# Ken Ward's Mathematics Pages

## Trigonometry Compound Angles

### Sine

[1.1]

By substituting −β for β in Equation 1.1, we get:
[1.2]

### Cosine

By noting that sin (π/2−θ)=cos θ, we can write Equation 1.1 (with θ=α+β) as:
sin (π/2-α-β)=
cos(α+β)=
sin(π/2-α)·cos(−β)+cos(π/2-α)·sin(−β)

=cosα·cosβ−sinα·sinβ

Hence:
[1.3]

Similarly, by substituting −β for β in Equation 1.3:
[1.4]

### Tangent

[1.4]
To prove this we divide Equation 1.1 by Equation 1.3:

Which yields:

After dividing throughout by cosα·cosβ.

By substituting −β for β in 1.4, we get:
[1.5]

## Double Angles

The formulae for double angles follow from those for compound angles.

### Sine

Using:
[1.1, repeated]
And setting β to α, we have:
sin(2·α)=sinα·cosα+sinα·cosα=
by [2.1]

### Cosine

As with sine, setting β to α in the following:
[1.3, repeated]

We have:
cos2α=cosα·cosα-sinα·sinα=
cos2α−sin2α=
2·cos2α−1
Because sin2α=1−cos2α

So:
[2.2]
Or:
[ [2.3]
[2.4]

### Tangent

As before, setting β to α in the following:
[1.4, repeated]

After simplifying, we obtain the formula:
[2.4]

Trigonometry Contents

Ken Ward's Mathematics Pages