Question: "I need to prove by induction that:∑k=1nsin⁡(kx)=sin⁡(n+12x)sin⁡(n2x)sin⁡(x2)"

haugmbd

Answered question

2022-02-26

I need to prove by induction that:

\sum_{k = 1}^{n} \sin (k x) = \frac{\sin (\frac{n + 1}{2} x) \sin (\frac{n}{2} x)}{\sin (\frac{x}{2})}

legertopdxa

Beginner2022-02-27Added 8 answers

You can actually use telescopy, which is just induction in disguise.
We have that

\cos b - \cos a = 2 \sin \frac{a + b}{2} \sin \frac{a - b}{2}

Now let

b = (k + \frac{1}{2}) x

b = (k - \frac{1}{2}) x

Then

\cos (k + \frac{1}{2}) x - \cos (k - \frac{1}{2}) x = 2 \sin k x \sin \frac{x}{2}

\cos (k + \frac{1}{2}) x - \cos (k - \frac{1}{2}) x = 2 \sin k x \sin \frac{x}{2}

Now sum through

k = 1, \dots, n

\sum_{k = 1}^{n} \cos (k + \frac{1}{2}) - \cos (k - \frac{1}{2}) x = 2 \sin \frac{x}{2} \sum_{k = 1}^{n} \sin k x \cos (n + \frac{1}{2}) x - \cos \frac{x}{2}

= 2 \sin \frac{x}{2} \sum_{k = 1} \sin k x

whence

\frac{\cos (n + \frac{1}{2}) x - \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \sum_{x = 1}^{n} \sin k x

But using our first formula once more, we have

\cos (n + \frac{1}{2}) x - \cos \frac{x}{2} = 2 \sin \frac{(n + 1) x}{2} \sin \frac{n x}{2}

so finally

\frac{\sin \frac{(n + 1) x}{2} \sin \frac{n x}{2}}{\sin \frac{x}{2}} = \sum_{k = 1}^{n} \sin k x

as desired.

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