Trigonometric Ratios for angles greater than 90 degrees and the Unit Circle I am confused about the Unit Circle explanation for the trigonometric ratios for angles greater than 90 degrees.
It seems that for the first (top right) quadrant, $\sin(\theta)$ is equivalent to the y-coordinate, because


*

*$\sin(\theta)$ = opposite / hypotenuse

*$\sin(\theta)$ = opposite       [hypotenuse is 1]

*$\sin(\theta)$ = y-coordinate   [length of opposite side is the
y-coordinate]


then, as theta extends in counter-clockwise manner to the top left quadrant, it is assumed that $\sin(\theta)$ is the still the value of the y-coordinate. 
From what I understand, the basis for this is that because $\sin(\theta)$ is equivalent to the y-coordinate in the first quadrant, this extends to all quadrants. But this does not make sense to me because the $\sin(\theta)$ = o/h equation was applicable in the first quadrant but not in the others.
It seems to me that there are two definitions for the sine function:


*

*The relationship between the opposite side and the hypotenuse for an
acute angle in a right-angled triangle

*The y-coordinate of a point along the unit circle, with angle theta (counter-clockwise from the x-axis)
The co-existence of these two definitions is making it confusing for me as it is not clear to me how we can get from the first to the second. 
 A: Extending to other quadrants
If trigonometric identity formulas hold true so long as we are within the first quadrant, shouldn't we be able to extend our trig functions to all quadrants in this manner?
Mainly, you'd want the following identities.
$$\cos(\alpha+\theta)=\cos(\alpha)\cos(\theta)-\sin(\alpha)\sin(\theta)$$
$$\sin(\alpha+\theta)=\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)$$
Since we know $\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{\sqrt2}2$, we can derive $\sin\left(\frac\pi2\right)=1$, $\cos\left(\frac\pi2\right)=0$, and so forth.
Similarly, we can derive what $\sin(-\theta)$ is by using $\sin(\theta-\theta),$ $\cos(\theta-\theta)$, and $\cos^2+\sin^2=1$, the Pythagorean identity.  Two equations, and you can solve for $\cos(-\theta),\sin(-\theta)$ by substitution.  Use Pythagorean identity for simplifying the answer.
It just happens to be that on the unit circle, the hypotenuse is by definition $1$, so...
$$\sin=\frac{\text{opp}}{\text{hyp}}=\text{opp}=y$$
$$\cos=\frac{\text{adj}}{\text{hyp}}=\text{adj}=x$$
And since both this definition and the one above derived by trig identities come out the same for $\theta>90\deg$ or $\pi$, then both are equally correct.
A: For all angles, on the unit circle $\cos\theta=x,\sin\theta=y$. So
$$\begin{matrix}&0°&&90°&&180°&&270°&&360°\\\cos&&+&&-&&-&&+\\\sin&&+&&+&&-&&-\end{matrix}$$
