Precalculus

Trigonometry

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Half-Angle Identities

Solved Problems

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Example 1

Find sin α/2, cos α/2, and tan α/2 if α = π/3.

Example 2

Find sin β/2, cos β/2, and tan β/2 if tan β = 24/7 and π < β < 3π/2.

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

Solution.

The angle \(\frac{\alpha }{2} = \frac{\pi }{6}\) 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 \(\sin \frac{\beta}{2},\) \(\cos \frac{\beta}{2},\) and \(\tan \frac{\beta}{2}\) if \(\tan \beta = \frac{{24}}{7}\) and \(\pi \lt \beta \lt \frac{{3\pi}}{2}.\)

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 \(\frac{\beta }{2}\) 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

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

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:

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

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

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

gives

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