This example demonstrates how to derive the trigonometric identities using the trigonometric functions and the geometry of the unit circle. This example contains how to:

- Derive Pythagorean Identity (Unit Circle)
- Derive Sum of Two Angles Identities (Unit Circle)
- Derive Difference of Two Angles Identities (Unit Circle)
- Derive Double Angle Identities (Unit Circle)
- Derive Half Angle Identities (Unit Circle)

The Pythagorean identity relates the squared sides of the right triangle on the unit circle together. However, it can be hard to see how the squared sides translate to the geometry of the right triangle on the unit circle. This example shows the steps to finding the relationship between the length of and the squared length of the and .

Start with the right triangle on the unit circle as shown below defined by the point . Label the lengths of the adjacent and opposite sides in terms of the angle of the triangle.

The lengths of the adjacent and opposite sides can be solved for by applying the definitions of the sine and cosine functions.

Next, divide the right triangle into two similar triangles by drawing a line from the corner of the right-triangle perpendicular to its hypotenuse. This is shown below:

Then find the length of the adjacent side, labeled with the variable , of the first similar triangle shown below:

Apply the definition of cosine and then substitute the length of in for the hypotenuse and the length of for the adjacent side and then solve for .

Repeat this process to find the length of the opposite side of the second similar triangle, labeled with the variable :

Apply the definition of sine and then substitute the length of in for the hypotenuse and the length of for the opposite side and then solve for .

Finally, we can observe that the hypotenuse the right triangle of lenght can be expressed as the sum of the lengths and which gives us pythagorean’s identity:

This example derives the sum of two angle identities using the circle definitions of the trigonometric functions sine and cosine on the unit circle.

Start by drawing the point corresponding to sum of two angles (alpha) and (beta) on the unit circle. From the circle definitions of the trigonometric functions we know the horizontal coordinate of this point is equal to and the vertical coordinate is equal to .

This gives us the initial setup to derive the identities, where the goal is to express and in terms of the trigonometry of the individual angles and .

Next, draw the trigonometry related to the angles and as two right triangles stacked on top of eachother. I’ll also draw the length on top of the figure to make room for other expressions.

The key insight here is to draw a vertical line through the right-corner vertex of the right-triangle corresponding to the angle . This forms two right-triangles that are

*similar*to the right triangle formed by the angle . The vertical and horizontal components of the point formed by then can be represented using the side lengths of these two right-triangles.From the unit circle we know that the adjacent side of the right triangle formed by is equal to and the opposite side is equal to .

Since we now know the hypotenuse of this first similar right triangle, we can solve for the adjacent and opposite sides using the circle definitions of the trig functions. Note, you can also imagine scaling the adjacent side and opposite side by the value .

Repeat the same process to find the lengths of the second similar right triangle. This time the side lengths are scaled by the value .

Finally, equate the vertical and horizontal lengths together to derive the identities.

This results in the sum of two angles identities shown below.

TODO:

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:

TODO: