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How do I get theta out of this equation?

  1. Sep 4, 2006 #1
    I've been working on a statics problem and I need to solve for theta, the equation I have right now is (4/5) = sin(theta) + cos(theta).

    I can't remember how to get theta out of this, my brain isn't functioning very well tonight... Been up too long :( :yuck:
     
  2. jcsd
  3. Sep 4, 2006 #2
    You can square both sides and after simplification you should get

    [tex]9/25 = 1 + 2\sin(\theta)\cos(\theta) = 1 + \sin(2\theta)[/tex]

    [tex] \Rightarrow -16/25 = \sin(2\theta) \Rightarrow \theta = 1/2\cdot\arcsin(-16/25) [/tex]
     
    Last edited: Sep 4, 2006
  4. Sep 4, 2006 #3
    Thanks Swapnil.
     
  5. Sep 4, 2006 #4

    VietDao29

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    Homework Helper

    One more way is to use the identity:
    [tex]\sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)[/tex]
    So:
    [tex]\frac{4}{5} = \sin x + \cos x = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)[/tex]
    [tex]\Leftrigharrow \frac{\sqrt{2 ^ 3}}{5} = \cos \left( x - \frac{\pi}{4} \right)[/tex]
    Can you go from here? :)
    By the way, when using the first mentioned method, you should check if the solution gives out really satisfies the problem, since if you square both sides, like you have (-2)2 = (2)2, but -2 is not equal to 2. Do you follow me? :)
    And another point is that, there are infinite numbers of solutions to the equation:
    [tex]\frac{-16}{25} = \sin (2 \theta)[/tex], not just one as Swapnil mentioned.
     
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