Sum of trigonometric function with argument in arithmetic progresion

Let

$$c_n=1+\cos x+\cos 2x+\cos3x..+\cos nx$$

and

$$s_n=\sin x+\sin 2x+\sin 3x+...+\sin nx$$

We have

$$c_n+is_n=1+\cos x+\cos 2x+\cos3x..+\cos nx+i(\sin x+\sin 2x+\sin 3x+...+\sin nx)=$$
$$1+(\cos x+i\sin x)+(\cos 2x++i\sin 2x)+(\cos3x+\sin 3x)+..+(\cos nx+i \sin nx)$$
$$=1+(\cos x+i\sin x)+(\cos x+i\sin x)^2+(\cos x+i\sin x)^3+...(\cos x+i\sin x)^n=\frac{1-(\cos x+i\sin x)^{n+1}}{1-(\cos x+i\sin x)}=$$
$$\frac{1-\cos (n+1)x-i\sin (n+1)x}{1-\cos x-i\sin x}=\frac{2\sin^2 \frac{(n+1)x}{2}-2i\sin \frac{(n+1)x}{2}\cos \frac{(n+1)x}{2}}{2\sin^2 \frac{x}{2}-i2\sin \frac{x}{2}\cos \frac{x}{2}}=$$
$$=\frac{-2i^2 \sin^2 \frac{(n+1)x}{2}-2i\sin \frac{(n+1)x}{2}\cos \frac{(n+1)x}{2}}{-2i^2\sin^2 \frac{x}{2}-i2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{-2i\sin \frac{(n+1)x}{2}(\cos \frac{(n+1)x}{2}+i\sin \frac{(n+1)x}{2})}{-2i\sin \frac{x}{2}(\cos \frac{x}{2}+i\sin \frac{x}{2})}=$$
$$=\frac{\sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}(\cos \frac{nx}{2}+i\sin \frac{nx}{2})$$

From this relation we have

$$c_n=\frac{\sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}\cos \frac{nx}{2}$$

and

$$s_n=\frac{\sin \frac{(n+1)x}{2}}{\sin \frac{x}{2}}\sin  \frac{nx}{2}$$
Average: 3.3 (3 votes)

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