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The problem asks you to use mathematical induction to prove the following trigonometric identity:

\[
\sin \theta + \sin 2\theta + \cdots + \sin n\theta = \frac{\sin \frac{n\theta}{2} \sin \frac{(n+1)\theta}{2}}{\sin \frac{\theta}{2}}
\]

### Step 1: Base Case
For \( n = 1 \), the left-hand side (LHS) is simply:
\[
\sin \theta
\]
The right-hand side (RHS) for \( n = 1 \) is:
\[
\frac{\sin \frac{1\theta}{2} \sin \frac{(1+1)\theta}{2}}{\sin \frac{\theta}{2}} = \frac{\sin \frac{\theta}{2} \sin \frac{2\theta}{2}}{\sin \frac{\theta}{2}} = \sin \theta
\]
So, the base case holds true.

### Step 2: Inductive Hypothesis
Assume the formula holds for \( n = k \), i.e.,
\[
\sin \theta + \sin 2\theta + \cdots + \sin k\theta = \frac{\sin \frac{k\theta}{2} \sin \frac{(k+1)\theta}{2}}{\sin \frac{\theta}{2}}
\]

### Step 3: Inductive Step
We need to prove that the formula holds for \( n = k+1 \), i.e., that:
\[
\sin \theta + \sin 2\theta + \cdots + \sin k\theta + \sin (k+1)\theta = \frac{\sin \frac{(k+1)\theta}{2} \sin \frac{(k+2)\theta}{2}}{\sin \frac{\theta}{2}}
\]

Using the inductive hypothesis:
\[
\sin \theta + \sin 2\theta + \cdots + \sin k\theta + \sin (k+1)\theta = \frac{\sin \frac{k\theta}{2} \sin \frac{(k+1)\theta}{2}}{\

Use mathematical induction to prove the trigonometric identity: sin(θ) + sin(2θ) + ... + sin(nθ) = (sin(nθ/2) * sin((n+1)θ/2)) / sin(θ/2).

Explore the mathematical induction proof for the trigonometric identity involving the summation of sinusoidal terms: sin(θ) + sin(2θ) + ... + sin(nθ). This detailed explanation covers key trigonometry concepts and the use of the mathematical induction principle. Suitable for advanced high school or college-level students.

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