Inverse Trig Function

  • #1
3,003
2
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?

Thanks!
 

Answers and Replies

  • #2
499
2
You know that Tan(90+theta) = Tan(theta), so the 90 goes away. The rest should be a half-angle identity.
 
  • #3
537
3
You know that Tan(90+theta) = Tan(theta), so the 90 goes away.
That's not true. [itex]\tan(x+\pi) = \tan x[/itex], but [itex]\tan(x+\pi/2)\not=\tan x[/itex].
 
  • #4
33,722
5,418
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?

Thanks!
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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