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C3 Help

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  1. Offline

    Having trouble with 2. b) in getting the last mark.

  2. Offline

     \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

    Was this the confusion?
  3. Offline

    (Original post by raheem94)
     \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

    Was this the confusion?
    Thanks, obvious!
  4. Offline

    Do you mean from:




    Because if that's the last mark then 6cos3x+6 can be re-written as 6+6k (let's say cos3x is k).

    so 6+6k/(1+k)^2 which is equal to:


    take away a (1+k) from top and bottom and you are left with 6/1+k which is:


    Somehow, if you made it that far, I doubt that is the step you were stuck at so apologies if I just explained something basic.
  5. Offline

    Can anyone help me with 7b

  6. Offline

    Try to rearrange the dy/dx into p(1+(tan4x)(tan4x))
    This can then be used in the product rule (or chain, i can't remember which is which.)

    Once you find that rearrange and remember y=tan4x


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