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: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$
Square everything:
$$\sin^2 y =4\cos^2(y/2)\sin^2(y/2)=4(1-\sin^2 (y/2))\sin^2(y/2)=4\sin^2(y/2)-4\sin^4(y/2)$$
So
$$
1-\sin^2 y = 1 - 4 \sin^2(y/2)+4\sin^4(y/2)=(1-2 \sin^2 (y/2))^2
$$
That is,
$$
\cos^2 y = (1-2 \sin^2 (y/2))^2$$
Unfortunately this only gives you
$$
\pm \cos y = 1 - 2 \sin^2(y/2)
$$
(the cost of squaring in the first place). But since both sides are smooth, we have to make the same choice of sign everywhere — and at $0$ it clearly is the positive one. So we have showed that:
$$
\cos y = 1 - 2 \sin^2(y/2)
$$
and so
$$
\sin (y/2)=\pm \sqrt{\frac{1-\cos y}{2}}
$$
Best Answer
You can find these proofs on Wikipedia.