Half Angle Formulas/Cosine/Proof 2
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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 $\text {III}$ |
Proof
Define:
- $u = \dfrac \theta 2$
Then:
\(\ds \cos^2 u\) | \(=\) | \(\ds \frac {1 + \cos 2 u} 2\) | Power Reduction Formulas | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cos \frac \theta 2\) | \(=\) | \(\ds \pm \sqrt {\frac {1 + \cos \theta} 2}\) |
We also have that:
- In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
- In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.
$\blacksquare$