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Silly question about trig past pi/2

  1. Apr 8, 2008 #1
    I know all the well know angles from 0 to [tex] \pi/2[/tex]. However, past pi/2 I am rather clueless and have no idea what angle corresponds to what value.

    For example, in trying to figure out [tex] tan^{-1}(-1)[/tex] How does one know it is 5pi/4 and not 7pi/8? I was debating between measuring pi/4 from the negative x axis (which would give me 7pi/8) and measuring pi/4 from the y axis,(which would give me 5pi/4) but as you can see, I choose the wrong value. Is there a trick to it? Thanks.
  2. jcsd
  3. Apr 8, 2008 #2

    [tex]arctan(-1)=-arctan(1)=-\frac{\pi}{4}[/tex] so it means that [tex] \frac{\pi}{4} [/tex] lies on the fourth quadrant of the unit circle. So you can get that value by subtracting it from [tex]2\pi[/tex] that is

    [tex]2\pi-\frac{\pi}{4}=\frac{8\pi-\pi}{4}=\frac{7\pi}{4}[/tex] because -pi/4 is the angle counted clockwise direction.
    YOu know that it does not lie on the 3rd quadrant because tan(x) there is positive. There are other tricks too.
    Last edited: Apr 8, 2008
  4. Apr 8, 2008 #3
    That's the "problem" with inverse trig functions--since, for example, there are an infinite number of solutions for y to the equation [itex]\tan{y} = x[/itex], in order for [itex]f(x) = \arctan{x}[/itex] to be a well-defined function, you have to consider it as mapping to a particular interval of angles. Without qualification, arctan typically returns values between [itex]\frac{-\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex], but any interval of length [itex]\pi[/itex] is valid as long as your usage is consistent. There are similar considerations for arcsin and arccos.

    Edit: Because I'm not sure if I made it clear, let me say this: [itex]\arctan{x}[/itex] is usually defined as the function from the real numbers to the interval [itex](\frac{-\pi}{2} , \frac{\pi}{2})[/itex] such that [itex]\tan{(\arctan{x})} = x[/itex] for any real x.
    Last edited: Apr 8, 2008
  5. Apr 8, 2008 #4
    Oh I forget that arctan is an odd function. Thanks. But what if the question asks for a value for arctan(-1) within pi? I would have tried the 7pi/4 approach if I wasn't restricted to 0 to pi. Thanks.
  6. Apr 8, 2008 #5
    If you have to find the value within an interval of a length [tex]\pi[/tex] that doent mean that your interval must be [tex][0,\pi][/tex] , moreover the common interval for this problem is chosen like JohnDuck stated [tex][-\frac{\pi}{2},\frac{\pi}{2}][/tex].

    This means that also [tex]
    \frac{7\pi}{2}[/tex] is within that interval.
    Last edited: Apr 8, 2008
  7. Apr 8, 2008 #6
  8. Apr 8, 2008 #7

    Blahhhhhh! Yup.
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