# How do you draw 45^circ and -45^circ in standard position and then show that sin(-45^circ)=-sin45^circ?

May 16, 2017

On the unit circle you can see that the two segments corresponding to the $\sin$ for the two angles and on the $\sin$ graph you can see the corresponding values where:
$\sin \left(- {45}^{\circ}\right) = - \sin \left({45}^{\circ}\right) = - 0.707$