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: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$
Best Answer
TL;DR: What you did was completely correct and does not rely on a right-triangle definition of the sine function.
You can use triangles to help you find values of the sine function while still using the unit circle to define the sine function.
For the angle $\frac{7\pi}{6},$ you have already shown how to use the unit-circle definition to demonstrate that $\sin\left(\frac{7\pi}{6}\right)$ is the $y$-coordinate of the end of a radial segment of the unit circle, making an angle $\frac\pi6$ below the negative $y$-axis.
As you observed, if you drop a perpendicular from the end of that radial segment to the $y$-axis, you form a $30$-$60$-$90$-degree right triangle. You are not using that triangle to define the sine, but you happen to know that the leg opposite the $30$-degree angle is half the length of the hypotenuse, and since the hypotenuse is the radial segment, already known to have length $1,$ the length of the opposite segment is $\frac12.$ That is, the end of the radius is $\frac12$ unit directly below the $y$-axis, so it has $y$-coordinate $-\frac12,$ and therefore, according to the unit-circle definition of the sine function, $$\sin\left(\frac{7\pi}{6}\right) = -\frac12.$$
If you did not already know the ratios of the sides of a $30$-$60$-$90$-degree right triangle, then (as observed already in another answer) you could derive them by cutting an equilateral triangle in half. Most angles do not have a geometric construction that gives us exact values of their trigonometric functions, but it happens that the angle $\frac\pi6$ does have such a construction and it gives a particularly simple result, so we have schoolchildren memorize that result soon after introducing the idea of measuring the angles of a triangle.
Indeed, for almost any other angle there is not such a simple way to find the sine; you can construct a triangle within the unit circle whose legs are the same magnitude as the angle's sine and cosine, but you will not be able to write the exact values of that sine and cosine as simple ratios of integers. You cannot even write those values in any finite expression involving integers, addition, subtraction, multiplication, and division. That is why we usually settle for an approximate value expressed as a decimal fraction.