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.
Best Answer
"The definition of sin of and angle is opposite/hypotenuse" - well that only works for a right-angled triangle, and is the beginning of the definition of the sine function.
In order to extend the definition draw a unit circle with centre at the origin. Measure the angle counterclockwise from the positive $x$-axis and take a point $(x,y)$ on the circle. The sine of the angle (equivalent to opposite/hypotenuse in the first quadrant, with the hypotenuse made equal to $1$) is $y$ and the cosine of the angle is $x$. You can go round the circle more than once, so you can see that the functions are periodic.
This is why the functions are sometimes known as circular functions and underlies why they come in so surprisingly useful.
In case you are interested ...
In more advanced work the sine function is sometimes defined very differently, with the angle measured in radians ($2\pi$ radians $=360^{\circ}$). Then $$\sin x = \sum_{r=1}^\infty (-1)^{r-1}\frac{x^{2r-1}}{(2r-1)!}=\frac {e^{ix}-e^{-ix}}{2i}$$This can be applied in more general circumstances still, and the series is convenient because it converges rapidly and enables the sine function to be computed accurately for practical use. The first few terms are $$x-\frac {x^3}6+\frac {x^5}{120}-\frac {x^7}{5040}+\dots$$