Using the identity 1/2 + cos x + cos 2x +... + cos nx =

Using the identity $\frac{1}{2} + \cos x + \cos 2x + \cdots + \cos nx = \frac{\sin\!\left(n+\tfrac{1}{2}\right)x}{2\sin\tfrac{x}{2}}, \qquad x \neq 2k\pi, \tag{1}$ find the sum $\sin x + 2\sin 2x + 3\sin 3x + \cdots + n\sin nx$.

Step 1: Differentiate the left-hand side

Term by term, $\bigl(\cos kx\bigr)' = -k\sin kx$, and the constant $\tfrac12$ contributes nothing: \[\frac{d}{dx}\!\left(\frac12 + \sum_{k=1}^{n}\cos kx\right) = -\sum_{k=1}^{n} k\sin kx.\]

So the sum we want is $-\dfrac{d}{dx}$ of the right-hand side of (1).

Step 2: Differentiate the right-hand side

Write $S = \sin\!\left(n+\tfrac12\right)x$, $C = \cos\!\left(n+\tfrac12\right)x$, $s = \sin\tfrac{x}{2}$, $c = \cos\tfrac{x}{2}$. By the quotient rule with $u = S$, $u' = \left(n+\tfrac12\right)C$, $v = 2s$, $v' = c$:

\[ \frac{d}{dx}\!\left(\frac{S}{2s}\right) = \frac{\left(n+\tfrac12\right)C\cdot 2s - S\,c}{4s^{2}}. \]

Therefore $\sum_{k=1}^{n}k\sin kx = \frac{S\,c - (2n+1)\,C\,s}{4s^{2}}$.

using $2\left(n+\tfrac12\right) = 2n+1$.

Step 3: Convert the products to sums

Two product-to-sum identities turn the numerator into ordinary sines: \[S\,c = \sin\!\left(n+\tfrac12\right)x\cos\tfrac{x}{2} = \tfrac12\Bigl[\sin(n+1)x + \sin nx\Bigr],\]

\[ C\,s = \cos\!\left(n+\tfrac12\right)x\sin\tfrac{x}{2} = \tfrac12\Bigl[\sin(n+1)x - \sin nx\Bigr]. \]

Step 4: Collect the numerator

\[ \begin{aligned} S\,c - (2n+1)C\,s &= \tfrac12\bigl[\sin(n+1)x+\sin nx\bigr] - \tfrac{2n+1}{2}\bigl[\sin(n+1)x-\sin nx\bigr] \\[4pt] &= \sin(n+1)x\left(\tfrac12 - \tfrac{2n+1}{2}\right) + \sin nx\left(\tfrac12 + \tfrac{2n+1}{2}\right) \\[4pt] &= -n\sin(n+1)x + (n+1)\sin nx. \end{aligned} \]

Step 5: Verification

Case $n = 1$. The right-hand side should reduce to $\sin x$:

\[\frac{2\sin x - \sin 2x}{4\sin^{2}\frac{x}{2}} = \frac{2\sin x\bigl(1-\cos x\bigr)}{4\sin^{2}\frac{x}{2}} = \frac{2\sin x\cdot 2\sin^{2}\frac{x}{2}}{4\sin^{2}\frac{x}{2}} = \sin x \;\;\checkmark\]

\[n = 2,\ x = \frac{\pi}{2}:\quad \text{left} = 1 + 2\cdot 0 = 1; \quad \text{right} = \frac{3\sin\pi - 2\sin\frac{3\pi}{2}}{4\sin^{2}\frac{\pi}{4}} = \frac{0+2}{2} = 1 \;\;\checkmark\]