11.2 Compound angle identities and applications

Trigonometry - grade 12

Compound angle identities

Derivation of $sin(\alpha +\beta )$

We have, for any angles $\alpha$ and $\beta$ , that

How do we derive this identity? It is tricky, so follow closely.

Suppose we have the unit circle shown below. The two points $L(a,b)$ and $K(x,y)$ are on the circle.

We can get the coordinates of $L$ and $K$ in terms of the angles $\alpha$ and $\beta$ . For the triangle $LOK$ , we have that

Thus the coordinates of $L$ are $(cos\beta ;sin\beta )$ . In the same way as above, we can see that the coordinates of $K$ are $(cos\alpha ;sin\alpha )$ . The identity for $cos(\alpha -\beta )$ is now determined by calculating $K{L}^2$ in two ways. Using the distance formula (i.e. $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ or $d^2={(x_2-x_1)}^2+{(y_2-y_1)}^2$ ), we can find $K{L}^2$ :

The second way we can determine $K{L}^2$ is by using the cosine rule for $▵KOL$ :

Equating our two values for $K{L}^2$ , we have

Now let $\alpha \to {90}^{\circ }-\alpha$ . Then

But $cos({90}^{\circ }-(\alpha +\beta ))=sin(\alpha +\beta )$ . Thus

Derivation of $sin(\alpha -\beta )$

We can use

to show that

We know that

and

Therefore,

Derivation of $cos(\alpha +\beta )$

We can use

to show that

We know that

Therefore,

Derivation of $cos(\alpha -\beta )$

We found this identity in our derivation of the $sin(\alpha +\beta )$ identity. We can also use the fact that

to derive that

As

we have that

Derivation of $sin2\alpha$

We know that

When $\alpha =\beta$ , we have that

Derivation of $cos2\alpha$

We know that

When $\alpha =\beta$ , we have that

However, we can also write

and

by using

The $cos2\alpha$ Identity

Use

to show that:

Problem-solving strategy for identities

The most important thing to remember when asked to prove identities is:

When proving trigonometric identities, never assume that the left hand side is equal to the right hand side. You need to show that both sides are equal.

A suggestion for proving identities: It is usually much easier simplifying the more complex side of an identity to get the simpler side than the other way round.