Let s : \{0, \frac{π}{6} , \frac{π}{4} , \frac{π}{3} , \frac{π}{2} \} → R, s(x) = sin(x). Note I have chosen the domain of s to be \{0, \frac{π}{6} , \frac{π}{4} , \frac{π}{3} , \frac{π}{2} \}. Hence the image of s is \{0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1\}.

Observe \{0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1\} = \{0, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\}. Note 0 < \frac{π}{6} < \frac{π}{4} < \frac{π}{3} < \frac{π}{2} . Note 0 < \frac{1}{2} < \frac{\sqrt{2}}{2} < \frac{\sqrt{3}}{2} < 1. Since sin is an increasing function on the interval [0, \frac{π}{2} ], we know that sin(0) = 0, sin(\frac{π}{6} ) = \frac{1}{2}, sin(\frac{π}{4} ) = \frac{\sqrt{2}}{2}, sin(\frac{π}{3} ) = \frac{\sqrt{3}}{2}, sin(\frac{π}{2} ) = 1.

Note \{0 + \frac{π}{2} , \frac{π}{6} + \frac{π}{2} , \frac{π}{4} + \frac{π}{2} , \frac{π}{3} + \frac{π}{2} , \frac{π}{2} + \frac{π}{2} \} = \{\frac{π}{2} , \frac{2π}{3} , \frac{3π}{4} , \frac{5π}{6} , π\} Let S : \{\frac{π}{2} , \frac{2π}{3} , \frac{3π}{4} , \frac{5π}{6} , π\} → R, S(x) = sin(x). Note the image of S is \{0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1\}. Since sin is a decreasing function on the interval [0, \frac{π}{2} ], we know that sin(\frac{π}{2} ) = 1, sin(\frac{2π}{3} ) = \frac{\sqrt{3}}{2}, sin(\frac{3π}{4} ) = \frac{\sqrt{2}}{2}, sin(\frac{5π}{6} ) = \frac{1}{2}, sin(π) = 0.

We can similarly determine other values of the sin function as well as values of the cosine function.