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Trigonometry - half angles problems

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Previous part of the question I have solved:
    Express cos(2x) in terms of sin(x):

    I got this answer:
    cos(2x)=1-2sin2(x)

    Hence or otherwise solve the equation
    cos(x) + 3sin(x/2) = 2


    2. Relevant equations
    Double angle formulae:
    cos(2x)=cos(2x) - sin(2x)
    sin(2x)=2sin(x)cos(x)


    3. The attempt at a solution
    So basically I'm doing revision on trig, and I know I've come across this problem before, where you are presented with a full angle and a half angle, but I have failed to find an example. It's something to do with halving a double angle formula I think but I can't even start it. How do I get rid of the half angle? I know I should show an attempt, but I can't even start it off. Just a refresher on how to remove half angles would be great thanks :)
    (Yes i have read the rules but this problem is difficult to put into that format properly)
     
  2. jcsd
  3. Oct 14, 2011 #2
    If the first part read cos(2z) in terms of sin(z), what would your answer be? Then substitute z for x/2 in the second question
     
  4. Oct 14, 2011 #3

    eumyang

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    Homework Helper

    Note that x = 2 times (x/2). It may be helpful if you substitute another variable, say y, in for x/2. What would x equal? What does this equation:
    cos(x) + 3sin(x/2) = 2
    look like with the substitutions?
     
  5. Oct 14, 2011 #4
    cos(2y) + 3sin(y) = 2.
    Thanks very much that's what I needed a refresher on :)
     
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