Answer
$\sin{(45^\circ)} =-\sin{(-45^\circ)}= \dfrac{\sqrt{2}}{2}$
Work Step by Step
The coordinates of points on the terminal side of $45^{\circ}$ $(x_1,y_1)$ can be given by $(a,a)$, where $a$ is a positive number.
Choosing $a=1$ arbitrarily, the point is $(1,1)$.
$r_1 = \sqrt{(x_1)^2+(y_1)^2} = \sqrt{1^2+1^2} = \sqrt{2}$
$\sin{45^\circ} = \dfrac{y_1}{r_1} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$
The coordinates of points on the terminal side of $-45^{\circ}$ $(x_1,-y_1)$ can be given by $(a,-a)$, where $a$ is a positive number.
Choosing $a=1$ arbitrarily, the point is $(1,-1)$.
$r_2 = \sqrt{(x_2)^2+(y_2)^2} = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$
$\sin{(-45^\circ)} = \dfrac{-y_1}{r_2} = -\dfrac{1}{\sqrt{2}} = -\dfrac{\sqrt{2}}{2}$
$\therefore \sin{(45^\circ)} =-\sin{(-45^\circ)}= \dfrac{\sqrt{2}}{2}$
