 # Special Angles in the Unit Circle

How to use the symmetry of a unit circle to find values of sine and cosine functions to the angles related by symmetry to the special angles π/6, π/4 and π/3?

## Special Angles in the Unit Circle

Any line through the center of a circle is a line of symmetry. The center of the circle is point of symmetry. We now consider a unit circle with center at the origin of a system of x and y axes. We are here interested in the symmetries with respect to the origin, to the x-axis and to the y-axis.

Knowing the values of the sine and cosine of the angles in the first quadrant, it is easier to use the symmetry of the unit circle to obtain the sine and cosine of the angles in the other quadrants.

The unit circle below shows the values of the cosine and sine functions (coordinates in blue, with the x-coordinate being the cosine and the y-coordinate is the sine) for the special angles:
0, π/6 (30 �), π/4 (45 �), π/3 (60 �), π/2 (90 �), 2π/3 (120 �), 5π/4 (135 �) ...

### Example 1

Consider the angles: 2π/3, 4π/3 and 5π/3. They all have a relationship with π/3:
2π/3 = π - π/3 (symmetry with respect to y axis, see unit circle above)
Hence:
cos(2π/3) = - cos(π/3) = -1/2 and sin(2π/3) = sin(π/3) = √3/2
4π/3 = π + π/3 (symmetry with respect to origin)
Hence:
cos(4π/3) = - cos(π/3) = -1/2 and sin(4π/3) = - sin(π/3) = - √3/2
5π/3 = 2π - π/3 (symmetry with respect to x axis)
Hence:
cos(5π/3) = cos(π/3) = 1/2 and sin(5π/3) = - sin(π/3) = - √3/2

NOTE that the sine and cosine of π/3, 2π/3, 4π/3 and 5π/3 are all equal in absolute value and their signs depend on the quadrant of their terminal sides. Therefore knowing that cos(π/3) = 1/2 and sin(π/3) = √3 / 2, the sine and cosine of 2π/3, 4π/3 and 5π/3 are easily obtained from cos(π/3) = 1/2 and sin(π/3) = √3 / 2 by using the symmetry of the unit circle and changing the sign.

### Example 2

Consider the angles: 5π/6, 7π/6 and 11π/6. They all have a relationship with π/6:
5π/6 = π - π/6 (symmetry with respect to y axis)
Hence:
cos(5π/6) = - cos(π/6) = -√3/2 and sin(5π/6) = sin(π/6) = 1/2
7π/6 = π + π/6 (symmetry with respect to origin)
Hence:
cos(7π/6) = - cos(π/6) = -√3/2 and sin(7π/6) = - sin(π/6) = -1/2
11π/6 = 2π - π/6 (symmetry with respect to x axis)
Hence:
cos(11π/6) = cos(π/6) = √3/2 and sin(11π/6) = - sin(π/6) = - 1/2
NOTE that the sine and cosine of π/6, 5π/6, 7π/6 and 11π/6 are all equal in absolute value and their signs depend on the quadrant of their terminal sides.

### Example 3

Consider the angles: 3π/4, 5π/4 and 7π/4. They all have a relationship with π/4:
3π/4 = π - π/4 (symmetry with respect to y axis)
Hence:
cos(3π/4) = - cos(π/4) = -√2/2 and sin(3π/4) = sin(π/4) = √2/2
5π/4 = π + π/4 (symmetry with respect to origin)
Hence:
cos(5π/4) = - cos(π/4) = -√2/2 and sin(5π/4) = - sin(π/4) = - √2/2
7π/4 = 2π - π/4 (symmetry with respect to x axis)
Hence:
cos(7π/4) = cos(π/4) = √2/2 and sin(7π/4) = - sin(π/4) = - √2/2
NOTE that the sine and cosine of π/4, 3π/4, 5π/4 and 7π/4 are all equal in absolute value and their signs depend on the quadrant of their terminal sides.