# 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^2(alpha/2)-1`

Reverse the equation:

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

Add 1 to both sides:

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

Divide both sides by `2`

`cos^2(alpha/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)`

Proof

First, we recall `tan x = (sin x) / (cos x)`.

`tan a/2=(sin a/2)/(cos a/2)`

Then we use the sine and cosine of a half angle, as given above:

`=sqrt((1-cos a)/2)/sqrt((1+cos a)/2)`

Next line is the result of multiplying top and bottom by `sqrt 2`.

`=sqrt((1-cos a)/(1+cos a))`

We then multiply top and bottom (under the square root) by `(1 − cos α)`

`=sqrt(((1-cos a)^2)/((1+cos a)(1-cos a)))`

Next is a difference of 2 squares.

`=sqrt((1-cos a)^2/(1-cos^2a))`

We then make use of the identity `sin^2theta+cos^2theta=1`

`=sqrt((1-cos a)^2/(sin^2a))`

We then find the square root:

`=(1-cos a)/(sin a)`

Of course, we would need to make allowance for positive and negative signs, depending on the quadrant in question.

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

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

Proof

We multiply numerator (top) and denominator (bottom) of the right hand side of our first result by `1+cos alpha`, and obtain:

`(1-cos alpha)/(sin alpha) xx (1+cos alpha)/(1+cos alpha)`

Next, we use the difference of 2 squares.

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

We recall `sin^2θ + cos^2θ = 1`, and use it to obtain:

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

Finally, we cancel out the sin α.

`=(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.

Answer

With *α* = 30° and the formula

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

we obtain:

`sin 15^text(o)=+-sqrt((1-cos 30^text(o))/2)` `=+-sqrt((1-0.866...)/2)` `=0.2588`

Here the positive value is chosen because 15° is in the first quadrant.

[Why bother doing this when we can use a calculator? This is just to illustrate that the formula works.]

### Example 2

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

Answer

We use *α* = 330°, and so `α/2 = 165^@`.

`cos 165^text(o)=+-sqrt((1+cos 330^text(o))/2)`

`=+-sqrt((1+0.866...)/2)`

`=-0.9659`

Here the minus sign is used because 165° is in the second quadrant.

### Example 3

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

Answer

Using the above formula to substitute for `(cos alpha)/2`, we get:

`"LHS"=2 cos^2(x/2)-cos x`

`=2(sqrt((1+cos x)/2))^2-cos x`

`=2((1+cos x)/2)-cos x`

`=1+cos x-cos x`

`=1`

`="RHS"`

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

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

Answer

`sin 75^text(o)=+-sqrt((1-cos 150^(text(o)))/2)`

`=+-sqrt(((1+0.866))/2)`

`=0.9659`

First Quadrant, so it's **positive.**

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

Answer

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

`=sqrt((1-12/13)/2)`

`=sqrt((1/13)/2)`

`=sqrt(1/26)`

`=0.1961`

We choose positive because we are in the first quadrant.

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

Answer

`"LHS"=2 sin^2(x/2)+cos x`

`=2(sqrt((1-cos x)/(2)))^2+cos x`

`=2((1-cos x)/2)+cos x`

`=1-cos x+cos x`

`=1`

`="RHS"`

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

Answer

`"LHS"=2\ cos^2(theta/2)sec theta`

`=2(sqrt((1+cos theta)/(2)))^2sec\ theta`

`=(1+cos theta)sec\ theta`

`=(1+cos theta)1/(cos theta)`

`=sec\ theta+1`

`="RHS"`