# Trigonometric Identities and the Unit Circle

Questions With Solutions

How to use the unit circle to find properties and trigonometric identities of the sine and
cosine functions? Grade 11 trigonometry questions are presented along with detailed solutions and explanations.

A circle has an infinite number of symmetries with respects to lines through the center and a symmetry with respect to its center. We are interested here on the symmetries with respect to its center, the x-axis, the y-axis an the line y = x. It will be shown how the use of these symmetries allows us to write several identities in trigonometry.

## Identities due to Symmetry of the Unit Circle on the origin, x and y axes

Four angles ( θ , π - θ , π + θ and 2π - θ ) are shown below in a unit circle. To each angle corresponds a point (

*A, B, C*or

*D*) on the unit circle.

.

The four angles have the same reference angle equal to θ. Because of the symmetry of the circle, the four points form a rectangle ABCD as shown above. Points A and B are reflection of each other of the y-axis. Points A and C are reflection of each other on the origin of the system of axis. Points A and D are reflection of each other on the x-axis. Given the coordinates

*a*and

*b*of point A and using the symmetries of the circle, the coordinates of A, B, C and D are given by:

A: (a , b) , B: (- a , b), C: (- a , - b) and D: (a , - b)

We now express the coordinates of each point in terms of the sine and cosine of the corresponding angle as follows.

A: (a , b) = (cos θ , sin θ)

B: (- a , b) = (cos(π - θ) , sin(π - θ))

C: (- a , - b) = (cos(π + θ) , sin(π + θ))

D: (a , - b) = (cos(2π - θ) , sin(2π - θ))

## Examples of Identities

Comparing the x and y-coordinates of points A and B, we can writecos(π - θ) = - cos θ

sin(π - θ) = sin θ

Comparing the x and y-coordinates of points A and C, we can write

cos(π + θ) = - cos θ

sin(π + θ) = - sin θ

Comparing the x and y-coordinates of points A and D, we can write

cos(2π - θ) = cos θ

sin(2π - θ) = - sin θ

## More Identities due to Symmetry of the Unit Circle on the x axis (Negative angles)

Two angles θ , and - θ are shown below in a unit circle to which correspond the points A and D on the unit circle..

Points A and D are reflection of each other on the x-axis. Given the coordinates

*a*and

*b*of point A, the coordinates of D are given by:

D: (a , - b)

We now express the coordinates of points A and D in terms of the sine and cosine of the corresponding angle as follows.

A: (a , b) = (cos θ , sin θ)

D: (a , - b) = (cos(- θ) , sin(- θ))

Examples of Identities that may be Deduced

cos(- θ) = cos θ

sin( - θ) = - sin θ

## Identities due to Symmetry of the Unit Circle on the line y = x

Points A and B shown in the unit circle below are reflection of each other on the line y = x. Because of the symmetry of the unit circle with respect to the line y = x, the corresponding angles to these points are θ and π/2 - θ as shown below..

Points A and B are reflection of each other on the line y = x. Given the coordinates

*a*and

*b*of point A, the coordinates of B are given by:

B: (b , a)

We now express the coordinates of points A and B in terms of the sine and cosine of the corresponding angle as follows.

A: (a , b) = (cos θ , sin θ)

B: (b , a) = (cos(π/2 - θ) , sin(π/2 - θ))

Examples of Identities that may be Deduced

cos(π/2 - θ) = sin θ

sin(π/2 - θ) = cos θ

## Questions with Solutions

Use the following general identities

1) cos (A + B) = cos A cos B - sin A sin B

2) cos (A - B) = cos A cos B + sin A sin B

3) sin(A + B) = sin A cos B + cos A sin B

4) sin(A - B) = sin A cos B - cos A sin B

to verify the identities listed below.

- cos(π - θ) = - cos θ
- sin(π - θ) = sin θ
- cos(π + θ) = - cos θ
- sin(π + θ) = - sin θ
- cos(2π - θ) = cos θ
- sin(2π - θ) = - sin θ
- cos(- θ) = cos θ
- sin( - θ) = - sin θ
- cos(π/2 - θ) = sin θ
- sin(π/2 - θ) = cos θ

## Solutions

- Use the general identity cos (A - B) = cos A cos B + sin A sin B to expand cos(π - θ) as follows:

cos(π - θ) = cos π cos θ + sin π sin θ

Use cos π = - 1 and sin π = 0 to simplify the above to

= - cos θ)

- Use the general identity sin(A - B) = sin A cos B - cos A sin B to expand sin(π - θ) as follows:

sin(π - θ) = sin π cos θ - cos π sin θ

Use sin π = 0 and cos π = - 1 to simplify the above to

= sin θ

- Use the general identity cos (A + B) = cos A cos B - sin A sin B to expand cos(π + θ) as follows:

cos(π + θ) = cos π cos θ - sin π sin θ

Use cos π = - 1 and sin π = 0 to simplify the above to

= - cos θ)

- Use the general identity sin(A + B) = sin A cos B + cos A sin B to expand sin(π + θ) as follows:

sin(π + θ) = sin π cos θ + cos π sin θ

Use sin π = 0 and cos π = - 1 to simplify the above to

= - sin θ

- Use the general identity cos (A - B) = cos A cos B + sin A sin B to expand cos(2π - θ) as follows:

cos(2π - θ) = cos 2π cos θ + sin 2π sin θ

Use cos 2π = 1 and sin 2π = 0 to simplify the above to

= cos θ)

- Use the general identity sin(A - B) = sin A cos B - cos A sin B to expand sin(2π - θ) as follows:

sin(2π - θ) = sin 2π cos θ - cos 2π sin θ

Use sin 2π = 0 and cos 2π = 1 to simplify the above to

= - sin θ

- We first write the left side of the identity to verify cos(- θ) = cos θ as follows:

cos(- θ) = cos(0 - θ)

We then use the general identity cos (A - B) = cos A cos B + sin A sin B to expand cos(0 - θ) as follows:

cos(- θ) = cos(0 - θ) = cos 0 cos θ + sin 0 sin θ

Use cos 0 = 1 and sin 0 = 0 to simplify the above to

= cos θ)

- We first write the left side of the given identity to verify sin( - θ) = - sin θ as follows:

sin( - θ) = sin (0 - θ)

We then use the general identity sin(A - B) = sin A cos B - cos A sin B to expand sin(0 - θ) as follows:

sin( - θ) = sin(0 - θ) = sin 0 cos θ - cos 0 sin θ

Use sin 0 = 0 and cos 0 = 1 to simplify the above to

= - sin θ

- Use the general identity cos (A - B) = cos A cos B + sin A sin B to expand cos(π/2 - θ) as follows:

cos(π/2 - θ) = cos π/2 cos θ + sin π/2 sin θ

Use cos π/2 = 0 and sin π/2 = 1 to simplify the above to

= sin θ)

- Use the general identity sin(A - B) = sin A cos B - cos A sin B to expand sin(π/2 - θ) as follows:

sin(π/2- θ) = sin π/2 cos θ - cos π/2 sin θ

Use sin π/2 = 1 and cos π/2 = 0 to simplify the above to

= cos θ

### More References and links

Trigonometric IdentitiesVerify Trigonometric Identities

Questions on Unit Circles in Trigonometry

HTML5 Applet to Explore the Unit Circle and Trigonometric Functions

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