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Trigonometry Question: Find the max and min.

  1. Dec 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the maximum and minimum values of 4cos[tex]\theta[/tex]-3sin[tex]\theta[/tex].


    2. Relevant equations
    I have no idea.


    3. The attempt at a solution
    I have no idea how to do this question

    Please help me!
     
  2. jcsd
  3. Dec 19, 2009 #2

    Borek

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    Staff: Mentor

    Try to draw a plot - even if it will not give you an exact answer, it may give you some hints.
     
  4. Dec 19, 2009 #3

    Mark44

    Staff: Mentor

    The trick to these kinds of problems is to write Asin(x) - Bcos(x) as Csin(x - [itex]\theta[/itex]). It can then be seen that the maximum value is C and the minimum value is -C (assuming that C > 0).
    4cos[itex]\theta[/itex] - 3sin[itex]\theta[/itex]
    = 5[(4/5)cos[itex]\theta[/itex] - (3/5)sin[itex]\theta[/itex]]

    Now what you need to do is find an angle [itex]\alpha[/itex] such that sin([itex]\theta[/itex]) = 4/5 and cos([itex]\theta[/itex]) = 3/5. Then you can use the identity sinAcosB - cosAsinB = sin(A-B).
     
  5. Dec 19, 2009 #4
    I'm just curious, can you go into more detail how you generated this from,

     
  6. Dec 19, 2009 #5

    ideasrule

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    Homework Helper

    Do you mean why it's true, or why Mark wrote it like that? It's easy to prove why it's true: just expand and you get 4cos x - 3 sin x. As for why it's useful, you want an equation of the form cos(a)cos(b)-sin(a)sin(b), because such an equation is equal to cos(a+b). Since cos(a) can't be 4 and sin(a) can't be 3, Mark decided to factor out a five. You can just as well factor out a ten, or a 100.
     
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