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Homework Help: Solve the equation (sin and cos problems)

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the equations:
    [tex]100sin(\alpha) -400cos(30^{o}-\alpha) = 210[/tex]


    2. Relevant equations
    [tex]cos(A-B)=cos A cos B + sin A sin B[/tex]


    3. The attempt at a solution

    I threw the identity into the original equation, simplified a bit now I'm stuck at the following:

    [tex]200sin(\alpha) -400\sqrt{3}cos(\alpha) - 400 sin(\alpha) = 420[/tex]

    [tex] \Rightarrow[/tex] [tex] 10sin(\alpha) -20\sqrt{3}cos(\alpha) - 20 sin(\alpha) = 21 [/tex]

    Any ideas?
     
    Last edited: Jan 13, 2010
  2. jcsd
  3. Jan 13, 2010 #2

    rock.freak667

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    200sinα-400sinα simplifies to?
     
  4. Jan 13, 2010 #3
    [tex] \Rightarrow[/tex] [tex] -10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21 [/tex]
    Slowly getting there... Anymore nudges in the right direction?

    I can make things into an uglier equation ;),

    [tex] \Rightarrow[/tex] [tex]sin(\alpha) + 2\sqrt{3}cos(\alpha) = \frac{-21}{10} [/tex]

    but that really doesn't get me anywhere either... I have a feeling I may have taken the wrong route from the get go.
     
    Last edited: Jan 13, 2010
  5. Jan 13, 2010 #4

    rock.freak667

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    Put the ledt side in the form Rsin(α+A) or Rcos(α-A)
     
  6. Jan 13, 2010 #5
    [tex] a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta[/tex] So,

    For [tex] sin\theta, 1=Rcos\alpha[/tex]

    For [tex] cos\theta, 2\sqrt{3}=Rsin\alpha[/tex]

    So,

    [tex]2\sqrt{3} = tan\alpha[/tex]

    So,

    [tex]\alpha = 49.1^{o}[/tex]

    Is this correct?

    EDIT: I'm lost in how to use the form [tex]Rsin(\theta\pm\alpha)[/tex], could you show me how to apply it ?
     
    Last edited: Jan 13, 2010
  7. Jan 13, 2010 #6

    rock.freak667

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    Yes that looks correct, and what is R=?
     
  8. Jan 13, 2010 #7
    R would simply be as follows,

    [tex] R = \sqrt{a^{2} + b^{2}} = \sqrt{13} [/tex]

    But,

    [tex] \alpha = 49.1^{o}[/tex], doesn't seem to be satisfying my original equation? Thoughts?
     
  9. Jan 14, 2010 #8

    rock.freak667

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    tanα=2√3 ⇒α=tan-1(2√3)=73.9° (re-check that)
     
  10. Jan 14, 2010 #9
    Even with [tex] \alpha = 73.9^{o} [/tex], my original equation still isn't being satisfied.
     
  11. Jan 14, 2010 #10

    rock.freak667

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    Sorry your notation confused me. I used 'A' for a reason. A=73.9°. Your equation is now

    √13sin(α+73.9°)= -21/10
     
  12. Jan 14, 2010 #11

    vela

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    You need to be more careful. You have

    [tex]sin(\alpha) + 2\sqrt{3}cos(\alpha) = \frac{-21}{10}[/tex]

    and the identity

    [tex]a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta[/tex]

    But the [itex]\alpha[/itex] in the first line isn't the same as the [itex]\alpha[/itex] in the identity. Compare what you have to the LHS of the identity to see how the various quantities match up.
     
  13. Jan 14, 2010 #12
    Sorry if I'm wrong, but won't this lead me in circles?

    [tex]sin(A+B)=sin A cos B + cos A sin B[/tex]

    Ill be back to having an equation with a
    [tex] sin\alpha,cos\alpha[/tex],
    making it just as hard as the original to solve no?
     
  14. Jan 14, 2010 #13

    vela

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    Solve for [itex]\alpha+73.9^\circ[/itex].
     
  15. Jan 14, 2010 #14

    rock.freak667

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    If you have

    √13sin(α+73.9°)= -21/10 and then divide by √13, then you will have


    sin(α+73.9°)= -21/10√13

    Now just do as vela suggests and solve for α+73.9° and get α.


    Also, your initial question did not state the domain of α. Are you to find a general solution or did you leave out the the range for your answer(s)?
     
  16. Jan 14, 2010 #15
    Can I simply use [tex]sin^{-1}[/tex], to move the sin to the other side of the equation?

    If so,

    [tex] \alpha = -109.52^{o}[/tex]

    ... Turns out this works! Pefect! Thank you both for all your help it's appreciated!

    The question only states "Solve the equations:" and nothing else, so I'm assuming it's a general solution.
     
    Last edited: Jan 14, 2010
  17. Jan 14, 2010 #16
    Is it valid to write this as an answer?

    [tex] \alpha = -109.5^{o} +360^{o}k,\forall k Z [/tex],

    I'm just not sure how to state the "for all integers k" part...
     
  18. Jan 14, 2010 #17
    Bump, just looking for some final clarification on my notation.
     
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