This example demonstrates how to derive the double angle identities using the inscribed angle theorem on the unit circle.

Start drawing a unit circle and dividing it into two equal parts by drawing its diameter .

Draw another point along the circumference of the circle which forms the angle labeled as (alpha). The inscribed angle theorem guarantees that this triangle will always be a right triangle even as the point moves freely.

Then, observe that there is another instance of the inscribed angle theorem with the angle formed by the points .

Apply the definitions of the functions cosine and sine to the point shown here by drawing the right triangle .

This gives us the initial setup for the derivation, because we can relate the trigonometry of the expression and to the trigonometry of the angle (alpha).

Solve for the unknown lengths of the right triangle

Apply the definition of the functions cosine and sine to find the lengths and . Note, the hypotenuse is of length .

- Because this is a right triangle we can find the lengths of its adjacent and opposite sides using the definitions of cosine and sine. Note, since the hypotenuse will be of length these are scaled by a factor of .
Observe that there is another instance of the inscribed angle theorem, this time with the right-triangle associated with the central angle :

Then we can apply the definitions of sine and cosine to find the lengths of the right-triangle . These lengths are what the identities describe.

Observe that the triangles and are similar triangles because they have the same three angles. Shown below are the two triangles highlighted with their angles labeled.

Since the triangles are similar the ratio of their corresponding sides are equal as shown in the equation below:

Then we can substitute the lengths into the equation which relates the ratio of adjacent sides together:

Using the diagram above which illustrates the lengths from previous steps and solve for the expression gives the first identity:

This gives us the double angle sine identity:

Then we can substitute the lengths into the equation which relates the ratio of opposite sides together:

Using the diagram above and solving for the expression: gives the second identity:

This gives us a form of the double angle cosine identity:

To get in terms of sine and cosine we can substitute Pythagorean’s Identity in for :

This gives us another form of the double angle cosine identity:

Finally, to get the last form of the cosine double angle identity, we can rearrange Pythagorean’s identity and substitute for the cosine expression.

This gives us the last form of the double angle cosine identity: