# Half-Angle Identities

## Solved Problems

Click or tap a problem to see the solution.

### 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

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