一个数列满足递推关系 对 $n = 1,2,3,\dots$，有 $U_{n+1}-U_n = 2\cos\left(\dfrac{(2n+1)x}{2}\right)\sin\dfrac{x}{2}$，且 $U_1 = \sin x$。 利用数学归纳法证明对所有正整数 $n$，均有 $U_n = \sin(nx)$。

1. State the proposition and verify the base case for $n=1$.

Let $P_n$ be the statement that $U_n = \sin(nx)$.
When $n=1$, $U_1 = \sin(1\cdot x) = \sin x$.
This matches the given information that $U_1 = \sin x$, so $P_1$ is true.

2. State the inductive hypothesis.

Assume $P_k$ is true for some positive integer $k$, which means $U_k = \sin(kx)$.

3. Prove that $P_{k+1}$ is true using the recurrence relation and trigonometric identities.

For $n=k+1$, we use the given recurrence relation:
$U_{k+1} - U_k = 2\cos\left(\frac{(2k+1)x}{2}\right)\sin\frac{x}{2}$
Substituting $U_k = \sin(kx)$:
$U_{k+1} = \sin(kx) + 2\cos\left(\frac{(2k+1)x}{2}\right)\sin\frac{x}{2}$
Using the factor formula $2\cos A\sin B = \sin(A+B) - \sin(A-B)$ where $A = \frac{2kx+x}{2}$ and $B = \frac{x}{2}$:
$2\cos\left(\frac{(2k+1)x}{2}\right)\sin\frac{x}{2} = \sin\left(\frac{2kx+x+x}{2}\right) - \sin\left(\frac{2kx+x-x}{2}\right)$
$= \sin\left(\frac{2(k+1)x}{2}\right) - \sin\left(\frac{2kx}{2}\right)$
$= \sin((k+1)x) - \sin(kx)$
Substituting this back into our equation for $U_{k+1}$:
$U_{k+1} = \sin(kx) + \sin((k+1)x) - \sin(kx) = \sin((k+1)x)$
Thus, $P_{k+1}$ is true.

4. State the conclusion of the induction.

Since $P_1$ is true, and $P_k$ being true implies that $P_{k+1}$ is true for any positive integer $k$, by the Principle of Mathematical Induction, $P_n$ is true for all positive integers $n$.