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lexazver203
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#1
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#1
How did you do it - for those who want to try it is

cot2x+cosec2x = cotx
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Hoofbeat
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(Original post by lexazver203)
How did you do it - for those who want to try it is

cot2x+cosec2x = cotx
convert everything into sin's and cos's then use double angle formula, everything cancels and voila, you prove that LHS=RHS!
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lgs98jonee
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(Original post by lexazver203)
How did you do it - for those who want to try it is

cot2x+cosec2x = cotx
i rearranged cot2x to cos2x/sin2x

so i had (cos2x+1)/sin2x=2cos^2x/2sinxcosx
=cotx
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musicbloke
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(Original post by lexazver203)
How did you do it - for those who want to try it is

cot2x+cosec2x = cotx
1/tan2x + 1/sin2x

=cos2x/sin2x + 1/sin2x
=(cos2x + 1)/sin2x
=(cos^2x - sin^2x + 1)/sin2x
=(cos^2x +(1 - sin^2x))/sin2x
=(2cos^2x)/2cosxssinx
=cosx/sinx
=cotx

MB
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Lithium
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(Original post by lexazver203)
How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

(Cos2x/sin2x) + 1/(sin2x)=CotX

[(Cos^2X-Sin^2x)+(1/Sin2x)] = Cotx

cos^2X-Sin^2x + ( Sin^2x + Cos^2x)=cot X

(2cos^2X/Sin2X)=cot X

(2cox^2 X/ 2sinXcoxX)=cotx

The 2's are eliminated, Cos^2X and CosX to leave (
(CosX/SinX) which equals CotX


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lexazver203
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ok then how about the answer for the bionomial expansion? what was the term with no x in i got it to be 495/512 and the question was

(x^3-1/2x)^12
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lgs98jonee
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(Original post by lexazver203)
ok then how about the answer for the bionomial expansion? what was the term with no x in i got it to be 495/512 and the question was

(x^3-1/2x)^12
no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)
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lexazver203
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(Original post by lgs98jonee)
no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)
wasnt it 9th term of the expansion?
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crana
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(Original post by musicboy)
1/tan2x + 1/sin2x

=cos2x/sin2x + 1/sin2x
=(cos2x + 1)/sin2x
=(cos^2x - sin^2x + 1)/sin2x
=(cos^2x +(1 - sin^2x))/sin2x
=(2cos^2x)/2cosxssinx
=cosx/sinx
=cotx

MB
I did mine a really different way, I started with inverting the identities for tan2x and sin2x and at the end I had something like

1/2 cot x + 1/2 sin x + 1/2 cot x - 1/2 sin x

that cancelled out

anyone else do it that way? (your way is nicer MB!)
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Hoofbeat
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(Original post by lgs98jonee)
no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)
OMG! I think I got something like -55/128! And I thought I'd done that whole question wrong! You have just made me sooo happy!!!!!
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Silly Sally
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YAY!!! I got 55/128!!!! Can't remember about the -ve sign - couldn't do the trig proof though
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lexazver203
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(Original post by Hoofbeat)
OMG! I think I got something like -55/128! And I thought I'd done that whole question wrong! You have just made me sooo happy!!!!!
how??
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BossLady
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hey guys, was ques 5 the binomials right with [x^3-1/2x]^12? and then 6 was that iteration stuff (zzz) and 7 was that trap rule stuff?
you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?
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toxi
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I just kept on expanding until i reached a part where the x's would be gone....
I prob had to use sum other method tho, dont really know
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JaF
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My method is a lot more fun. Just to awkward I used the RHS - makes it more interesting for the examiner anyway.

cotx = cosx/sinx

= 2(cosx)^2 / 2sinxcosx

From the double angle formula:

cos2x = 1 - 2(cosx)^2
2(cosx)^2 = 1 - cos2x

cotx = 1- cos2x / 2sinxcosx
= 1/2sinxcosx -cos2x/2sinxcosx
= 1/sin2x - cos2x/sin2x
= cosec2x - cot2x
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lexazver203
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(Original post by toxi)
I just kept on expanding until i reached a part where the x's would be gone....
I prob had to use sum other method tho, dont really know
well what i did is took out x^3 so i got left with x^36(1-1/2x^4)^12
so then x will cancel at

x^36(12*11*10*9*8*7*6*5/8! * 1/((2^9)(x^36)) and it cancelled down to 495/512 and it is -ve
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Barny
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Can anyone show me what the graph of ln|3x-6| looks like?
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crana
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(Original post by imasillynarb)
Can anyone show me what the graph of ln|3x-6| looks like?
If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie
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lexazver203
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(Original post by crana)
If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie
beast i got 2 out of 3 points but the grapf is completely messed up
i thought it would look like that
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infekt
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#20
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the easisest way to do the question was tht x was decreasing by

X^(36-4x)
when 36-4x = 0
x=9
hence it was the 10th term

since its the tenth term use 12C9 * (-1/512)
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