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Inverse Trig Function

  1. Oct 18, 2009 #1
    I have reached a part of an example physics problem in which they go from one line to the next and I cannot figure out how.

    So my task is to show that

    [tex]\tan^{-1}\frac{sin\theta}{\cos\theta - 1} = 90^{\circ} + \frac{\theta}{2}[/tex]

    Can I get a hint in the right direction?

  2. jcsd
  3. Oct 18, 2009 #2
    You know that Tan(90+theta) = Tan(theta), so the 90 goes away. The rest should be a half-angle identity.
  4. Oct 18, 2009 #3
    That's not true. [itex]\tan(x+\pi) = \tan x[/itex], but [itex]\tan(x+\pi/2)\not=\tan x[/itex].
  5. Oct 19, 2009 #4


    Staff: Mentor

    Your equation is equivalent to [tex]\frac{sin\theta}{\cos\theta - 1}~=~tan(90^{\circ} + \frac{\theta}{2})~=~tan(1/2(\theta + \pi))[/tex]

    You can use one of the half-angle formulas for tangent to work with the expression on the right.
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