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Homework Help: Sine in algebra

  1. Apr 10, 2008 #1
    in bsin30-85.7sin(theta)=0
    How can I make theta the subject?
     
  2. jcsd
  3. Apr 10, 2008 #2

    Dick

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    Solve for sin(theta) and use arcsin.
     
  4. Apr 10, 2008 #3
    how is that done?
     
  5. Apr 10, 2008 #4
    It's the inverse.

    Recall ...

    [tex]y=x^2,x\geq0[/tex]

    [tex]f^{-1}(x)=\sqrt x[/tex]

    So similarly ...

    [tex]\sin x=y[/tex]

    [tex]\sin^{-1}y=x[/tex]
     
  6. Apr 10, 2008 #5
    so is this correct
    sin(theta)=-bsin30/-85.7
    1/sin(theta)=(-85.7)/(-binsin30)


    sin^-1(-bsin30/-85.7)=(theta)
     
    Last edited: Apr 10, 2008
  7. Apr 10, 2008 #6

    Dick

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    That doesn't make it the subject, now does it? If sin(theta)=a, then theta=arcsin(a). That's the definition of arcsin.
     
  8. Apr 10, 2008 #7
    I forgot to add that arcsine is defined in Quadrants I & IV. It's domain is: [tex]-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}[/tex].
     
  9. Apr 10, 2008 #8
    I think what you did was implied something like ... [tex]\frac{1}{x}=x^{-1}[/tex]

    It's not the same, the reciprocal of sine is cosecant. Refer to Dick's post.
     
  10. Apr 10, 2008 #9

    Dick

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    Yes. That's it. arcsin(b*sin(30)/85.7)=theta.
     
  11. Apr 11, 2008 #10
    That's it. arcsin(b*sin(30)/85.7)=theta.
    How do you work this out on a calculator?
     
  12. Apr 11, 2008 #11

    Dick

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    To work it out on a calculator you need to know b, right? The arcsin function is usually labeled sin^(-1).
     
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