# Ken Ward's Mathematics Pages

## Trigonometry

Trigonometry Contents

## Half Angle Formulae

### Sine

One form of the double-angle formula for the cosine is:
[1.1]

If we rewrite α as α/2
cos(2·α/2)=1-2·sin2(α/2)-1
As:
cos(2·α/2)=cos(α),
We have:
cos(α)=1−2·sin2(α/2) [1.2]
By rearranging :
2·sin2(α/2)=1-cos(α)

So:
sin2(α/2)=(1-cos(α))/2

And, by taking the square root, the sine half alpha is:
[1.3]
or
[1.4]

### Cosine

We again start with the double angle for cosine, but this time use this version:
[1.5]

So, by rearranging,
2·cos2α=cos2α+1

Rewriting α as α/2:
2·cos2(α)/2=cosα+1
Rearranging:
2cos2(α/2)=cosα+1
cos2(α/2)=(cosα+1)/2
And by taking the square root, we have two formulae (one for the + and one for the minus root):
[1.6]

and
[1.7]

### Tangent

To get the tangent of half an angle, knowing the angle, we divide the sine by the cosine formulae:
tan(α/2)=sin(α/2)/cos(α/2)
=√((1-cosα)/(1+cosα))

Multiply top and bottom by 1-cosα
=√((1-cosα)2/(1-cos2α))
As 1-cos2α=sin2α, we have:
= √((1-cosα)2/(sin2α))
Taking the square root:
= ±((1-cosα)/(sinα))
[1.8]
or
[1.9]

## Tangent Half Angle Formulae

### Sine

Starting with the double angle formula for sine:

Writing α as α/2
sinα=2·sin(α/2)·cos(α/2)

Dividing by cos2(α/2)·sec2(α/2) (which is equal to 1):
sin(α)=2·tan(α/2)/sec2(α/2)
=2·tan(α/2)/1+tan2(α/2)
Because sec2(α/2)=1+tan2(α/2)

[2.1]

### Cosine

Starting with the double angle formula for cosine: [1.1, repeated]
Writing α as α/2
cos α=1−2·sin2(α/2)

Divide by cos2(α/2)·sec2(α/2) (which is equal to 1):
cos(α)=[sec2(α/2)−2·tan2(α/2)]/sec2(α/2)

We note
sec2(α/2)=1+tan2(α/2), so
[2.2]

### Tangent

Dividing sinα by cosα (Equation 2.1 by 2.2) gives us:
[2.3]

Alternatively, we note the tangent double angle formula is:

By setting α as α/2, we immediately get the half angle formula (Equation 2.3)

### Relationship Between Tangent of Half Angles

The three values that occur in the half tangent formula are sides of a right angled triangle, so writing t=tan(α/2), and the hypotenuse, h=(1+t2), base, b=(1-t2), and perpendicular, p=2t, so
h2=b2+p2,

and substituting the half tangent, t, values, we get:
(1+t2)2=(1-t2)2+(2t)2

Trigonometry Contents

Ken Ward's Mathematics Pages