How to define trigonometry functions in a non unit circle? I was reading trigonometry functions and how they are defined for non acute angles in a unit circle. My question is that in the unit circle definition the sine of the ray is said to be it's y axis coordinate. What if the circle is not unitary? Why do we even use unit circle? How to define sine in a circle which is not unitary but lets say its radius is r = 2 units. How to define sine in such scenario and show that it will still remain same no matter whatever the radius is
Thanks
 A: Let $P'(x',y')$ be a point on a circle centred at the origin $O,$ and $\theta$ be the angle swept out by $OP'$ as it rotates anticlockwise from the positive $x$-axis.
By definition, $$\cos\theta:=\frac {x'}{OP'}\,;\\\sin\theta:=\frac {y'}{OP'}\,;\\\tan\theta:=\frac {y'}{x'}\,.$$
When the circle is scaled into a unit circle with new general point $P(x,y),$ due to triangle similarity, these trigonometric ratios are preserved. In other words, the following set of definitions is equivalent to the above: $$\cos\theta:=x\,;\\\sin\theta:=y\,;\\\tan\theta:=\frac y x.$$
The unit-circle definition of the trigonometric functions, rather than the general-circle definition, is the standard/conventional presentation simply because it is simpler. But they are equivalent to each other.
A: Why do we even use unit circle? 
As @RyanG has indicated, a radius of 1 unit is as much a convenience as anything else.
The reason for defining trig functions in terms of a unit circle is that it allows us to move away from ratio  definitions based on right-triangles, and this in turn allows us to think about non-acute angles. For example, in a triangle based definition of $sin\ \theta$, what does $sin\ 90°$ mean? Or $sin\ 0$. In both cases there is no longer a triangle. And then what about right-triangles that have the $\theta$ angle larger than 90°?
So, the unit circle let's us define $sin\ \theta$ etc for any angle in a simple and consistent way. And the beauty of it is that it preserves the ratio definitions by incorporating a right-triangle in the first quadrant if we choose to see it.
