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 \(\beta = \large{\frac{\alpha }{2}}\normalsize.\) 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 \(\large{\frac{\alpha }{2}}\normalsize\) 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 \(\beta \to \large{\frac{\alpha }{2}}\normalsize\) 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 \(\large{\frac{\alpha }{2}}\normalsize\) lies.

Tangent of a Half Angle

Now we can derive the formula for calculating \(\tan \large{\frac{\alpha }{2}}\normalsize.\) 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 \(\large{\frac{\alpha }{2}}\normalsize\) falls into.

We can also obtain an expression for \(\tan \large{\frac{\alpha }{2}}\normalsize\) 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 \({\cos \large{\frac{\alpha }{2}}\normalsize}\) 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 \({\sin \large{\frac{\alpha }{2}}\normalsize},\) 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).

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find \(\sin \large{\frac{\alpha}{2}}\normalsize,\) \(\cos \large{\frac{\alpha}{2}}\normalsize,\) and \(\tan \large{\frac{\alpha}{2}}\normalsize\) if \(\alpha = \large{\frac{\pi}{3}}\normalsize.\)

Example 2

Find \(\sin \large{\frac{\beta}{2}}\normalsize,\) \(\cos \large{\frac{\beta}{2}}\normalsize,\) and \(\tan \large{\frac{\beta}{2}}\normalsize\) if \(\tan \beta = \large{\frac{{24}}{7}}\normalsize\) and \(\pi \lt \beta \lt \large{\frac{{3\pi}}{2}}\normalsize.\)

Example 3

Simplify the expression \[2\,{\sin ^2}\frac{\alpha }{2} + \cos \alpha .\]

Example 4

Simplify the expression \[\frac{{1 – \cos 2\alpha }}{{\sin 2\alpha }}.\]

Example 5

Simplify the expression \[\frac{{1 – {{\tan }^2}\frac{\beta }{2}}}{{1 + {{\tan }^2}\frac{\beta }{2}}}.\]

Example 6

Simplify the expression \[\tan \frac{\alpha }{2} + 2\,{\sin ^2}\frac{\alpha }{2}\cot \alpha .\]

Example 7

Prove the identity \[\cot \frac{\alpha }{2} – \tan \frac{\alpha }{2} = 2\cot \alpha .\]

Example 8

Prove the identity \[\frac{{{{\sin }^2}\alpha }}{{1 + \cos \alpha }} + 2\,{\cos ^2}\frac{\alpha }{2} = 2.\]

Example 9

Prove the identity \[1 + \sin \beta = 2\,{\cos ^2}\left( {\frac{\pi }{4} – \frac{\beta }{2}} \right).\]

Example 10

Prove the identity \[1 – \sin \alpha = 2\,{\sin ^2}\left( {\frac{\pi }{4} – \frac{\alpha }{2}} \right).\]

Example 1.

Find \(\sin \large{\frac{\alpha}{2}}\normalsize,\) \(\cos \large{\frac{\alpha}{2}}\normalsize,\) and \(\tan \large{\frac{\alpha}{2}}\normalsize\) if \(\alpha = \large{\frac{\pi}{3}}\normalsize.\)

Solution.

The angle \(\large{\frac{\alpha }{2}}\normalsize = \large{\frac{\pi }{6}}\normalsize\) lies in the \(1\text{st}\) quadrant, in which all the functions have a positive sign. Therefore,

\[{\sin \frac{\alpha }{2} = \sqrt {\frac{{1 – \cos \alpha }}{2}} }={ \sqrt {\frac{{1 – \cos \frac{\pi }{3}}}{2}} }={ \sqrt {\frac{{1 – \frac{1}{2}}}{2}} }={ \sqrt {\frac{1}{4}} }={ \frac{1}{2};}\]

\[{\cos \frac{\alpha }{2} = \sqrt {\frac{{1 + \cos \alpha }}{2}} }={ \sqrt {\frac{{1 + \cos \frac{\pi }{3}}}{2}} }={ \sqrt {\frac{{1 + \frac{1}{2}}}{2}} }={ \sqrt {\frac{3}{4}} }={ \frac{{\sqrt 3 }}{2};}\]

\[{\tan \frac{\alpha }{2} = \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} }={ \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} }={ \frac{1}{{\sqrt 3 }}.}\]

Example 2.

Find \(\sin \large{\frac{\beta}{2}}\normalsize,\) \(\cos \large{\frac{\beta}{2}}\normalsize,\) and \(\tan \large{\frac{\beta}{2}}\normalsize\) if \(\tan \beta = \large{\frac{{24}}{7}}\normalsize\) and \(\pi \lt \beta \lt \large{\frac{{3\pi}}{2}}\normalsize.\)

Solution.

