Σ_(k=0) ^n sin(kx)=Σ_(k=0) ^n Im(e^(ikx) ) =Im(Σ_(k=0) ^n (e^(ix) )^k ) =Im(((1−e^(i(n+1)x) )/(1−e^(ix) ))) ((1−e^(i(n+1)x) )/(1−e^(ix) ))=((e^(i(n+1)(x/2)) (e^(−i(n+1)(x/2)) −e^(i(n+1)(x/2)) ))/(e^(i(x/2)) (e^(−i(x/2)) −e^(i(x/2)) ))) =e^(in(x/2)) ×((−2sin((n+1)(x/2)))/(−2sin((x/2)))) =e^(in(x/2)) ((sin((n+1)(x/2)))/(sin((x/2)))) Σ_(k=0) ^n sin(kx)=((sin((n+1)(x/2) ))/(sin((x/2))))Im(e^(in(x/2)) ) =((sin((n+1)(x/2)))/(sin((x/2))))sin(((nx)/2))


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Question Number 185012 by aba last updated on 15/Jan/23

$\underset{k = 0}{\sum^{n}} \sin (kx) = \underset{k = 0}{\sum^{n}} Im (e^{ikx})$ $= Im (\underset{k = 0}{\sum^{n}} {(e^{ix})}^{k})$ $= Im (\frac{1 - e^{i (n + 1) x}}{1 - e^{ix}})$ $\frac{1 - e^{i (n + 1) x}}{1 - e^{ix}} = \frac{e^{i (n + 1) \frac{x}{2}} (e^{- i (n + 1) \frac{x}{2}} - e^{i (n + 1) \frac{x}{2}})}{e^{i \frac{x}{2}} (e^{- i \frac{x}{2}} - e^{i \frac{x}{2}})}$ $= e^{in \frac{x}{2}} \times \frac{- 2 \sin ((n + 1) \frac{x}{2})}{- 2 \sin (\frac{x}{2})}$ $= e^{in \frac{x}{2}} \frac{\sin ((n + 1) \frac{x}{2})}{\sin (\frac{x}{2})}$ $\underset{k = 0}{\sum^{n}} \sin (kx) = \frac{\sin ((n + 1) \frac{x}{2})}{\sin (\frac{x}{2})} Im (e^{in \frac{x}{2}})$ $= \frac{\sin ((n + 1) \frac{x}{2})}{\sin (\frac{x}{2})} \sin (\frac{nx}{2})$
Commented by Frix last updated on 15/Jan/23

$!!!!$ $Yes n o w {it}^{'} s right .$
Commented by aba last updated on 15/Jan/23

$thank$
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