  # 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

# Faster Arithmetic - by Ken Ward

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: 