Deriving the sine or cosine of an arbitrary angle takes a bit more math than is covered here. However, deriving them for a few special angles is relatively straightforward.
Note first of all that a right-angled triangle contains two angles. Since all three angles (in any triangle) add up to 180°, the two sharp angles add up to 90°. It follows that if one of the angles is of A degrees, the other one (its "complementary angle") is (90°- A).
The sine and cosine were defined as the following ratios:
sin A = (side opposite to A)/(long side)
cos A = (side adjacent to A)/(long side)
Because the side opposite to A is the one adjacent to (90°- A), it follows that the sine of one angle is the cosine of the other, and vice versa:
sin A = a/c = cos (90° - A)
cos A = b/c = sin (90° - A)
This is a great help: deriving (for instance) the sine and cosine of 30°also gives us, as a bonus, the sine and cosine of 60°.
(1) A = 45°
If A = 45°, then also (90° - A) = 45°, and therefore
However, it was found earlier that for any angle A
2 sin2 45° = 1
and if SQRT means "square root of"
Pushing a button on your calculator gives
sin2 45° = 1/2
sin 45° = 0.707107... = cos 45°
Another way, slightly more transparent, is to write
sin2 45° = 1/2 = 2/4
The square root of 2 is 1.4142135..., dividing it in two recovers 0.707107 as before.
sin 45° = SQRT(2)/SQRT(4) = SQRT(2)/2
(2) A = 30°, (90° - A) = 60°
Consider the triangle PQR (drawing) with all three angles equal to 60°. By symmetry, all three sides are equal too (a more rigorous proof exists, but we skip it). Drop a line QS perpendicular to PR: it divides the big angle into two right-angled triangles with sharp angles of (30°, 60°), which is the kind we are interested in. By symmetry, the triangles are of equal size and shape ("congruent") and therefore (skipping another proof)
In the notation of the drawing
a = (1/2) c
a/c = 1/2 = sin 30° = cos 60°
Subtract 1/4 from both sides
cos2 30° = 3/4
cos 30° = SQRT(3)/ SQRT(4) = SQRT(3)/2 = 1.7320508/2
cos 30° = 0.8660254 = sin 60°
(3) A = 90° , (90° - A) = 0
It would be rather hard to draw a right-angled triangle with a second angle of 90° inside it, because the third angle then has to be 0°. But we can visualize this strange triangle as the limiting case of long skinny triangles with an angle A that is very steep and and its complement (90° - A) very small (drawing). In the limiting case
cos A = b/c = 0
1 = sin2A + cos2A = sin2A + 0
it follows that
sin2A = 1
sin A = 1
cos 90° = sin 0° = 0
sin 90° = cos 0° = 1