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
If \(0 < \theta < \frac{\pi}{2}\) and \(\sin \theta=\frac{24}{25},\) what is the value of
\[\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}+\cos \frac{\theta}{2}}?\]
If \(\tan^2 \frac{3\pi}{8}\) is a root of the polynomial with rational coefficients \(2x^2-3ax+b\), what is the value of \(a+b?\)