Why is the $\tan(\frac{\theta}{2})$ not $\pm$?

Question

Given an angle $\theta$, the formulas for $\sin(\frac{\theta}{2})$ and $\cos(\frac{\theta}{2})$ are either positive or negative, which is why $\pm$ is used in their formulas. So, if $\tan(\frac{\theta}{2})=\frac{\sin(\theta/2)}{\cos(\theta/2)}$, and both the numerator and denominator are either positive or negative (and unless $\theta$ is defined, we must assume they can be both), shouldn't the formula for $\tan(\frac{\theta}{2})$ also be $\pm$?

Below are the equations I am referring to.

$$\sin \left( \frac{u}{2} \right) = \pm \sqrt{\frac{1-\cos(u)}{2}}$$

$$\cos \left( \frac{u}{2} \right) = \pm \sqrt{\frac{1+\cos(u)}{2}}$$ $$\tan \left( \frac{u}{2} \right)=\frac{1-\cos(u)}{\sin(u)}=\frac{\sin(u)}{1+\cos(u)}$$

Answer 1

Basically, your first two equations involve square roots, and the third doesn't. The first requires a $\pm$ as $\sin u/2$ may be positive or negative, but by convention square roots are positive.

But (using duplication formulas for sine and cosine) $$\frac{1-\cos u}{\sin u}=\frac{1-(1-2\sin^2(u/2))}{2\sin(u/2)\cos(u/2)} =\frac{2\sin^2(u/2)}{2\sin(u/2)\cos(u/2)}=\frac{\sin(u/2)}{\cos(u/2)}=\tan(u/2)$$ etc.

Answer 2

I believe you are referring to the identity often stated as:

$$\tan(\theta/2) = \pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}$$

In some instances, such as solutions to an equation, the $\pm$ sign indicates both are valid.

In the case of values of functions only up to one of the two signs are valid for a specific angle $\theta$. You need to investigate which.

The most similar "correct" version of this identity is:

$$\tan^2(\theta/2) = \dfrac{1-\cos\theta}{1+\cos\theta}$$

There are other versions of the tangent half-angle identity which do not contain this ambiguity, such as

$$\tan(\theta/2) = \dfrac{\sin(\theta)}{1+\cos(\theta)}$$

Answer 3

Depend on $u$- the sign of $\cos \left( \frac{u}{2} \right)$ matter