Half Angle Formulas/Cosine

Theorem

$\ds \cos \frac \theta 2$

$=$

$\ds +\sqrt {\frac {1 + \cos \theta} 2}$

for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {IV}$

$\ds \cos \frac \theta 2$

$=$

$\ds -\sqrt {\frac {1 + \cos \theta} 2}$

for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {III}$

where $\cos$ denotes cosine.

$\ds \cos \theta$

$=$

$\ds 2 \cos^2 \frac \theta 2 - 1$

$\ds \leadsto \ \ $

$\ds 2 \cos^2 \frac \theta 2$

$=$

$\ds 1 + \cos \theta$

$\ds \leadsto \ \ $

$\ds \cos \frac \theta 2$

$=$

$\ds \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$

In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$

Define:

$u = \dfrac \theta 2$

Then:

$\ds \cos^2 u$

$=$

$\ds \frac {1 + \cos 2 u} 2$

$\ds \leadsto \ \ $

$\ds \cos \frac \theta 2$

$=$

$\ds \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$

In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.42$
1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): half-angle formula
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): half-angle formula