Precalculus

Trigonometry

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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:

Sine half-angle formula

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,

Cosine half-angle formula

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,

Tangent half-angle formula

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,

Tangent half-angle identity in terms of sine and cosine (1st formula).

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

Tangent half-angle identity in terms of sine and cosine (2nd formula).

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:

Cotangent half-angle identity
Cotangent half-angle identity in terms of sine and cosine (1st formula).
Cotangent half-angle identity in terms of sine and cosine (2nd formula).

See solved problems on Page 2.

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