# Half-Angle Identities

## Key Questions

• Common Half angle identity:
1. $\sin a = 2 \sin \left(\frac{a}{2}\right) \cdot \cos \left(\frac{a}{2}\right)$

Half angle Identities in term of t = tan a/2.
2. $\sin a = \frac{2 t}{1 + {t}^{2}}$

3.$\cos a = \frac{1 - {t}^{2}}{1 + {t}^{2}}$

1. $\tan a = \frac{2 t}{1 - {t}^{2}} .$

Use of half angle identities to solve trig equations.

Example. Solve $\cos x + 2 \cdot \sin x = 1 + \tan \left(\frac{x}{2}\right) .$
Solution. Call $t = \tan \left(\frac{x}{2}\right)$. Use half angle identities (2) and (3) to transform the equation.

$\frac{1 - {t}^{2}}{4} + \frac{1 + {t}^{2}}{4} = 1 + t .$

$1 - {t}^{2} + 4 t = \left(1 + t\right) \left(1 + {t}^{2}\right)$

${t}^{3} + 2 {t}^{2} - 3 t = t \cdot \left({t}^{2} + 2 t - 3\right) = 0.$

Next, solve the $3$ basic trig equations: tan (x/2) = t = 0; tan (x/2) = -3; and $\tan \left(\frac{x}{2}\right) = 1.$

• The half-angle identities are defined as follows:

$\setminus m a t h b f \left(\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}\right)$

$\left(+\right)$ for quadrants I and II
$\left(-\right)$ for quadrants III and IV

$\setminus m a t h b f \left(\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}\right)$

$\left(+\right)$ for quadrants I and IV
$\left(-\right)$ for quadrants II and III

$\setminus m a t h b f \left(\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}\right)$

$\left(+\right)$ for quadrants I and III
$\left(-\right)$ for quadrants II and IV

We can derive them from the following identities:

${\sin}^{2} x = \frac{1 - \cos \left(2 x\right)}{2}$

${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1 - \cos \left(x\right)}{2}$

$\textcolor{b l u e}{\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}}$

Knowing how $\sin x$ is positive for $0 - {180}^{\circ}$ and negative for $180 - {360}^{\circ}$, we know that it is positive for quadrants I and II and negative for III and IV.

${\cos}^{2} x = \frac{1 + \cos \left(2 x\right)}{2}$

${\cos}^{2} \left(\frac{x}{2}\right) = \frac{1 + \cos \left(x\right)}{2}$

$\textcolor{b l u e}{\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(x\right)}{2}}}$

Knowing how $\cos x$ is positive for $0 - {90}^{\circ}$ and $270 - {360}^{\circ}$, and negative for $90 - {270}^{\circ}$, we know that it is positive for quadrants I and IV and negative for II and III.

$\tan \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\cos \left(\frac{x}{2}\right)} = \frac{\pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}}{\pm \sqrt{\frac{1 + \cos \left(x\right)}{2}}}$

$\textcolor{b l u e}{\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{1 + \cos \left(x\right)}}}$

We can see that if we take the conditions for positive and negative values from $\sin x$ and $\cos x$ and divide them, we get that this is positive for quadrants I and III and negative for II and IV.