trigonometric summation with arithmetic progression of angles and coefficients

Question

How can we sum up sin and cos series when the angles and also coefficients are in arithmetic progression? Such as cos(x)+2cos(2x)+3cos(3x)+.......+ncos(nx)

Answer 1

Hint

Consider $$C_n=\sum_{k=1}^n k\, \cos(kx)\qquad \text{and}\qquad S_n=\sum_{k=1}^n k\, \sin(kx) $$ Integrate with respect to $x$ $$\int C_n\,dx=\sum_{k=1}^n \sin(kx)\qquad \text{and}\qquad \int S_n\,dx=-\sum_{k=1}^n \cos(kx)$$ which take you back to a much simpler and classical problem.

When done, differneiate with respect to $x$.

Answer 2

You can show that, using complex numbers, or using trigonometric identities:

$$\sum_{k=1}^{n} \sin(kx) = \frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})} \sin(x+(n-1)\tfrac{x}{2})$$

Now differentiate wrt $x$ to get:

$$\sum_{k=1}^{n} k\cos(kx) = \frac{d}{dx}\left(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})} \sin(x+(n-1)\tfrac{x}{2})\right) $$

Answer 3

$$\cos x+2\cos 2x+...+n\cos nx = S\\ S=\sum_{k=1}^n \cos nx+\sum_{k=2}^n \cos nx... \\=\frac{\sin(n\frac{x}{2})}{\sin ( \frac{x}{2} )} \cos \biggl( \frac{ 2 x + (n-1)\cdot x}{2}\biggr) + \frac{\sin((n-1)\frac{x}{2})}{\sin ( \frac{x}{2} )} \times \cos \biggl( \frac{ 4x + (n-2)\cdot x}{2}\biggr) \\ +\frac{\sin((n-2) \frac{x}{2})}{\sin ( \frac{x}{2} )} \times \cos \biggl( \frac{ 6x + (n-3)\cdot x}{2}\biggr) \cdots \\ =\frac{1}{\sin\frac x2}\left[\sin(n\frac x2)\cos((n+1)\frac x2)+\sin((n-1)\frac x2)\cos((n+2)\frac x2)+\\ \sin((n-2)\frac x2)\cos((n+3)\frac x2) \cdots \right] \\ = \frac{1}{2\sin \frac x2}\left[\sin(nx+x/2)-\sin(x/2)+ \sin(nx+x/2)-\sin(x/2)\cdots\right] \\ =\frac{n\left[\sin(nx+x/2)-\sin(x/2) \right]}{2\sin \frac x2} = \frac{n\cos((n+1)x/2)\cos(nx/2)}{\sin \frac x2}$$