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Homework Help: Trig cosx question

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    solve 5sinx +12cosx =6.5 between 0 and 180 degrees

    2. Relevant equations

    3. The attempt at a solution

    i tried squaring both sides. (5sinx +12cosx)^2= (6.5)^2

    25sinx +60sinxcosx +60sinxcosx + 144cosx^2 =42.25
  2. jcsd
  3. May 14, 2012 #2
    Squaring isn't really a good idea since it might yield extraneous roots. Try writing the given expression as a single trigonometric ratio.

    Hint : try converting it into a form of sin(x+T)
    Last edited: May 14, 2012
  4. May 14, 2012 #3


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    Sine is always postive between 0 and 180 degrees so you can write that as [itex]5\sqrt{1- cos^2(x)}+ 12cos(x)= 6.5[/itex].

    Rewrite it as [itex]5\sqrt{1- cos^2(x)}= 6.5- 12 cos(x)[/itex] and square both sides.
  5. May 14, 2012 #4
    Or, you can divide equation by [itex]\sqrt{5^2+12^2} = 13[/itex], and you will have
    [itex]\frac{5}{13}\sin x + \frac{12}{13}\cos x = \frac{6.5}{13} = \frac{1}{2}[/itex]

    And you know that [itex]\cos \arccos \frac{5}{13} = \frac{5}{13}[/itex] and [itex]\sin \arccos \frac{5}{13} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \frac{12}{13}[/itex]

    And then, use formula for [itex]\sin(x+y)[/itex]

    sorry for bad English
  6. May 14, 2012 #5


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    Excellent Engliosh, excellent mathematics!
  7. May 15, 2012 #6
    This is exactly what I was suggesting. :smile:
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