Everything Maths Grade 12

CHAPTER 10. TRIGONOMETRY 10.2

1

1

x

y

a

b

�

�

�

(α−β)

α β

L(a;b)

K(x;y)

O M (x;y)

We can get the coordinates of L and K in terms of the angles α and β. For the triangle LOM , we have
that

sinβ =

b

1

=⇒ b = sinβ

cosβ =

a

1

=⇒ a = cosβ

Thus the coordinates of L are (cosβ; sinβ). In the same way as above, we can see that the coordinates
of K are (cosα; sinα). The identity for cos(α−β) is now determined by calculating KL^2 in two ways.
Using the distance formula (i.e. d =

�

(x 2 −x 1 )^2 + (y 2 −y 1 )^2 or d^2 = (x 2 −x 1 )^2 + (y 2 −y 1 )^2 ), we
can find KL^2 :

KL^2 = (cosα− cosβ)^2 + (sinα− sinβ)^2

= cos^2 α− 2 cosα cosβ + cos^2 β + sin^2 α− 2 sinα sinβ + sin^2 β

= (cos^2 α + sin^2 α) + (cos^2 β + sin^2 β)− 2 cosα cosβ− 2 sinα sinβ

= 1 + 1− 2(cosα cosβ + sinα sinβ)

= 2− 2(cosα cosβ + sinα sinβ)

The second way we candetermine KL^2 is by using the cosine rule for�KOL:

KL^2 = KO^2 +LO^2 − 2 .KO.LO. cos(α−β)

= 1^2 + 1^2 − 2(1)(1) cos(α−β)

= 2− 2. cos(α−β)

Equating our two valuesfor KL^2 , we have

2 − 2. cos(α−β) = 2− 2(cosα cosβ + sinα. sinβ)

=⇒ cos(α−β) = cosα. cosβ + sinα. sinβ

Now let α→ 90 ◦−α. Then

cos(90◦−α−β) = cos(90◦−α) cosβ + sin(90◦−α) sinβ

= sinα. cosβ + cosα. sinβ

But cos(90◦− (α +β)) = sin(α +β). Thus

sin(α +β) = sinα. cosβ + cosα. sinβ