Half Angle Formulas/Sine/Proof 2

Theorem

\(\ds \sin \frac \theta 2\)

\(=\)

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

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

\(\ds \sin \frac \theta 2\)

\(=\)

\(\ds -\sqrt {\dfrac {1 - \cos \theta} 2}\)

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

Proof

Define:

$u = \dfrac \theta 2$

Then:

\(\ds \sin^2 u\)

\(=\)

\(\ds \frac {1 - \cos 2 u} 2\)

Power Reduction Formulas

\(\ds \leadsto \ \ \)

\(\ds \sin \frac \theta 2\)

\(=\)

\(\ds \pm \sqrt {\frac {1 - \cos \theta} 2}\)

We also have that:

In quadrant $\text I$, and quadrant $\text {II}$, $\sin \theta > 0$

In quadrant $\text {III}$, and quadrant $\text {IV}$, $\sin \theta < 0$.

$\blacksquare$