Σ_(k = 0) ^n cos(a + kb)


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Question Number 72483 by TawaTawa last updated on 29/Oct/19

$\underset{k = 0}{\sum^{n}} \cos (a + kb)$
	Answered by Tanmay chaudhury last updated on 29/Oct/19

	$T_{k} = c o s (a + k b)$ $S = T_{0} + T_{1} + T_{2} + . . . + T_{n}$ $T_{0} \times 2 s i n \frac{b}{2} = 2 c o s a \times s i n \frac{b}{2} = s i n (a + \frac{b}{2}) - s i n (a - \frac{b}{2})$ $T_{1} \times 2 s i n \frac{b}{2} = 2 c o s (a + b) \times s i n \frac{b}{2} = s i n (a + \frac{3 b}{2}) - s i n (a + \frac{b}{2})$ $T_{2} \times 2 s i n \frac{b}{2} = 2 c o s (a + 2 b) \times s i n \frac{b}{2} = s i n (a + \frac{5 b}{2}) - s i n (a + \frac{3 b}{2})$ $. . .$ $. . . .$ $T_{n} \times 2 s i n \frac{b}{2} = 2 c o s (a + n b) \times s i n \frac{b}{2} = s i n (a + \frac{(2 n + 1) b}{2}) - s i n (a + \frac{(2 n - 1) b}{2})$ $a d d L H S a n d R H S$ $S \times 2 s i n \frac{b}{2} = s i n (a + \frac{(2 n + 1) b}{2}) - s i n (a - \frac{b}{2})$ $s o S = \frac{s i n (a + \frac{(2 n + 1) b}{2}) - s i n (a - \frac{b}{2})}{2 s i n \frac{b}{2}}$
	Commented by TawaTawa last updated on 29/Oct/19

	$God bless you sir$
	Answered by mind is power last updated on 29/Oct/19
	$cos(a+kb)=Re(e^(i(a+kb)) ) Σ_(k=0) ^n cos(a+kb)=Re(Σ_(k=0) ^n e^(i(a+kb)) )=Re {e^(ia) .((1−e^(i(n+1)b) )/(1−e^(ib) ))} =Re{((e^(i(((nb+2a))/2)) .(e^(−i((n+1)/2)) −e^(i((n+1)/2)) ))/(e^(−((ib)/2)) −e^((ib)/2) ))}=Re{((e^(i(((nb+2a)/2))) .sin(((n+1)/2)b))/(sin((b/2))))} =((sin(((n+1)/2)b))/(sin((b/2)))).cos(((nb+2a)/2))$
	$\cos (a + kb) = Re (e^{i (a + kb)})$ $\underset{k = 0}{\sum^{n}} \cos (a + kb) = Re (\underset{k = 0}{\sum^{n}} e^{i (a + kb)}) = Re {e^{ia} . \frac{1 - e^{i (n + 1) b}}{1 - e^{ib}}}$ $= Re {\frac{e^{i \frac{(nb + 2 a)}{2}} . (e^{- i \frac{n + 1}{2}} - e^{i \frac{n + 1}{2}})}{e^{- \frac{ib}{2}} - e^{\frac{ib}{2}}}} = Re {\frac{e^{i (\frac{nb + 2 a}{2})} . \sin (\frac{n + 1}{2} b)}{\sin (\frac{b}{2})}}$ $= \frac{\sin (\frac{n + 1}{2} b)}{\sin (\frac{b}{2})} . \cos (\frac{nb + 2 a}{2})$
	Commented by TawaTawa last updated on 29/Oct/19

	$God bless you sir$
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