How does the unit circle relate to triangles with a hypotenuse larger than 1?

Question

Why is a unit circle representative of all triangles, even triangles larger than the circle itself? I understand that $\sin\theta= Y$ and $\cos\theta = X$, but that is only because the hypotenuse is $1$. What if the hypotenuse is $2$ for example? How is $\sin\theta = Y$ still? How does the unit circle form a basis for all angles?

Answer 1

If a triangle has the same proportions as one that fits a unit circle, it is called similar. Under these conditions, if you take the ratios of any $2$ sides, you will come up with the same $number(s)$ which mean the same angle(s) as though it fit the unit circle.

Answer 2

For a directed angle $\theta$ with vertex at the origin and initial side on the positive $x$-axis, the point at which the terminal side of the angle $\theta$ intersects the unit circle $x^2 + y^2 = 1$ is defined to be $(\cos\theta, \sin\theta)$.

If the terminal side of angle $\theta$ intersects the circle $x^2 + y^2 = r^2$ at the point $(x, y)$, then by similar triangles, \begin{align*} \frac{|x|}{r} & = \frac{|\cos\theta|}{1} = |\cos\theta|\\ \frac{|y|}{r} & = \frac{|\sin\theta|}{1} = |\sin\theta| \end{align*} Moreover, since $r > 0$, the sign of $x$ agrees with the sign of $\cos\theta$ and the sign of $y$ agrees with $\sin\theta$. Therefore, \begin{align*} \cos\theta & = \frac{x}{r}\\ \sin\theta & = \frac{y}{r} \end{align*} Consequently, \begin{align*} \tan\theta & = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}\\ \csc\theta & = \frac{1}{\sin\theta} = \frac{r}{y}\\ \sec\theta & = \frac{1}{\cos\theta} = \frac{r}{x}\\ \cot\theta & = \frac{\cos\theta}{\sin\theta} = \frac{x}{y} \end{align*}