Half-Angle Identities
Sine of a Half Angle
The cosine of a double angle is given by
\[\cos 2\beta = 1 - 2\,{\sin ^2}\beta .\]
It follows from this formula that
\[{\sin ^2}\beta = \frac{{1 - \cos 2\beta }}{2}.\]
Let \(\beta = \frac{\alpha }{2}.\) Then we have
\[{\sin ^2}\frac{\alpha }{2} = \frac{{1 - \cos \alpha }}{2},\;\; \Rightarrow \left| {\sin \frac{\alpha }{2}} \right| = \sqrt {\frac{{1 - \cos \alpha }}{2}} ,\;\; \Rightarrow \sin \frac{\alpha }{2} = \pm \sqrt {\frac{{1 - \cos \alpha }}{2}} .\]
We have derived the sine half-angle identity:
\[\sin \frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}\]
The \(\pm\) sign at the beginning of the right-hand side means that the square root can be positive or negative - depending on the quadrant in which the angle \(\frac{\alpha }{2}\) lies.
Cosine of a Half Angle
Using the double-angle identity for cosine in the form
\[\cos 2\beta = 2{\cos ^2}\beta - 1,\]
we can express \({\cos ^2}\beta \) in terms of \(\cos 2\beta :\)
\[{\cos ^2}\beta = \frac{{1 + \cos 2\beta }}{2}.\]
By replacing \(\beta \to \frac{\alpha }{2}\) we get the cosine half-angle identity:
\[{\cos ^2}\frac{\alpha }{2} = \frac{{1 + \cos \alpha }}{2},\;\; \Rightarrow \left| {\cos \frac{\alpha }{2}} \right| = \sqrt {\frac{{1 + \cos \alpha }}{2}} ,\;\; \Rightarrow \cos \frac{\alpha }{2} = \pm \sqrt {\frac{{1 + \cos \alpha }}{2}} .\]
Hence,
\[\cos \frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}\]
The sign depends on the quadrant in which \(\frac{\alpha }{2}\) lies.
Tangent of a Half Angle
Now we can derive the formula for calculating \(\tan \frac{\alpha }{2}.\) Using the above identities, we obtain
\[{\tan ^2}\frac{\alpha }{2} = \frac{{{{\sin }^2}\frac{\alpha }{2}}}{{{{\cos }^2}\frac{\alpha }{2}}} = \frac{{1 - \cos \alpha }}{{1 + \cos \alpha }}.\]
Therefore,
\[\tan \frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}}\]
where the \(\pm\) sign depends on what circle quadrants the angle \(\frac{\alpha }{2}\) falls into.
We can also obtain an expression for \(\tan \frac{\alpha }{2}\) without taking the square root.
Multiplying both the numerator and denominator in the right-hand side of the formula
\[\tan \frac{\alpha }{2} = \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}\]
by \({\cos \frac{\alpha }{2}}\) yields:
\[\tan \frac{\alpha }{2} = \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} = \frac{{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}}{{2\cos \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = \frac{{\sin \alpha }}{{2{{\cos }^2}\frac{\alpha }{2}}} = \frac{{\sin \alpha }}{{1 + \cos \alpha }},\]
that is,
\[\tan \frac{\alpha}{2} = \frac{\sin\alpha}{1 + \cos\alpha}\]
Similarly, multiplying the numerator and denominator by \({\sin \frac{\alpha }{2}},\) we can derive the tangent half-angle identity in the form
\[\tan \frac{\alpha }{2} = \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} = \frac{{2\sin \frac{\alpha }{2}\sin \frac{\alpha }{2}}}{{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}} = \frac{{2{{\sin }^2}\frac{\alpha }{2}}}{{\sin \alpha }} = \frac{{1 - \cos \alpha }}{{\sin \alpha }},\]
or
\[\tan \frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha}\]
Cotangent of a Half Angle
By definition,
\[\cot \frac{\alpha }{2} = \frac{1}{{\tan \frac{\alpha }{2}}}.\]
Therefore, we have the following half-angle identities for cotangent:
\[\cot \frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}}\]
\[\cot \frac{\alpha}{2} = \frac{\sin\alpha}{1 - \cos\alpha}\]
\[\cot \frac{\alpha}{2} = \frac{1 + \cos\alpha}{\sin\alpha}\]
See solved problems on Page 2.
