How are trigonometric ratios of specific angles derived? How do I derive the trigonometric ratios of angles like $0°,30°,45°,60°\text{and }90°$. Can I use the same method to find ratios of other angles too? 
 A: For your given angles, you need to examine special right triangles with sides $1,1,\sqrt2$ and $1,\sqrt3,2$. The first triangle has two $45$ angles (obviously) and the second one has $30,60$ in it (cut an equilateral in half to see why). By definition of sine and cosine from the old age, by placing these triangles with an hypotenuse set to $1$ in the unit circle, you find the trig ratios you are looking for 
A: First, recall that the sum of the three angles of a triangle is always $180^\circ$.  If you don't know how to prove that, look it up or post the question here.  It's something that zillions of students claim to know, but they have no idea why it's true, although it's not really hard to explain.
That implies that if a triangle has a $90^\circ$ angle and the other two angles both have the same measure, then they're both $45^\circ$.  A simple theorem of geometry tells us that if the measures of two angles of a triangle are the same, then the lengths of the sides opposite those two angles are the same.  That means this triangle has two sides of lengths that we can call $1$, and then the Pythagorean theorem tells us the hypotenuse has length $\sqrt 2$.  Then we can say
$$
\sin45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac 1 {\sqrt 2}.
$$
And similarly we can find the cosine, tangent, etc.
Next consider an equilateral triangle.  Call the lengths of the three sides $1$.  A simple theorem of geometry similar to the one mentioned above tells us that if it's equilateral then it's equi-angular: all three angle measures are equal.  Since they must add up to $180^\circ$, each must be $60^\circ$.
Call the three corners $A,B,C$.  Drop a perpendicular line from $A$ to the base $BC$.  It hits $BC$ exactly at the midpoint of $BC$.  Draw the picture and you'll see why that is so.  That means it splits the equilateral triangle into two right triangles.  The hypotenuse has length $1$ (refer to the picture you drew) and one leg, the one that is half of the base $BC$, has length $1/2$.  The Pythagorean theorem then tells us that the other leg has length $\dfrac{\sqrt 3} 2$.  Thus we get
$$
\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1/2} 1 = \frac 1 2
$$
and
$$
\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt 3/2} 1 = \frac {\sqrt 3} 2.
$$
See also How Ptolemy computed chords for methods as elementary as all this but far more complicated, that can be used to compute, with considerable work, things like $\sin29^\circ$, etc.  Nowadays more sophisticated methods are used but they are still quite labor-intensive, but that labor is done by computer.
