# Why does cos θ, which normally equals x, become 0 when you evaluate this function

1. Jul 11, 2010

### land_of_ice

Why does cos θ, which normally equals x, become 0 when you evaluate this function for π/r radians?
(π/r means pi over r radians)
Why does cos θ, which normally equals x, become 0 when you evaluate this function for π/r radians?
It says that the point on the circle for this question is (0,1) so zero = x and y = 1 here.

2. Jul 11, 2010

### Mentallic

For $$x=cos\theta$$, when x=0, $$\theta=\pi/2$$ (it can also be $$-\pi/2,3\pi/2$$ etc. but I don't think you need to deal with that right now).

When you're looking at the circle and any point on the circle (x,y) remember that $$cos\theta$$ is adjacent/hypotenuse or x/1=x. So since the point on the circle is (0,1), the x value is 0 so $$cos(\pi/2)=0$$