# Use the principle of mathematical induction to prove that?: See picture

Apr 8, 2017

#### Explanation:

Prove:

$\sin \theta + \sin 2 \theta + \cdots + \sin \left(n \theta\right) = \frac{\sin \left(\frac{\left(n + 1\right) \theta}{2}\right) \sin \left(\frac{n \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

Base Case (n=1)

(LHS)$= \sin \theta$

(RHS)$= \frac{\sin \left(\frac{\left(1 + 1\right) \theta}{2}\right) \cancel{\sin} \left(\frac{1 \theta}{2}\right)}{\cancel{\sin \left(\frac{\theta}{2}\right)}} = \sin \theta$

Hence, (RHS)=(LHS).

Induction Hypothesis (n=k)

Assume that
$\sin \theta + \sin 2 \theta + \cdots + \sin k \theta = \frac{\sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{k \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

Induction Step (n=k+1)

Let us show that

$\sin \theta + \sin 2 \theta + \cdots + \sin k \theta + \sin \left(k + 1\right) \theta = \frac{\sin \left(\frac{\left(k + 2\right) \theta}{2}\right) \sin \left(\frac{\left(k + 1\right) \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

By I.H.,

(LHS)$= \frac{\sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{k \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)} + \sin \left(k + 1\right) \theta$

By taking the common denominator,

$= \frac{\sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{k \theta}{2}\right) + \sin \left(k + 1\right) \theta \cdot \sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

By $\sin \left(k + 1\right) \theta = 2 \sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \cos \left(\frac{\left(k + 1\right) \theta}{2}\right)$,

$= \frac{\sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{k \theta}{2}\right) + 2 \sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \cos \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

By factoring out $\sin \left(\frac{\left(k + 1\right) \theta}{2}\right)$,

$= \frac{\left[\sin \left(\frac{k \theta}{2}\right) + 2 \cos \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)\right] \sin \left(\frac{\left(k + 1\right) \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

By $\sin \left(\frac{k \theta}{2}\right) = \sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \cos \left(\frac{\theta}{2}\right) - \cos \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{\theta}{2}\right) ,$

$= \frac{\left[\sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \cos \left(\frac{\theta}{2}\right) + \cos \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)\right] \sin \left(\frac{\left(k + 1\right) \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)}$

By $\sin \left(\frac{\left(k + 2\right) \theta}{2}\right) = \sin \left(\frac{\left(k + 1\right) \theta}{2}\right) \cos \left(\frac{\theta}{2}\right) + \cos \left(\frac{\left(k + 1\right) \theta}{2}\right) \sin \left(\frac{\theta}{2}\right) ,$

$= \frac{\sin \left(\frac{\left(k + 2\right) \theta}{2}\right) \sin \left(\frac{\left(k + 1\right) \theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right)} =$(RHS)

Hence, the equation is true for all natural numer $n$.