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Trig - Addition Formula Question

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the equation cos(x-60)=sinx

    2. Relevant equations

    cosAcosB+sinAsinB

    3. The attempt at a solution

    cos(x-60)=sinx
    cosxcos60+sinxsin60=sinx
    1/2cosx+(√3)/2sinx=sinx

    How do I then solve to find x for0<x<360
     
  2. jcsd
  3. Dec 4, 2011 #2

    SammyS

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    Subtract sin(x) from both sides.

    Divide by cos(x).
     
  4. Dec 4, 2011 #3

    Curious3141

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    Suggestion: don't even do it this way.

    You know that sin x = cos (90-x)

    So re-express RHS like that:

    cos (x-60) = cos (90-x)

    x-60 = 90-x + 360n (where n is an integer)

    2x = 150 + 360n

    x = 75 + 180n

    So x = 75 or 255 for n = 0 and 1 respectively. Those are the only two solutions in the required range.
     
  5. Dec 4, 2011 #4

    SammyS

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    Did you try what I suggested ?

    What did you get ?

    (1/2)cos(x)+((√3)/2-1)sin(x)=0

    Now, divide by cos(x) .

    [itex]\displaystyle \frac{\sin(x)}{\cos(x)}=\tan(x)[/itex] ---- Right?
     
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