find ((2 sin 2°+4 sin 4°+...+180 sin 180°)/(90))=? [an unsolved old question]


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Question Number 199226 by mr W last updated on 29/Oct/23

$f i n d$ $\frac{2 \sin 2 ° + 4 \sin 4 ° + . . . + 180 \sin 180 °}{90} = ?$ $[a n u n s o l v e d o l d q u e s t i o n]$
	Answered by mr W last updated on 29/Oct/23

	${l e t}^{'} s f i n d a t f i r s t$ $S = \underset{k = 1}{\sum^{n}} {k q}^{k} = q + 2 q^{2} + 3 q^{3} + . . . + {n q}^{n}$ $q S = q^{2} + 2 q^{3} + 3 q^{4} + . . . + (n - 1) q^{n} + {n q}^{n + 1}$ $(1 - q) S = q + q^{2} + q^{3} + . . . + q^{n} - {n q}^{n + 1}$ $(1 - q) S = \frac{q - q^{n + 1}}{1 - q} - {n q}^{n + 1} = \frac{q - (n + 1) q^{n + 1} + {n q}^{n + 2}}{1 - q}$ $\Rightarrow S = \frac{q - (n + 1) q^{n + 1} + {n q}^{n + 2}}{{(1 - q)}^{2}}$ $s a y$ $A = \underset{k = 1}{\sum^{n}} k \cos k α$ $B = \underset{k = 1}{\sum^{n}} k \sin k α$ $A + B i = \underset{k = 1}{\sum^{n}} k (\cos k α + i \sin k α) = \underset{k = 1}{\sum^{n}} {k e}^{k α i}$ $A + B i = \underset{k = 1}{\sum^{n}} k {(e^{α i})}^{k} = \frac{e^{α i} - (n + 1) e^{(n + 1) α i} + {n e}^{(n + 2) α i}}{{(1 - e^{α i})}^{2}}$ $A + B i = \frac{e^{α i} - (n + 1) e^{(n + 1) α i} + {n e}^{(n + 2) α i}}{1 + e^{2 α i} - 2 e^{α i}}$ $A + B i = \frac{\cos α + i \sin α - (n + 1) \cos (n + 1) α - i (n + 1) \sin (n + 1) α + n \cos (n + 2) α + i n \sin (n + 2) α}{1 + \cos 2 α + i \sin 2 α - 2 \cos α - i 2 \sin α}$ $A + B i = \frac{[\cos α - (n + 1) \cos (n + 1) α + n \cos (n + 2) α] + i [\sin α - (n + 1) \sin (n + 1) α + n \sin (n + 2) α]}{[1 + \cos 2 α - 2 \cos α] + i [\sin 2 α - 2 \sin α]}$ $A + B i = \frac{p + i q}{u + i v}$ $A + B i = \frac{(p + i q) (u - i v)}{(u + i v) (u - i v)} = \frac{(p u + q v) + (q u - p v) i}{u^{2} + v^{2}}$ $\Rightarrow A = \frac{p u + q v}{u^{2} + v^{2}}$ $= \frac{[\cos α - (n + 1) \cos (n + 1) α + n \cos (n + 2) α] [1 + \cos 2 α - 2 \cos α] + [\sin α - (n + 1) \sin (n + 1) α + n \sin (n + 2) α] [\sin 2 α - 2 \sin α]}{{(1 + \cos 2 α - 2 \cos α)}^{2} + {(\sin 2 α - 2 \sin α)}^{2}}$ $= \frac{\cos α [\cos α - (n + 1) \cos (n + 1) α + n \cos (n + 2) α] + \sin α [\sin α - (n + 1) \sin (n + 1) α + n \sin (n + 2) α]}{2 (\cos α - 1)}$ $= \frac{(n + 1) \cos n α - n \cos (n + 1) α - 1}{2 (1 - \cos α)}$ $\Rightarrow B = \frac{q u - p v}{u^{2} + v^{2}}$ $= \frac{[\sin α - (n + 1) \sin (n + 1) α + n \sin (n + 2) α] [1 + \cos 2 α - 2 \cos α] - [\cos α - (n + 1) \cos (n + 1) α + n \cos (n + 2) α] [\sin 2 α - 2 \sin α]}{{(1 + \cos 2 α - 2 \cos α)}^{2} + {(\sin 2 α - 2 \sin α)}^{2}}$ $= \frac{\cos α [(n + 1) \sin (n + 1) α - n \sin (n + 2) α] + \sin α [n \cos (n + 2) α - (n + 1) \cos (n + 1) α]}{2 (1 - \cos α)}$ $= \frac{(n + 1) \sin n α - n \sin (n + 1) α}{2 (1 - \cos α)}$ $\frac{2 \sin 2 ° + 4 \sin 4 ° + . . . \times 180 \sin 180 °}{90}$ $= \frac{\underset{k = 1}{\sum^{90}} (2 k) \sin (2 k) °}{90} = \frac{2 \underset{k = 1}{\sum^{90}} k \sin (2 k) °}{90}$ $= \frac{2}{90} \times B_{α = 2 °, n = 90}$ $= \frac{2}{90} \times \frac{91 \sin (90 \times 2 °) - 90 \sin (91 \times 2 °)}{2 (1 - \cos 2 °)}$ $= \frac{\sin 2 °}{1 - \cos 2 °} = \frac{2 \sin 1 ° \cos 1 °}{1 - 1 + 2 \sin^{2} 1 °} = \frac{1}{\tan 1 °}$ $b t w . w e g e t$ $\frac{2 \cos 2 ° + 4 \cos 4 ° + . . . + 180 \cos 180 °}{90}$ $= \frac{2}{90} \times A_{α = 2 °, n = 90}$ $= \frac{2}{90} \times \frac{91 \cos (90 \times 2 °) - 90 \cos (91 \times 2 °) - 1}{2 (1 - \cos 2 °)}$ $= - 1 - \frac{1}{45 (1 - \cos 2 °)}$
	Commented by hardmath last updated on 29/Oct/23

	$cool dear professor, your solutions are$ $great . . .$
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