Find the value of \(\cos\beta\) given that \(\beta\) is in the \(3\text{rd}\) quadrant where cosine is negative:

\[{{\tan ^2}\beta + 1 = \frac{1}{{{{\cos }^2}\beta }},}\;\; \Rightarrow {{\cos ^2}\beta = \frac{1}{{{{\tan }^2}\beta + 1}},}\;\; \Rightarrow {\cos \beta = – \sqrt {\frac{1}{{{{\tan }^2}\beta + 1}}} }={ – \sqrt {\frac{1}{{{{\left( {\frac{{24}}{7}} \right)}^2} + 1}}} }={ – \sqrt {\frac{1}{{\frac{{576}}{{49}} + 1}}} }={ – \sqrt {\frac{{49}}{{625}}} }={ – \frac{7}{{25}}.}\]

Notice that

\[{\pi \lt \beta \lt \frac{{3\pi }}{2},}\;\; \Rightarrow {\frac{\pi }{2} \lt \frac{\beta }{2} \lt \frac{{3\pi }}{4},}\]

that is, the angle \(\large{\frac{\beta }{2}}\normalsize\) falls into the \(2\text{nd}\) quadrant. In this quadrant, the sine is positive, and cosine and tangent are negative. Hence,

\[{\sin \frac{\beta }{2} = \sqrt {\frac{{1 – \cos \beta }}{2}} }={ \sqrt {\frac{{1 – \left( { – \frac{7}{{25}}} \right)}}{2}} }={ \sqrt {\frac{{16}}{{25}}} }={ \frac{4}{5};}\]

\[{\cos \frac{\beta }{2} = – \sqrt {\frac{{1 + \cos \beta }}{2}} }={ – \sqrt {\frac{{1 + \left( { – \frac{7}{{25}}} \right)}}{2}} }={- \sqrt {\frac{9}{{25}}} }={ -\frac{3}{5};}\]

\[{\tan \frac{\beta }{2} = \frac{{\sin \frac{\beta }{2}}}{{\cos \frac{\beta }{2}}} }={ \frac{{\frac{4}{5}}}{{ – \frac{3}{5}}} }={ – \frac{4}{3}.}\]

Example 3.

Simplify the expression \[2\,{\sin ^2}\frac{\alpha }{2} + \cos \alpha .\]

Solution.

Using the sine half-angle identity, we have

\[\require{cancel}{2\,{\sin ^2}\frac{\alpha }{2} + \cos \alpha }={ 2 \cdot \frac{{1 – \cos \alpha }}{2} + \cos \alpha }={ 1 – \cancel{\cos \alpha} + \cancel{\cos \alpha} }={ 1.}\]

Example 4.

Simplify the expression \[\frac{{1 – \cos 2\alpha }}{{\sin 2\alpha }}.\]

Solution.

Using the identities

\[{1 – \cos 2\alpha = 2\,{\sin ^2}\alpha\;\;\text{and}\;\;}\kern0pt{\sin 2\alpha = 2\sin \alpha \cos \alpha ,}\]

we obtain

\[\frac{{1 – \cos 2\alpha }}{{\sin 2\alpha }} = \frac{{\cancel{2}\,{{\sin }^\cancel{2}}\alpha }}{{\cancel{2}\cancel{\sin \alpha} \cos \alpha }} = \frac{{\sin \alpha }}{{\cos \alpha }} = \tan \alpha .\]

Example 5.

Simplify the expression \[\frac{{1 – {{\tan }^2}\frac{\beta }{2}}}{{1 + {{\tan }^2}\frac{\beta }{2}}}.\]

Solution.

We use the tangent half-angle identity

\[\tan \frac{\beta }{2} = \pm \sqrt {\frac{{1 – \cos \beta }}{{1 + \cos \beta }}} .\]

Then

\[{\frac{{1 – {{\tan }^2}\frac{\beta }{2}}}{{1 + {{\tan }^2}\frac{\beta }{2}}} }={ \frac{{1 – \frac{{1 – \cos \beta }}{{1 + \cos \beta }}}}{{1 + \frac{{1 – \cos \beta }}{{1 + \cos \beta }}}} }={ \frac{{\frac{{1 + \cos \beta – \left( {1 – \cos \beta } \right)}}{{1 + \cos \beta }}}}{{\frac{{1 + \cos \beta + 1 – \cos \beta }}{{1 + \cos \beta }}}} }={ \frac{{\cancel{1} + \cos \beta – \cancel{1} + \cos \beta }}{{1 + \cancel{\cos \beta} + 1 – \cancel{\cos \beta} }} }={ \frac{{\cancel{2}\cos \beta }}{\cancel{2}} }={ \cos \beta .}\]

Example 6.

Simplify the expression \[\tan \frac{\alpha }{2} + 2\,{\sin ^2}\frac{\alpha }{2}\cot \alpha .\]

Solution.

