How does the unit circle definition of trigonometric functions scale? The unit circle definition assumes that we are using a circle with unit radius. However we are also using the trigonometric functions out of this context. How does that definition scale in order to hold for circles of any radius?
 A: The answer is that calling a circle a unit circle simply means that the length unit is identical to the length of the radius. Of course, they are many possible length units such as feet and meters. It is always possible to convert linear lengths measured using different units by multiplying by a conversion factor depending on both units. Thus, there is no loss of generality in restricting to a unit circle because we can choose the unit.
Also, you should notice that trigonometric functions are defined as quotients (or ratios) of line segment lengths and thus do not depend on the measuring unit used. We are implicitly using the scale invariance of Euclidean geometry which allows for the existence of similar triangles. This property fails for other geometries such elliptic or hyperbolic geometry. There the trigonometric relations are more complicated, but reduce to Eucldiean relations in the limit of zero curvature.
A: Well, $x^2+y^2=1$ with $r_1=1$ , the sine of the angle that the radius makes in first quadrant is $\frac{opposite}{r_1}$ with the opposite being f(x) so, $y=\pm \sqrt{1-x^2}$ so it's $\frac{\pm \sqrt{1-x^2}}{1}$, in another circle, let's say $x^2+y^2=4 $ ,so $\frac{opposite}{r_2}$ is $\frac{\pm \sqrt{4-x^2}}{2}$, let's see,  dividing the first by the second we get
$$\frac{2 \times(\pm \sqrt{1-x^2)}}{\sqrt{4-x^2}}$$
getting that 2 inside the radical as $4$, we get
$$\frac{\pm\sqrt{4-4x^2}}{\pm\sqrt{4-4x^2}} =1$$
no matter how you scale your circle, trigonometric functions will give the same result, you can do the same with other functions.
A: Suppose you have a circle of radius $r$ centred at the origin. This is given by $$x^2+y^2=r^2.$$ But this can be put in the form $$\left(\frac xr\right)^2+\left(\frac yr\right)^2=1,$$ by dividing through by $r^2.$ This also is the unit circle $$x'^2+y'^2=1,$$ with $x'=x/r$ and $y'=y/r.$ Thus, the circular functions are well-defined; they don't depend on the radius of the circle to use, as shown above. The explanation also shows why they are sometimes called ratios.
