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)
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Can you specify a little bit, by writing an equation of the sum that you want to calculate? – Matti P. Apr 18 '18 at 05:39
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related question: https://math.stackexchange.com/questions/1172449/find-formula-of-sum-sin-nx – farruhota Apr 18 '18 at 06:20
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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 $x$.
Claude Leibovici
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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:
$$\sum_{k=1}^{n} k\cos(kx) = \frac{d}{dx}\left(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})} \sin(x+(n-1)\tfrac{x}{2})\right) $$
jonsno
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$$\cos x+2\cos 2x+...+n\cos nx = S\\ S=\sum_{k=1}^n \cos nx+\sum_{k=2}^n \cos nx... \\=\frac{\sin(n\frac{x}{2})}{\sin ( \frac{x}{2} )} \cos \biggl( \frac{ 2 x + (n-1)\cdot x}{2}\biggr) + \frac{\sin((n-1)\frac{x}{2})}{\sin ( \frac{x}{2} )} \times \cos \biggl( \frac{ 4x + (n-2)\cdot x}{2}\biggr) \\ +\frac{\sin((n-2) \frac{x}{2})}{\sin ( \frac{x}{2} )} \times \cos \biggl( \frac{ 6x + (n-3)\cdot x}{2}\biggr) \cdots \\ =\frac{1}{\sin\frac x2}\left[\sin(n\frac x2)\cos((n+1)\frac x2)+\sin((n-1)\frac x2)\cos((n+2)\frac x2)+\\ \sin((n-2)\frac x2)\cos((n+3)\frac x2) \cdots \right] \\ = \frac{1}{2\sin \frac x2}\left[\sin(nx+x/2)-\sin(x/2)+ \sin(nx+x/2)-\sin(x/2)\cdots\right] \\ =\frac{n\left[\sin(nx+x/2)-\sin(x/2) \right]}{2\sin \frac x2} = \frac{n\cos((n+1)x/2)\cos(nx/2)}{\sin \frac x2}$$
Sonal_sqrt
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