0
13 years ago
#118
(Original post by mockel)
[6(5x+3)1/2 - 15x(5x+3)-1/2]/(5x+3)

Multiply top and bottom by (5x+3)1/2:

= [6(5x+3) - 15x]/(5x+3)3/2

= (30x + 18 - 15x)/(5x+3)3/2

= (15x+18)/(5x+3)3/2

= 3(5x+6)/(5x+3)3/2
thanks for your reply, I can understand fully how you managed to get the numerator to be 3(5x+6), however I am slightly confused with the ^3/2 in the denominator. Thanks.
0
13 years ago
#119
(5x+3)(5x+3)1/2 = (5x+3)3/2 , going back to your work on indices.

In general:
(ab)(ac) = ab+c

So similarly, the power is 3/2 here, since (1+½) = 3/2
0
13 years ago
#120
(Original post by mockel)
(5x+3)(5x+3)1/2 = (5x+3)3/2 , going back to your work on indices.

In general:
(ab)(ac) = ab+c

So similarly, the power is 3/2 here, since (1+½) = 3/2
Indeed. What originally confused me was according to the quotient rule, the denominator must be udv/dx-vdu/dx/v^2, so I was unsure whether for the denominator should have a v^2 or hence /(5x+3)^2 , as I have noticed that the ^2 was not there, hence the only way to get ^3/2 is through multiplying the bracket with ^1/2.

Also, when we cross multiplied to make the denominators the same, normally I would do the following:

take our example here:

[6(5x+3)1/2 - 15x(5x+3)-1/2]/(5x+3)

we multiplied the top and bottom (right hand side) by (5x+3)^1/2, however I noticed that the right hand side was not multiplied by the bracket (5x+3)? why is this the case. Thanks!
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cardiff Metropolitan University
Sat, 27 Apr '19
• University of East Anglia
Could you inspire the next generation? Find out more about becoming a Primary teacher with UEA… Postgraduate
Sat, 27 Apr '19
• Anglia Ruskin University
Health, Education, Medicine and Social Care; Arts, Humanities and Social Sciences; Business and Law; Science and Engineering Undergraduate
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (551)
37.82%
No - but I will (114)
7.82%
No - I don't want to (102)
7%
No - I can't vote (<18, not in UK, etc) (690)
47.36%