- #1

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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!

- Thread starter Saladsamurai
- Start date

- #1

- 3,003

- 2

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!

- #2

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- #3

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That's not true. [itex]\tan(x+\pi) = \tan x[/itex], but [itex]\tan(x+\pi/2)\not=\tan x[/itex].You know that Tan(90+theta) = Tan(theta), so the 90 goes away.

- #4

Mark44

Mentor

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

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!

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

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