Half Angle Formula - Practice Questions & MCQ

Half Angle Formula
1. $\sin \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1-\cos \alpha}{2}}$
2. $\cos \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1+\cos \alpha}{2}}$
3. $\tan \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}$

These formulae can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle.

The half-angle formula for sine is derived as follows:

$
\begin{aligned}
\sin ^2 \theta & =\frac{1-\cos (2 \theta)}{2} \\
\sin ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \left(2 \cdot \frac{\alpha}{2}\right)}{2} \\
& =\frac{1-\cos \alpha}{2} \\
\sin \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1-\cos \alpha}{2}}
\end{aligned}
$
To derive the half-angle formula for cosine, we have

$
\begin{aligned}
\cos ^2 \theta & =\frac{1+\cos (2 \theta)}{2} \\
\cos ^2\left(\frac{\alpha}{2}\right) & =\frac{1+\cos \left(2 \cdot \frac{\alpha}{2}\right)}{2} \\
& =\frac{1+\cos \alpha}{2} \\
\cos \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1+\cos \alpha}{2}}
\end{aligned}
$

For the tangent identity, we have

$\begin{aligned} \tan ^2 \theta & =\frac{1-\cos (2 \theta)}{1+\cos (2 \theta)} \\ \tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \left(2 \cdot \frac{\alpha}{2}\right)}{1+\cos \left(2 \cdot \frac{\alpha}{2}\right)} \\ & =\frac{1-\cos \alpha}{1+\cos \alpha} \\ \tan \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}\end{aligned}$