Using unit circle to explain $\cos(0) = 1$ and $\sin(90) = 1$ We have been taught $\cos(0) = 1$  and $\sin(90) = 1$.
But, how do I visualize these angles on the unit circle? 


 A: Suppose you have an angle $\theta$ in the unit circle. Then, the functions $\cos\theta$ and $\sin\theta$ represent the $x$ and $y$ components respectively. See the image below.

Plugging the values $0$ and $90$ you will get it.
A: Here is one way to do it:

Here is another one regarding the sine function:

A: $\theta$ is the angle measured from the $x$-axis in a counter clockwise direction.
We know that 
$$\sin\theta=\frac{OPPOSITE}{HYPOTENUSE}$$
$$\cos\theta=\frac{ADJACENT}{HYPOTENUSE}$$

For a unit circle $r=1$
$$x=\cos \theta$$
$$y=\sin \theta$$
When $\theta=0$
The length is of $ADJACENT=HYPOTENUSE$
Adjacent means horizontal distance measure from the origin to the point.
Meaning if we ratio them 
$$\cos 0=1$$
Opposite is the vertical distance. We don't have any vertical distance when $\theta=0$
$$\sin 0=0$$
Let's us see for the opposite length
As $\theta=\frac{\pi}{2}$
We see that the horizontal distance is disappearing for $\theta=\frac{\pi}{2}$
and the vertical distance is increasing.At $\theta=\frac{\pi}{2}$  the vertical distance equals to the hypotenuse.
Taking 
$$\sin\frac{\pi}{2}=1$$
$$\cos\frac{\pi}{2}=0$$
A: The standard way is with the right angle leg along the positive $x$-axis, so that e.g. acute angles are exactly the ones with the other angle leg within the first quadrant.
Then look at the point where the second angle leg intersects the unit circle. It will have coordinates $(\cos x, \sin x)$.
For instance, $\cos 0^\circ = 1$ because the angle $0$ has its second leg along the positive $x$-axis, and it intersects the unit circle in $(\cos0^\circ, \sin 0^\circ)$. But that point is also $(1,0)$, since it's along the positive $x$-axis. Thus the two sets of coordinates are equal, and we get $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$. The case for $\cos90^\circ = 0$ and $\sin 90^\circ$ is sone similarily, but this time the second angle leg goes straight up along the positive $y$-axis.
