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
A $30-60-90$ is one of the must basic triangles known in geometry and you are expected to understand and grasp it very easily.
In an equilateral triangle, angles are equal. As they add to $180$ then angles are are all $\frac {180}{3} = 60$. And as the sides are equal all sides are equal. (see image)
So that is a $60-60-60$ triangle.
If we draw the perpendicular bisector of the the base you cut the triangle into two congruent triangles. (see image)
$\triangle CAD$ (and $\triangle CBD$) are two congruent triangles that have angles $30- 60-90$. As it is half of an equilateral triangle, the short leg, is $\frac 12$ of the side of the equilateral, while the hypotenuse is the full side of the equilateral. SO $AD$ (and $DB$) $= \frac 12$ and $AC$ (and $BC$) $= 1$.
And because it is a right triangle $AD^2 + DC^2 = AC^2$ or $(\frac 12)^2+ DC^2 = 1$ so $DC^2 = 1 - (\frac 12)^2=1-\frac 14 = \frac 34$ so $DC = \sqrt {\frac 34} = \frac {\sqrt 3}{\sqrt 4} = \frac {\sqrt 3}2$.
And like all right triangles:
We can get that $AD = \cos 60^\circ *hypotenuse$ so $\cos 60^\circ = \frac{1}{2}$ and $DC = \sin 60^\circ*hypotenuse$ so $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
You can always do this for any right triangle. $AD = \cos \angle CAD*AC$ and $DC = \sin \angle CAD*AC$. Always!
But this is one of the BASIC possible right triangles were the values of $\sin 60^{\circ}$ and $\cos 60^{\circ}$ should follow immediately and !!!VERY!!! easily directly from geometry.