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

    cot2x+cosec2x = cotx
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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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    (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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    (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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    (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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    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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    (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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    (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?

    (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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    (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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    YAY!!! I got 55/128!!!! Can't remember about the -ve sign - couldn't do the trig proof though
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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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    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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    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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    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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    (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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    Can anyone show me what the graph of ln|3x-6| looks like?

    (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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    (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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    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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