To prove the statement using mathematical induction:

$\sin x + \sin 3x + \sin 5x + \dots + \sin(2n-1)x = \frac{1 - \cos(2nx)}{2 \sin x}, \quad \text{for all} \, n \geq 1,$ we follow the standard steps of induction.

Step 1: Base Case (n = 1)

For $n = 1$, the left-hand side (LHS) of the equation is: $\sin x.$ The right-hand side (RHS) for $n = 1$ is: $\frac{1 - \cos(2 \cdot 1 \cdot x)}{2 \sin x} = \frac{1 - \cos(2x)}{2 \sin x}.$ Now, using the identity: $1 - \cos(2x) = 2 \sin^2 x,$ we substitute into the RHS: $\frac{1 - \cos(2x)}{2 \sin x} = \frac{2 \sin^2 x}{2 \sin x} = \sin x.$ Thus, for $n = 1$, both sides are equal, confirming the base case.

Step 2: Inductive Hypothesis

Assume that the formula holds for some $n = k$, i.e., $\sin x + \sin 3x + \sin 5x + \dots + \sin(2k-1)x = \frac{1 - \cos(2kx)}{2 \sin x}.$ We need to prove that the formula also holds for $n = k + 1$.

Step 3: Inductive Step

For $n = k+1$, the LHS becomes: $\sin x + \sin 3x + \sin 5x + \dots + \sin(2k-1)x + \sin(2(k+1)-1)x = \sin x + \sin 3x + \dots + \sin(2k-1)x + \sin(2k+1)x.$ By the inductive hypothesis, we can substitute for the sum of the first $k$ terms: $\frac{1 - \cos(2kx)}{2 \sin x} + \sin(2k+1)x.$

We now combine these terms. Using the sum identity for sine: $\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right),$ let $A = (2k+1)x$ and $B = (2kx)$, but this step will involve exploring direct term transformations as suitable for your problem