Shouldn't the projection on the y-axis, of a radial vector in a circle making an angle theta with the x axis, also be the radius? 
I don't understand this diagram... the projection is meant to be a shadow, right? So shouldn't the shadow of the radial vector as shown in the circle on the y-axis, be the entire y-axis (since the radius is the same, everywhere)? Why is the projection shown to be smaller than the radius?
For instance, let the origin be O. So shouldn't the projection of the phasor, on y-axis, be OA and not Asin(theta)?
 A: If think that shadow is not going to be the proper term here, since if you want to talk about shadows you have to say from where the light rays are coming. What we're using here is the idea of orthogonal projection in this, if you want still to think about shadows, the light rays come orthogonal to one of the axis.
For example, if you want to find the orthogonal projection of the radius on the $y$ axis, your light rays will come at $90^\circ$ on the $y$ axis. In this case is clear that the whole $y$ axis will be covered by the shadow of the radius only when the radius is exactly aligned with the $y$ axis. Same goes for the projection on the $x$ axis where now the light rays come at $90^\circ$ to the $x$ axis. One of this "light rays" is even shown in the circle by the dashed lines.
It should be clear with this discussion that different angles that the radius makes with the $x$ axis, will have different length for the orthogonal projection on the two axis. The idea of using orthogonal projection boils down to the possibility of defining, geometrically and analytically, the $\sin\theta$ and $\cos\theta$ functions which are defined on right triangles. The right triangle in the figure is the one which has the origin, the tip of the radius and the  point o the projection on the $x$ axis as vertices.
