Half Angle Identities

These are just rearrangements of the power reduction identities. I rarely have seen much use for them.

Cosine

\[\cos^2 (\theta / 2) = \frac{1}{2}(1 + \cos(\theta)) \quad\implies\quad \cos (\theta/2) = \pm \sqrt{\frac{1 + \cos \theta}{2}}\]

It is positive if $\theta/2$ is in quadrant I or IV and negative if $\theta/2$ is in quadrant II or III.


Sine

\[\sin^2 (\theta / 2) = \frac{1}{2}(1 - \cos(\theta)) \quad\implies\quad \sin (\theta/2) = \pm \sqrt{\frac{1 - \cos \theta}{2}}\]

It is positive if $\theta/2$ is in quadrant I or II and negative if $\theta/2$ is in quadrant III or IV.


Tangent

\[\tan^2 (\theta/2) = \frac{1 - \cos \theta}{1 + \cos \theta} \quad\implies\quad \tan (\theta/2) = \pm \sqrt{ \frac{1 - \cos \theta}{1 + \cos \theta} }\]

It is positive if $\theta/2$ is in quadrant I or III and negative if $\theta/2$ is in quadrant II or IV.


There are actually three more variations of this half angle identity. Here is the first.

\[\begin{align} \tan (\theta/2) &= \pm \sqrt{ \frac{1 - \cos \theta}{1 + \cos \theta} } \\[10pt] &= \pm \sqrt{ \frac{1 - \cos \theta}{1 + \cos \theta} \cdot \frac{1 + \cos \theta}{1 + \cos \theta} } \\[10pt] &= \pm \sqrt{ \frac{1 - \cos^2 \theta}{(1 + \cos \theta)^2} } \\[10pt] &= \pm \sqrt{ \frac{\sin^2 \theta}{(1 + \cos \theta)^2} } \\[10pt] &= \pm \frac{\sin \theta}{1 + \cos \theta} \end{align}\]

Here is the second.

\[\begin{align} \tan (\theta/2) &= \pm \frac{\sin \theta}{1 + \cos \theta} \\[10pt] &= \pm \frac{\sin \theta}{1 + \cos \theta} \cdot \frac{1 - \cos \theta}{1 - \cos \theta} \\[10pt] &= \pm \frac{\sin \theta \cdot (1 - \cos \theta)}{1 - \cos^2 \theta} \\[10pt] &= \pm \frac{\sin \theta \cdot (1 - \cos \theta)}{\sin^2 \theta} \\[10pt] &= \pm \frac{1 - \cos \theta}{\sin \theta} \end{align}\]

Here is the final identity.

\[\begin{align} \tan (\theta/2) &= \pm \frac{1 - \cos \theta}{\sin \theta} \\[10pt] &= \pm \left ( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right ) \\[10pt] &= \pm (\csc \theta - \cot \theta) \end{align}\]

Secant, Cosecant, and Cotangent

There are no half angle formulas for these functions other than just taking the reciprocal of the above identities.

Trigonometry Series