# What are the Compound Angle Identities?

In addition to the basic trigonometric identities and the reciprocal identities there are the compound angle identities including the double angle identities. The compound angle identities (sometimes called the addition angle identities) are as follows:

Proof of the first two identities follows from considering two compound triangles and proof of the third comes from using the first two identities. By replacing with and noting that , and we also have:

See Example 1 and videos for examples using the compound angle formulae.Â

## Double Angle Formulae

The double angle formulae (or identities) follow from the compound angle identities. By setting in the first set of identities above we obtain:

Note that there are two additional identities for here. These come from using (see more trigonometric identities). It is worth memorising the double angle identities as it can save time when answering exam questions. See Example 2.

## Expressions of the form

We can use the compound angle identities above to solve equations that are given in the form , for example. We do this by converting the trigonometric expression into the form , for example. Using the identity , we can write:

This expression now has separate and terms, the coefficients of which are and . Comparing this with the expression we must equate the coefficients and so:

You might recognise these expressions from the right-angled triangle with hypoteneuse and angle . The sides can be found using SOHCAHTOA (what is SOHCAHTOA?). The opposite is (which is ) and the adjacent is (which is ). It follows that expressions of the form can be converted to expressions of the form by setting and . See Example 3. Note that the above process can be applied to any of the compound angle identities.

## Examples

1. Given that for obtuse and for acute , find the value of .
2. Show that .

Solution:

1. Using the identity we can write:

To get expressions for and we use the identity . More on trigonometric identities. Hence, and . Since is obtuse is negative and so . Also, since is acute is positive and so . It follows that .

2. By splitting into two familiar angles such as and , we can use the identity to write We already know these trigonometric ratios and so

as required.

1. Solve for
2. Given the parametric equations and , find the Cartesian equation. (See more on parametric equations).

Solution:

1. Since there is a terms, it suggests that we use the identity . Hence, the equation becomes . At this point it is tempting to divide both sides by but this is assuming that which of course it could be. Rather, subtract from both sides to get and factorise . It follows that the solutions come from solving the two equations and . Hence, the solutions on the given interval are (exact) and (to 2 d.p.).
2.  We can rewrite the equations as and . Using the double angle identity , we can write or . This can be left implicitly or we can obtain an explicit equation by making the subject: .

1. Express in the form . Hence solve the equation on the interval
2. Write in the form . Hence, sketch the curve of .

Solution:

1. Setting and to 2 decimal places, we can write . Hence, we solve the equation or . It follows that and so the solutions in the given interval are and to 2 d.p.
2. We follow the steps of the notes but this time we look at the identity . It follows that . Equating the coefficients gives and . It follows that and to 3 d.p. Hence, . It follows that graph of is the graph of shifted to the right by 0.955 and stretched in the -direction by a factor of . See more on transformations.

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