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Homework Help: Refresh my memory , Inverse Trig

  1. Mar 26, 2008 #1
    Refresh my memory plz, Inverse Trig

    [tex]x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)[/tex]

    Inverse sine is defined from [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] which lies in the 1st and 4th quadrant.

    So [tex]x=-\frac{\pi}{4}[/tex]
     
  2. jcsd
  3. Mar 26, 2008 #2

    dynamicsolo

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    Correct -- and your calculator and mine will back you up on that...
     
  4. Mar 26, 2008 #3
    That's the thing, it gives me different answers. (in radians)

    [tex]\sin^{-1}\left(-\frac{1}{\sqrt 2}\right)\approx -0.785[/tex]

    [tex]\sin\left(-\frac{\pi}{4}\right)\approx -0.707[/tex]
     
  5. Mar 26, 2008 #4
    Those aren't different answers. -pi / 4 is -0.785 and -1/ root 2 is -0.707.

    The inverse sin of the ratio will give you the angle, the sin of the angle will give you the ratio.
     
  6. Mar 26, 2008 #5

    dynamicsolo

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    And -45º converted to radians is? (In other words, what's [tex]\frac{-\pi}{4}[/tex]?)

    [I expect we're going to hear a "D'oh!" in 3, 2, 1, ...]
     
    Last edited: Mar 26, 2008
  7. Mar 26, 2008 #6
    LOL, I know that ... but what's with the calculator? Am I putting it in right?
     
  8. Mar 26, 2008 #7

    dynamicsolo

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    Of course you are! The sine of -45º or -pi/4 radians (which is approximately -0.785398... radians) gives you -1/sqrt(2) =
    -[sqrt(2)]/2 = -0.70710678... ,
    so taking the arcsine of (-0.70710678...) should give you
    -0.785398....

    I guess we're puzzled why you're puzzled. There's no reason you should get the same number in both directions. The magnitude of arcsin(x) doesn't match the magnitude of sin(x) at x = (pi)/4 ...
     
  9. Mar 26, 2008 #8
    LMAO ... omg, I crack myself up :)
     
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