# Half Angle Tangent Substitution

The **half-angle tangent substitution** consists of substituting some or all ratios of a given expression by a formula made up of only tangents of half the angles.

#### Contents

## Definition and Proof

These formulas are as follows:

\[\begin{array} &\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}}, &\cos\theta=\frac{1-\tan^{2}\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}}, &\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^{2}\frac{\theta}{2}}.\end{array}\]

We will use the double-angle formula to prove this:

\[\begin{align} \sin \theta & = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\\ & = \frac{ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2}} \\ & = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac {\theta}{2}},\\ \\ \tan \theta &= \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}},\\ \\ \cos \theta & = \frac {\sin \theta}{\tan \theta} \\ & = \frac{\frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac {\theta}{2}}}{\hspace{5mm} \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}}\hspace{5mm} } \\ & = \frac{1-\tan^{2} \frac{\theta}{2}}{1+\tan^{2} \frac{\theta}{2}}. \ _\square \end{align}\]

## Example Problems

## See Also

**Cite as:**Half Angle Tangent Substitution.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/half-angle-tangent-substitution/