I got a question about trig on the unit circle The unit circle is defined to be $x^2 + y^2 = 1$. Makes sense. Its an equation of a circle. Now from here if we think about cosine as the $x$ value and sine as the $y$ value then we get a trig identity most of use know. There was something about this that always bothered me. And maybe this sounds crazy, but why can we use trigonometric functions to define a point on the unit circle?
$$\cos x  = \frac{ \text{adj} }{ \text{hyp} } \quad \text{and} \quad \sin x  = \frac{ \text{opp} }{ \text{hyp} }$$
So I can see why $\cos(x)$ would be defined in terms of an $x$ value since the adjacent side is the leg of the triangle that is on the $x$ axis but where is the hypotenuse going at in this argument? Same for sine. 
Sorry I over think things but I like to really know what something is saying.  
 A: If we see $( \cos \theta, \sin \theta)$ as a point in $\mathbb{R}^{2}$, then we know that the distance from $(0, 0)$ to $( \sin \theta, \cos \theta)$ is (according to the Pythagorean Theorem)
\begin{align*}
\sqrt{(\cos \theta - 0)^{2} + (\sin \theta - 0)^{2}} & = \sqrt{(\cos \theta)^{2} + (\sin \theta)^{2}} \\
& = \sqrt{1} \\
& = 1
\end{align*}
Of course, the definition of a circle with center $(h, k)$ and radius $r$ is the set of points which are a distance $r$ from the center $(h, k)$. We see that every point $(\cos \theta, \sin \theta)$ is a distance $1$ from a center $(0, 0)$. That is to say, $(h, k) = (0, 0)$ and $r = 1$.
A: Another way to say this is that the map $\theta \mapsto (\cos \theta, \sin \theta)$ parameterizes the unit circle. (The fact that the image of this map is contained in the unit circle is exactly a consequence of the identity you mention.)
Geometrically, the point corresponding to $\theta$ is the intersection of the unit circle with the ray that makes a (signed) angle $\theta$ with the positive $x$-axis, so that the hypotenuse is the line segment with endpoints this intersection and the origin. (As usual in the setting, we allow adj and opp to take on negative values according the quadrant in which the indicated ray sits.)
A: Please see this reference : 
in this please read section 4.14.. This is a very good source of basic trigonometry.
http://gujarat-education.gov.in/textbook/Images/11sem1/maths11-eng/chap4.pdf
