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Express 3cosx+3sinx in the form Rcos(x-a)

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    (i) Express 3cosx+3sinx in the form Rcos(x-a) where R>0 and 0<a<(1/2)∏

    (ii) The expression T(x) is definded by T(x)=8/(3cosx+3sinx)

    (a) Determine a value of x for which T(x) is not defined

    (b) Find the smallest positive value of x satisfying T(3x)=(8/9)√6 giving your answer in exact form.

    3. The attempt at a solution

    for (i) 3√2cos(x-(1/4)∏)

    (a) 8/(3√2cos(x-(1/4)∏)). Could someone explain how to do (a) and (b) please in terms I can understand easy. Thanks.
     
  2. jcsd
  3. Dec 8, 2011 #2

    I like Serena

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    Hi studentxlol! :smile:

    To do (a) you need to consider when a function is not defined.
    In particular, when is that the case when dividing?

    To do (b) you first need to substitute T(3x) in the equation.
    Can you do that?
     
  4. Dec 8, 2011 #3
    (a) You can't divide a function when its denominator is 0 so 8/ 3cosx+3sinx can't equal 0 right?

    (b) Substitute T(3x)=(8/9)√6 into 3(8/(3cosx+3sinx) right?

    ???
     
  5. Dec 8, 2011 #4

    I like Serena

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    Yes, but only the denominator can't equal 0 (which is not the expression you just wrote).


    It's the other way around.
    Start with T(x)=8/(3cosx+3sinx).
    Now replace every occurrence of "x" by "3x".
     
    Last edited: Dec 8, 2011
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