# Precalculus

## Trigonometry # 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:

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,

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,

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,

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

## 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:

See solved problems on Page 2.