Since

\[\tan \frac{\alpha }{2} = \frac{{1 – \cos \alpha }}{{\sin \alpha }}\]

and

\[2\,{\sin ^2}\frac{\alpha }{2} = 1 – \cos \alpha ,\]

we have

\[{\tan \frac{\alpha }{2} + 2\,{\sin ^2}\frac{\alpha }{2}\cot \alpha }={ \frac{{1 – \cos \alpha }}{{\sin \alpha }} + \left( {1 – \cos \alpha } \right)\cot \alpha }={ \frac{{1 – \cos \alpha }}{{\sin \alpha }} + \left( {1 – \cos \alpha } \right)\frac{{\cos \alpha }}{{\sin \alpha }} }={ \frac{{1 – \cos \alpha + \left( {1 – \cos \alpha } \right)\cos \alpha }}{{\sin \alpha }} }={ \frac{{\left( {1 – \cos \alpha } \right)\left( {1 + \cos \alpha } \right)}}{{\sin \alpha }} }={ \frac{{1 – {{\cos }^2}\alpha }}{{\sin \alpha }} }={ \frac{{{{\sin }^\cancel{2}}\alpha }}{\cancel{\sin \alpha }} }={ \sin \alpha .}\]

Example 7.

Prove the identity \[\cot \frac{\alpha }{2} – \tan \frac{\alpha }{2} = 2\cot \alpha .\]

Solution.

We substitute the expressions for tangent and cotangent of a half angle:

\[{\cot \frac{\alpha }{2} = \frac{{1 + \cos \alpha }}{{\sin \alpha }},\;\;}\kern0pt{\tan \frac{\alpha }{2} = \frac{{1 – \cos \alpha }}{{\sin \alpha }}.}\]

This yields:

\[{\cot \frac{\alpha }{2} – \tan \frac{\alpha }{2} }={ \frac{{1 + \cos \alpha }}{{\sin \alpha }} – \frac{{1 – \cos \alpha }}{{\sin \alpha }} }={ \frac{{1 + \cos \alpha – \left( {1 – \cos \alpha } \right)}}{{\sin \alpha }} }={ \frac{{\cancel{1} + \cos \alpha – \cancel{1} + \cos \alpha }}{{\sin \alpha }} }={ \frac{{2\cos \alpha }}{{\sin \alpha }} }={ 2\cot \alpha .}\]

Example 8.

Prove the identity \[\frac{{{{\sin }^2}\alpha }}{{1 + \cos \alpha }} + 2\,{\cos ^2}\frac{\alpha }{2} = 2.\]

Solution.

Recall that

\[2\,{\cos ^2}\frac{\alpha }{2} = 1 + \cos \alpha .\]

Hence, the left-hand side of the original expression is given by

\[{\frac{{{{\sin }^2}\alpha }}{{1 + \cos \alpha }} + 2\,{\cos ^2}\frac{\alpha }{2} }={ \frac{{{{\sin }^2}\alpha }}{{1 + \cos \alpha }} + 1 + \cos \alpha }={ \frac{{{{\sin }^2}\alpha }}{{1 + \cos \alpha }} + \frac{{{{\left( {1 + \cos \alpha } \right)}^2}}}{{1 + \cos \alpha }} }={ \frac{{{{\sin }^2}\alpha + 1 + 2\cos \alpha + {{\cos }^2}\alpha }}{{1 + \cos \alpha }} }={ \frac{{2 + 2\cos \alpha }}{{1 + \cos \alpha }} }={ \frac{{2\cancel{\left( {1 + \cos \alpha } \right)}}}{\cancel{1 + \cos \alpha }} }={ 2.}\]

Example 9.

Prove the identity \[1 + \sin \beta = 2\,{\cos ^2}\left( {\frac{\pi }{4} – \frac{\beta }{2}} \right).\]

Solution.

Using the cosine half-angle identity

\[2\,{\cos ^2}\frac{\theta }{2} = 1 + \cos \theta\]

and the cofunction formula

\[\cos \left( {\frac{\pi }{2} – \theta } \right) = \sin \theta ,\]

we can write the right hand-side \(\left({RHS}\right)\) in the form

\[{2\,{\cos ^2}\left( {\frac{\pi }{4} – \frac{\beta }{2}} \right) }={ 1 + \cos \left( {\frac{\pi }{2} – \beta } \right) }={ 1 + \sin \beta .}\]

Hence, \(LHS=RHS.\)

Example 10.

Prove the identity \[1 – \sin \alpha = 2\,{\sin ^2}\left( {\frac{\pi }{4} – \frac{\alpha }{2}} \right).\]

Solution.

Applying the identities

\[{2\,{\sin ^2}\frac{\theta}{2} = 1 – \cos \theta\;\;\text{and}\;\;}\kern0pt{ \cos \left( {\frac{\pi }{2} – \theta } \right) = \sin \theta }\]

gives

\[{RHS }={ 2\,{\sin ^2}\left( {\frac{\pi }{4} – \frac{\alpha }{2}} \right) }={ 1 – \cos \left( {\frac{\pi }{2} – \alpha } \right) }={ 1 – \sin \alpha }={ LHS.}\]