Half Angle Formulas for Sine and Cosine

Theorem

$\displaystyle (1): \quad \sin \frac \theta 2 = \pm \sqrt {\frac {1 - \cos \theta} {2}}$

$\displaystyle (2): \quad \cos \frac \theta 2 = \pm \sqrt {\frac {1 + \cos \theta} {2}}$

$\displaystyle (3): \quad \tan \frac \theta 2 = \frac {\sin \theta} {1 + \cos \theta} = \frac {1 - \cos \theta} {\sin \theta}$

Proof

Define:

$u = \dfrac \theta 2$

\((1):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin^2 u\)	\(=\)	\(\displaystyle \frac {1 - \cos2u} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Power Reduction Formulas
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sin \frac \theta 2\)	\(=\)	\(\displaystyle \pm \sqrt {\frac {1 - \cos \theta} 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

\((2):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos^2 u\)	\(=\)	\(\displaystyle \frac {1 + \cos2u} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Power Reduction Formulas
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \cos \frac \theta 2\)	\(=\)	\(\displaystyle \pm \sqrt {\frac {1 + \cos \theta} 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

\((3):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \tan \frac \theta 2\)	\(=\)	\(\displaystyle \frac {\sin \frac \theta 2} {\cos \frac \theta 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From the definition of tangent
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\pm \sqrt {\frac {1 - \cos \theta} {2} } } {\pm \sqrt {\frac {1 + \cos \theta} {2} } }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Results $(1)$ and $(2)$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pm \sqrt {\frac {\left({1 - \cos \theta}\right) \left({1 + \cos \theta}\right)} {\left({1 + \cos \theta}\right)^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pm \sqrt {\frac {1 - \cos^2 \theta} {\left({1 + \cos \theta}\right)^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pm \sqrt {\frac {\sin^2 \theta} {\left({1 + \cos \theta}\right)^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pm \frac {\sin \theta} {1 + \cos \theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Since $\cos\theta \ge -1$, it follows that $\cos\theta + 1 \ge 0$.

We also have $\sin\theta > 0$ if $\theta$ is in the first or second quadrants and $\sin \theta < 0$ if $\theta$ is in the third or fourth quadrants

But $\displaystyle \tan \frac \theta 2 \geq 0$ if $\theta$ is in the first or second quadrants (because $\tan\theta > 0$ if $\theta$ is in the first quadrant), and $\displaystyle \tan \frac \theta 2 < 0$ if $\theta$ is in the third or fourth quadrants (because $\tan \theta < 0$ if $\theta$ is in the second quadrant).

Thus, $\displaystyle \tan \frac \theta 2$ and $\sin \theta$ have the same sign, so we can drop the $\pm$, and we obtain:

$\displaystyle \tan \frac \theta 2 = \frac {\sin \theta} {1 + \cos \theta}$

If we had proceeded by writing $\displaystyle \tan \frac \theta 2 = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta}} = \pm \sqrt {\frac {\left({1 - \cos \theta}\right)^2} {\left({1 + \cos \theta}\right) \left({1 - \cos \theta}\right)} }$, we would have ended up with:

$\displaystyle \tan \frac \theta 2 = \frac {1 - \cos \theta} {\sin \theta}$

Note

Technically, we should also check the boundaries between the first and fourth quadrants and the second and third quadrants.

If $\theta = \pi + 2k \pi, k \in \Z$, then $\displaystyle \tan\frac{\theta}{2}$ is undefined, $\displaystyle \frac{\sin\theta}{1+\cos\theta}$ is undefined, $\displaystyle \frac{1-\cos\theta}{\sin\theta}$ is undefined.

If $\theta = 2k \pi, k \in \Z$, then $\displaystyle \tan\frac{\theta}{2} = 0$, $\displaystyle \frac{\sin\theta}{1+\cos\theta} = 0$, and $\displaystyle \frac{1-\cos\theta}{\sin\theta}$ is undefined (although by L'Hôpital's Rule, $\displaystyle \lim_{\theta \to 0}\frac{1-\cos\theta}{\sin\theta} = 0$).

Thus, $\displaystyle \frac {1 - \cos \theta} {\sin \theta}$ is not a perfect formula for $\displaystyle \tan \frac \theta 2$.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $5.41, \ 5.42, \ 5.43$