Solve Inverse Trig Function: tan^-1(sinθ/(cosθ-1)) = 90° + θ/2

[Saladsamurai]

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

$$\tan^{-1}\frac{sin\theta}{\cos\theta - 1} = 90^{\circ} + \frac{\theta}{2}$$

Can I get a hint in the right direction?

Thanks!

 

[flatmaster]

You know that Tan(90+theta) = Tan(theta), so the 90 goes away. The rest should be a half-angle identity.

 

[n!kofeyn]

You know that Tan(90+theta) = Tan(theta), so the 90 goes away.

That's not true. $\tan(x+\pi) = \tan x$, but $\tan(x+\pi/2)\not=\tan x$.

 

[Mark44]

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

$$\tan^{-1}\frac{sin\theta}{\cos\theta - 1} = 90^{\circ} + \frac{\theta}{2}$$

Can I get a hint in the right direction?

Thanks!

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

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