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 \( \dfrac{\pi}{6} \), \( \dfrac{\pi}{4} \), and \( \dfrac{\pi}{3} \) ?
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:
Angles in radians and degrees: \( 0 \), \( \dfrac{\pi}{6} \) (30°), \( \dfrac{\pi}{4} \) (45°), \( \dfrac{\pi}{3} \) (60°), \( \dfrac{\pi}{2} \) (90°), \( \dfrac{2\pi}{3} \) (120°), \( \dfrac{5\pi}{4} \) (135°) ...
Consider the angles: \( \dfrac{2\pi}{3} \), \( \dfrac{4\pi}{3} \), and \( \dfrac{5\pi}{3} \). They all relate to \( \dfrac{\pi}{3} \):
1 -Hence:
\[ \cos\left(\dfrac{2\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = -\dfrac{1}{2}, \quad \sin\left(\dfrac{2\pi}{3}\right) = \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \] 2 -Hence:
\[ \cos\left(\dfrac{4\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = -\dfrac{1}{2}, \quad \sin\left(\dfrac{4\pi}{3}\right) = -\sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} \] 3 -Hence:
\[ \cos\left(\dfrac{5\pi}{3}\right) = \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}, \quad \sin\left(\dfrac{5\pi}{3}\right) = -\sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2} \]Note: The sine and cosine of \( \dfrac{\pi}{3} \), \( \dfrac{2\pi}{3} \), \( \dfrac{4\pi}{3} \), and \( \dfrac{5\pi}{3} \) all have the same absolute values. The signs depend on the quadrant in which their terminal sides lie. Therefore, knowing that:
\[ \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}, \quad \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \]We can easily determine the sine and cosine of \( \dfrac{2\pi}{3} \), \( \dfrac{4\pi}{3} \), and \( \dfrac{5\pi}{3} \) by applying the symmetry of the unit circle and adjusting the signs appropriately.
Consider the angles: \( \dfrac{5\pi}{6} \), \( \dfrac{7\pi}{6} \), and \( \dfrac{11\pi}{6} \). They all have a relationship with \( \dfrac{\pi}{6} \):
1 -Hence:
\[ \cos\left(\dfrac{5\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}, \quad \sin\left(\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2} \] 2 -Hence:
\[ \cos\left(\dfrac{7\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}, \quad \sin\left(\dfrac{7\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2} \] 3 -Hence:
\[ \cos\left(\dfrac{11\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}, \quad \sin\left(\dfrac{11\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2} \]NOTE: The sine and cosine of \( \dfrac{\pi}{6} \), \( \dfrac{5\pi}{6} \), \( \dfrac{7\pi}{6} \), and \( \dfrac{11\pi}{6} \) all have equal absolute values. Their signs depend on the quadrant where the terminal sides of the angles lie.
Consider the angles \( \dfrac{3\pi}{4} \), \( \dfrac{5\pi}{4} \), and \( \dfrac{7\pi}{4} \). Each of these angles is related to \( \dfrac{\pi}{4} \) by symmetry:
1 -Hence:
\[ \cos\left(\dfrac{3\pi}{4}\right) = -\cos\left(\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}, \quad \sin\left(\dfrac{3\pi}{4}\right) = \sin\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \] 2 -Hence:
\[ \cos\left(\dfrac{5\pi}{4}\right) = -\cos\left(\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}, \quad \sin\left(\dfrac{5\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2} \] 3 -Hence:
\[ \cos\left(\dfrac{7\pi}{4}\right) = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}, \quad \sin\left(\dfrac{7\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2} \]Note: The sine and cosine values of the angles \( \dfrac{\pi}{4} \), \( \dfrac{3\pi}{4} \), \( \dfrac{5\pi}{4} \), and \( \dfrac{7\pi}{4} \) have the same absolute value. Their signs depend on the quadrant in which the terminal side of each angle lies.