What you really want for the Riemann sum of $\int_a^b \sin x \, dx$ is
to take $\sin x$ at $n$ uniform steps within the interval $[a,b].$
So you want $x_i$ to be something like $a + i\Delta x,$
or even better,
$$x_i = a + i\Delta x - \tfrac12\Delta x \quad \text{where}
\quad \Delta x = \frac{b-a}{n},$$
so that $x_1 = a + \tfrac12\Delta x$ and $x_n = b - \tfrac12\Delta x.$
If you define $x_i$ and $\Delta x$ in that way,
the summation you're looking for is
$$ \sum_{i=1}^n \sin\left(x_i\right) \Delta x. $$
Here's a handy trigonometric identity you can use for this problem:
$$\sin A \sin B = \tfrac12 \cos(A−B) − \tfrac12 \cos(A+B).$$
We can apply it as follows: let $A = x_i$ and let $B = \frac12 \Delta x.$
Then
$$\sin (x_i) \sin \left(\tfrac12 \Delta x\right) =
\tfrac12 \cos\left(x_i - \tfrac12 \Delta x\right)
− \tfrac12 \cos\left(x_i + \tfrac12 \Delta x\right). \tag1$$
Since you want to compute a summation over $\sin (x_i) \Delta x$ rather than
$\sin (x_i) \sin \left(\tfrac12 \Delta x\right),$
let's multiply both sides of equation$\ (1)$ by
$\dfrac{\Delta x}{\sin \left(\tfrac12 \Delta x\right)}$ to obtain
$$\sin (x_i) \Delta x =
\frac{\Delta x}{2\sin\left(\tfrac12\Delta x\right)}
\cos\left(x_i - \tfrac12\Delta x\right)
− \frac{\Delta x}{2\sin \left(\tfrac12 \Delta x\right)}
\cos\left(x_i + \tfrac12 \Delta x\right). $$
The big fraction on the right side of this equation will occur at least once in every
equation we write after this; to reduce clutter, let
$k = \dfrac{\Delta x}{2\sin\left(\tfrac12\Delta x\right)}$ so that we can write
$$\sin (x_i) \Delta x = k \cos\left(x_i - \tfrac12\Delta x\right)
− k \cos\left(x_i + \tfrac12 \Delta x\right).$$
Now let's take a look at the next term in the summation, $\sin (x_{i+1}) \Delta x.$
Since $x_{i+1} = x_i + \Delta x,$
$$\begin{eqnarray}
\sin (x_{i+1}) \Delta x
&=& k \cos\left(x_{i+1} - \tfrac12 \Delta x\right)
− k \cos\left(x_{i+1} + \tfrac12 \Delta x\right)\\
&=& k \cos\left(x_i + \tfrac12 \Delta x\right)
− k \cos\left(x_{i+1} + \tfrac12 \Delta x\right).
\end{eqnarray}$$
Now notice what happens if we add $\sin (x_{i+1})$ to $\sin (x_i).$
The two terms $k \cos\left(x_i + \tfrac12 \Delta x\right)$ cancel,
and we're left with
$$\sin (x_i) \Delta x + \sin (x_{i+1}) \Delta x =
k \cos\left(x_i - \tfrac12 \Delta x\right)
− k \cos\left(x_{i+1} + \tfrac12 \Delta x\right). $$
This is what we call a "telescoping sum," and it simplifies the summation wonderfully:
if we write each term $\sin (x_i) \Delta x$ as a difference of two cosines
(times a constant), as in the equations above, a pair of cosines cancel each other
each time we add another term to the sum, and we're left with just a difference
of two cosines (times a constant) at the end.
More formally,
$$\begin{eqnarray}
\sum_{i=1}^n \sin\left(x_i\right) \Delta x
&=& \sum_{i=1}^n \left( k \cos\left(x_i - \tfrac12 \Delta x\right)
− k \cos\left(x_i + \tfrac12 \Delta x\right) \right) \\
&=& k \left( \sum_{i=1}^n \cos\left(x_i - \tfrac12 \Delta x\right)
− \sum_{i=1}^n \cos\left(x_i + \tfrac12 \Delta x\right) \right).
\end{eqnarray}$$
The last $n - 1$ terms of
$\sum_{i=1}^n \cos\left(x_i - \tfrac12 \Delta x\right)$
are the same as the first $n - 1$ terms of
$\sum_{i=1}^n \cos\left(x_i + \tfrac12 \Delta x\right),$
so these cancel, leaving
$$\begin{eqnarray}
\sum_{i=1}^n \sin\left(x_i\right) \Delta x
&=& k \left( \cos\left(x_1 - \tfrac12 \Delta x\right)
− \cos\left(x_n + \tfrac12 \Delta x\right) \right) \\
&=& \frac{\Delta x}{2 \sin \left(\tfrac12 \Delta x\right)} (\cos a − \cos b). \tag2
\end{eqnarray}$$
That's the sum. To derive a definite integral from this, take the limit
of the right-hand side of equation$\ (2)$ as $\Delta x \to 0.$
You'll find it's the same answer as you get from the antiderivative.
