Calculators either use the Taylor Series for $\sin / \cos$ or the CORDIC algorithm. A lot of information is available on Taylor Series, so I'll explain CORDIC instead.
The input required is a number in radians $\theta$, which is between $-\pi / 2$ and $\pi / 2$ (from this, we can get all of the other angles).
First, we must create a table of $\arctan 2^{-k}$ for $k=0,1,2,\ldots, N-1$. This is usually precomputed using the Taylor Series and then included with the calculator. Let $t_i = \arctan 2^{-i}$.
Consider the point in the plane $(1, 0)$. Draw the unit circle. Now if we can somehow get the point to make an angle $\theta$ with the $x$-axis, then the $x$ coordinate is the $\cos \theta$ and the $y$-coordinate is the $\sin \theta$.
Now we need to somehow get the point to have angle $\theta$. Let's do that now.
Consider three sequences $\{ x_i, y_i, z_i \}$. $z_i$ will tell us which way to rotate the point (counter-clockwise or clockwise). $x_i$ and $y_i$ are the coordinates of the point after the $i$th rotation.
Let $z_0 = \theta$, $x_0 = 1/A_{40} \approx 0.607252935008881 $, $y_0 = 0$. $A_{40}$ is a constant, and we use $40$ because we have $40$ iterations, which will give us $10$ decimal digits of accuracy. This constant is also precomputed1.
Now let:
$$ z_{i+1} = z_i - d_i t_i $$
$$ x_{i+1} = x_i - y_i d_i 2^{-i} $$
$$ y_i = y_i + x_i d_i 2^{-i} $$
$$ d_i = \text{1 if } z_i \ge 0 \text{ and -1 otherwise}$$
From this, it can be shown that $x_N$ and $y_N$ eventually become $\cos \theta$ and $\sin \theta$, respectively.
1: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$
One defines the circular trigonometric functions in three stages.
Stage 1: This uses actual triangles. Then all the lengths are positive, and the ratios are of positive numbers, but angles only make sense in the interval $[0,\pi/2]$ radians. In this setting, we can place a right triangle so that one vertex is at the origin of our coordinate system, the leg it is on runs along the $x$-axis to the right angle, then the other leg runs parallel to the $y$-axis to the other vertex, and the hypotenuse is a line segment from the origin to this last vertex, lying entirely in the first quadrant. (Or, in the degenerate case where the second leg is zero, on the $x$-axis, which is not part of the first quadrant.) In this stage, we say things like "the sine of an angle is the ratio of the opposite to the hypotenuse".
Stage 2: We allow the hypotenuse to lie in any quadrant. Then we still construct right triangles ("reference triangles") in "standard position" as: put one of the vertices not at the right angle at the origin, put one leg along the $x$-axis, turn at right angles from the $x$-axis and run parallel to the $y$-axis for the other leg. This allows positive and negative values for the $x$- and $y$-coordinates of the hypotenuse and hence for the signed lengths of the first and second legs. Then we say things like "the sine of the angle is the ratio of the $y$-coordinate to the hypotenuse". (The length of the hypotenuse is always positive because it is always a length, not a coordinate.)
Stage 3: We generalize to angles less than $0$ and greater than $2\pi$ radians. This is where we get "coterminal angles". Everything still happens with right triangles, signed leg lengths, and on one copy of the coordinate plane, but we imagine that we have spun our angle around the origin many, many times in either direction. (This more accurately models real rotating systems -- a wheel doesn't rotate exactly once and then stop...)
Having made these generalizations, we are then ready to go back and apply what we have learned to non-right (degenerate, acute, and obtuse) triangles. Your language suggests you have a gap between Stage 1 and this generalization.
Best Answer
This depends on your definition of $\sin$. It is not abnormal to define $\sin$ so that your equation holds. So either it is by definition, or you probably have to use some simple formula, like:
the formula $\sin (180 - x) = \sin(x)$. So $$ \frac{M'P'}{OP'} = \sin M'OP' = \sin (180 - M'OP') = \sin(90 + \theta). $$