+79 I know how to prove this equivalent expression through an identity but how do i show this on a unit circle visually?
+9481 First let's look at the two angles drawn on a unit circle separately:
Now....let's look at each triangle...
Look at the first triangle. Since there are pi radians in every triangle
∠ABC = pi - pi/2 - θ = pi/2 - θ
Look at the second triangle. Since there are 2pi radians in every circle
∠DEF = 2pi - (θ + 3pi/2) = 2pi - 3pi/2 - θ = pi/2 - θ
And both triangles have a side of length 1 . So....
△ABC is congruent to △DEF by the Ange-Angle-Side rule. And
side EF = side BC
cos( θ + 3pi/2 ) = sin θ
I hope this made sense! 
+9481 First let's look at the two angles drawn on a unit circle separately:
Now....let's look at each triangle...
Look at the first triangle. Since there are pi radians in every triangle
∠ABC = pi - pi/2 - θ = pi/2 - θ
Look at the second triangle. Since there are 2pi radians in every circle
∠DEF = 2pi - (θ + 3pi/2) = 2pi - 3pi/2 - θ = pi/2 - θ
And both triangles have a side of length 1 . So....
△ABC is congruent to △DEF by the Ange-Angle-Side rule. And
side EF = side BC
cos( θ + 3pi/2 ) = sin θ
I hope this made sense!
