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Solving for a variable inside a sin()

  1. Apr 4, 2006 #1
    sin( degtorad( (180 - (180 - 360/x))/2 ) ) = y/z

    degtorad(degrees) means the the degrees inside the parenthesis are converted to radians.

    How do you solve for x?

    Thank you.
     
  2. jcsd
  3. Apr 4, 2006 #2
    you can eliminate the 180's by using properties of sine
    sin (180+ x) = -sin x = sin(-x)

    also,
    sin(90+x) = cos(x)
     
  4. Apr 4, 2006 #3
    sin( degtorad( (180 - (180 - 360/x))/2 ) ) = y/z

    Is this right?

    sin( (180 - (180 - 360/x))/2 )
    = sin( ( (180 - 360/x))/2 )
    = sin( (180 - 360/x) /2 )

    sin( (180 - 360/x) /2 )
    = sin( ( 360/x) /2 )
    = sin( ( 180/x) )
    = sin( 180/x )

    sin( 180/x ) = y/z

    um, then what?
     
  5. Apr 4, 2006 #4
    sin( (180 - 360/x) /2 ) = sin (90 - 180/x) = cos(-180/x) = cos (180/x)

    Are you familiar with arccos (or cos^{-1})?

    By the way, what exactly are y and z?
     
  6. Apr 5, 2006 #5

    HallsofIvy

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    Science Advisor

    Shorn of all the other things, arcsin( ) (also written sin-1( )) is defined as the inverse of sin( ) and arccos() (also written cos-1( )) is defined as the inverse of cos( ) (well, principal value). That is, arcsin(sin(x))= x and arccos(cos(x))= x.

    You have to be a bit careful about that: since sin(x) and cos(x) are not "one-to-one" they don't have inverses, strictly speaking. Given an x between -1 and 1, there exist an infinite number of y such that sin(y)= x or cos(y)= x. Arcsin(x) always gives the value, y, between -pi/2 and pi/2 such that sin(y)= x and arccos(x) always gives the value, y, between 0 and pi such that cos(y)= x.
     
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