# 4. Half-Angle Formulas

by M. Bourne

We will develop formulas for the sine, cosine and tangent of a half angle.

## Half Angle Formula - Sine

We start with the formula for the cosine of a double angle that we met in the last section.

cos 2

θ= 1− 2sin2θ

### Formula Summary

We derive the following formulas on this page:

`sin\ alpha/2=+-sqrt((1-cos\ alpha)/2`

`cos\ alpha/2=+-sqrt((1+cos\ alpha)/2`

`tan\ alpha/2=(1-cos\ alpha)/(sin\ alpha`

Now, if we let

`theta=alpha/2`

then 2*θ* = *α* and our formula becomes:

`cos\ α = 1 − 2\ sin^2(α/2)`

We now solve for

`sin(alpha/2)`

(That is, we get `sin(alpha/2)` on the left of the equation and everything else on the right):

`2\ sin^2(α/2) = 1 − cos\ α`

`sin^2(α/2) = (1 − cos\ α)/2`

Solving gives us the following **sine of a half-angle** identity:

`sin\ alpha/2=+-sqrt((1-cos\ alpha)/2`

The sign (positive or negative) of `sin(alpha/2)` depends on the quadrant in which `α/2` lies.

If `α/2` is in the **first or second quadrants**, the formula uses the positive case:

`sin\ alpha/2=sqrt(1-cos\ alpha)/2`

If `α/2` is in the **third or fourth quadrants**, the formula uses the negative case:

`sin\ alpha/2=-sqrt(1-cos\ alpha)/2`

## Half Angle Formula - Cosine

Using a similar process, with the same substitution of `theta=alpha/2` (so 2*θ* = *α*) we subsitute into
the identity

cos 2θ= 2cos^{2}θ− 1 (see cosine of a double angle)

We obtain

`cos\ alpha=2\ cos^2alpha/2-1`

Reverse the equation:

`2\ cos^2alpha/2-1=cos\ alpha`

Add 1 to both sides:

`2cos^2alpha/2=1+cos\ alpha`

Divide both sides by `2`

`cos^2alpha/2=(1+cos\ alpha)/2`

Solving for `cos(α/2)`, we obtain:

`cos\ alpha/2=+-sqrt((1+cos\ alpha)/2`

As before, the sign we need depends on the quadrant.

If `α/2` is in the **first or fourth quadrants**, the formula uses the positive case:

`cos\ alpha/2=sqrt((1+cos\ alpha)/2`

If `α/2` is in the **second or third quadrants**, the formula uses the negative case:

`cos\ alpha/2=-sqrt((1+cos\ alpha)/2`

## Half Angle Formula - Tangent

The tangent of a half angle is given by:

`tan\ alpha/2=(1-cos\ alpha)/(sin\ alpha)`

We can also write the tangent of a half angle as follows:

`tan\ alpha/2=(sin\ alpha)/(1+cos\ alpha)`

### Summary of Tan of a Half Angle

`tan\ alpha/2=(1-cos\ alpha)/(sin\ alpha)=(sin\ alpha)/(1+cos\ alpha`

### Using *t*

It is sometimes useful to define *t* as the tan of a half angle:

`t=tan\ alpha/2`

This gives us the results:

`sin\ a=(2t)/(1+t^2)`

`cos\ alpha=(1-t^2)/(1+t^2)`

`tan\ alpha=(2t)/(1-t^2)`

### Tan of the Average of 2 Angles

With some algebraic manipulation, we can obtain:

`tan\ (alpha+beta)/2=(sin\ alpha+sin\ beta)/(cos\ alpha+cos\ beta)`

### Example 1

Find the value of `sin\ 15^@` using the sine half-angle relationship given above.

### Example 2

Find the value of `cos\ 165^@` using the cosine half-angle relationship given above.

### Example 3

Show that `2\ cos^2x/2-cos\ x=1`

### Exercises: Evaluating and Proving Half-Angle Identities

1. Use the half angle formula to evaluate `sin\ 75^@`.

2. Find the value of `sin(alpha/2)` if `cos\ alpha=12/13` where 0° < *α* < 90°.

3. Prove the identity:** **`2\ sin^2x/2+cos\ x=1`

4. Prove the identity: `2\ cos^2theta/2sec\ theta=sec\ theta+1`

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