# Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$

$\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$

Base case: For $n=1$

$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$

Induction hypothesis: For $n=m$

$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$

Induction step: $n=m+1$

$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$

Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$

Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x} {sin\frac{x}{2}}$

How to prove this equality? I used $sin(u)sin(v)$ identity but that didn’t help.

#### Solutions Collecting From Web of "Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$"

Notice, in the third step of induction, you should follow the right procedure

substituting $n=m+1$ in the equality, we get
$$\sum_{k=1}^{m+1}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
$$\sum_{k=1}^{m}\sin(kx)+\sin(m+1)x=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$

$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}-\sin(m+1)x$$
$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)-\sin(m+1)x\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$
Applying $\color{red}{\sin 2A=2\sin A\cos A}$
$$=\frac{\sin\left(\frac{(m+2)x}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)-2\sin\left(\frac{(m+1)x}{2}\right)\cos\left(\frac{(m+1)x}{2}\right)\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$

$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-2\cos\left(\frac{mx}{2}+\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)\right)$$
Applying $\color{red}{2\cos A\sin B=\sin(A+B)-\sin(A-B)}$
$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-\left(\sin\left(\frac{mx}{2}+\frac{x}{2}+\frac{x}{2}\right)-\sin\left(\frac{mx}{2}+\frac{x}{2}-\frac{x}{2}\right)\right)\right)$$
$$=\frac{\sin\left(\frac{(m+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\left(\sin\left(\frac{mx}{2}+x\right)-\sin\left(\frac{mx}{2}+x\right)+\sin\left(\frac{mx}{2}\right)\right)$$
$$\sum_{k=1}^{m}\sin(kx)=\frac{\sin\left(\frac{(m+1)x}{2}\right)\sin\left(\frac{mx}{2}\right)}{\sin\left(\frac{x}{2}\right)}$$

Which is true by the hypothesis.

HINT:

Double angle formula:

$$\sin(m+1)x=2\sin\dfrac{(m+1)x}2\cos\dfrac{(m+1)x}2$$

Werner Formula: $$2\sin\dfrac x2\cos\dfrac{(m+1)x}2=\sin\dfrac{(m+2)x}2-\sin\dfrac{mx}2$$

It’s simply a change product into sum or difference of trigonometric function.Note that we have double angle formula and apply it to $$\sin(m+1)x$$,we’ll get $$sin\frac{(m+1)x}2 *\sin \frac{mx}2+2 \sin\frac{ x}2 *\sin\frac{(m+1)x}2*\cos\frac{(m+1)x}2=\sin\frac{(m+1)x}2*(\sin\frac{mx}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2)$$. Apply the “change product into sum of trigonometric function” twice and we’ll get the answer:$$\sin\frac{mx}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2=\sin\frac{(m+1)x-x}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2=\sin\frac{(m+1)x+x}2$$

Hint: If you want another approach you could probably do a series of deconvolutions in the Fourier domain. The euclidean algorithm should work.