Math Problem Statement

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).

Solution

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

sinθ+sin2θ++sinnθ=sinnθ2sin(n+1)θ2sinθ2\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=1n = 1, the left-hand side (LHS) is simply: sinθ\sin \theta The right-hand side (RHS) for n=1n = 1 is: sin1θ2sin(1+1)θ2sinθ2=sinθ2sin2θ2sinθ2=sinθ\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=kn = k, i.e., sinθ+sin2θ++sinkθ=sinkθ2sin(k+1)θ2sinθ2\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+1n = k+1, i.e., that: sinθ+sin2θ++sinkθ+sin(k+1)θ=sin(k+1)θ2sin(k+2)θ2sinθ2\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}}{\

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Mathematical Induction
Summation

Formulas

Summation of sinusoids: sin(θ) + sin(2θ) + ... + sin(nθ) = (sin(nθ/2) * sin((n+1)θ/2)) / sin(θ/2)

Theorems

Principle of Mathematical Induction

Suitable Grade Level

College level or advanced high school (Grades 11-12)