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Question Number 57770 by Tawa1 last updated on 11/Apr/19

	Answered by Kunal12588 last updated on 11/Apr/19
	$Σ_(k=1) ^(13) (1/(sin((π/4)+(((k−1)π)/6))sin((π/4)+((kπ)/6)))) sin((π/4)+(((k−1)π)/6))sin((π/4)+((kπ)/6)) =sin(((3π+2(k−1)π)/(12)))sin(((3π+2kπ)/(12))) =sin((((2k+1)π)/(12)))sin((((2k+3)π)/(12))) =(1/2){cos((((2k+3)π−(2k+1)π)/(12)))−cos((((2k+3)π+(2k+1)π)/(12)))} =(1/2){cos(((2π)/(12)))−cos(((4k+4)π)/(12))} =(1/2){cos(π/6)−cos((k+1)/3)π} =(1/2){((√3)/2)−cos((k+1)/3)π} k=1,2,3,4,5,6,7,8,9,10,11,12,13 ((k+1)/3)=(2/3),1,1(1/3),1(2/3),2,2(1/3),2(2/3),3,3(1/3),3(2/3),4,4(1/3),4(2/3) cos((k+1)/3)π=c_(120°) ,c_(180) ,c_(240) ,c_(300) ,c_(360) ,c_(60) ,c_(120) ,... =−(1/2),−1,−(1/2),(1/2),1,(1/2),−(1/2),... Σ_(k=1) ^(13) (1/(sin((π/4)+(((k−1)π)/6))sin((π/4)+((kπ)/6)))) =2((4/((√3)+1))+(4/((√3)+2))+(4/((√3)+1))+(4/((√3)−1))+(4/((√3)−2))+(4/((√3)−1)))+(4/((√3)+1)) =8((((√3)−1)/2)+(((√3)−2)/(−1))+(((√3)−1)/2)+(((√3)+1)/2)+(((√3)+2)/(−1))+(((√3)+1)/2))+((4((√3)−1))/2) =4((√3)−1−2(√3)+4+(√3)−1+(√3)+1−2(√3)−4+(√3)+1)+2((√3)−1) =4(0)+2((√3)−1) =2((√3)−1) Ans: C$
	$\underset{k = 1}{\sum^{13}} \frac{1}{s i n (\frac{π}{4} + \frac{(k - 1) π}{6}) s i n (\frac{π}{4} + \frac{k π}{6})}$ $s i n (\frac{π}{4} + \frac{(k - 1) π}{6}) s i n (\frac{π}{4} + \frac{k π}{6})$ $= s i n (\frac{3 π + 2 (k - 1) π}{12}) s i n (\frac{3 π + 2 k π}{12})$ $= s i n (\frac{(2 k + 1) π}{12}) s i n (\frac{(2 k + 3) π}{12})$ $= \frac{1}{2} {c o s (\frac{(2 k + 3) π - (2 k + 1) π}{12}) - c o s (\frac{(2 k + 3) π + (2 k + 1) π}{12})}$ $= \frac{1}{2} {c o s (\frac{2 π}{12}) - c o s \frac{(4 k + 4) π}{12}}$ $= \frac{1}{2} {c o s \frac{π}{6} - c o s \frac{k + 1}{3} π}$ $= \frac{1}{2} {\frac{\sqrt{3}}{2} - c o s \frac{k + 1}{3} π}$ $k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13$ $\frac{k + 1}{3} = \frac{2}{3}, 1, 1 \frac{1}{3}, 1 \frac{2}{3}, 2, 2 \frac{1}{3}, 2 \frac{2}{3}, 3, 3 \frac{1}{3}, 3 \frac{2}{3}, 4, 4 \frac{1}{3}, 4 \frac{2}{3}$ $c o s \frac{k + 1}{3} π = c_{120 °}, c_{180}, c_{240}, c_{300}, c_{360}, c_{60}, c_{120}, . . .$ $= - \frac{1}{2}, - 1, - \frac{1}{2}, \frac{1}{2}, 1, \frac{1}{2}, - \frac{1}{2}, . . .$ $\underset{k = 1}{\sum^{13}} \frac{1}{s i n (\frac{π}{4} + \frac{(k - 1) π}{6}) s i n (\frac{π}{4} + \frac{k π}{6})}$ $= 2 (\frac{4}{\sqrt{3} + 1} + \frac{4}{\sqrt{3} + 2} + \frac{4}{\sqrt{3} + 1} + \frac{4}{\sqrt{3} - 1} + \frac{4}{\sqrt{3} - 2} + \frac{4}{\sqrt{3} - 1}) + \frac{4}{\sqrt{3} + 1}$ $= 8 (\frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} - 2}{- 1} + \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} + \frac{\sqrt{3} + 2}{- 1} + \frac{\sqrt{3} + 1}{2}) + \frac{4 (\sqrt{3} - 1)}{2}$ $= 4 (\sqrt{3} - 1 - 2 \sqrt{3} + 4 + \sqrt{3} - 1 + \sqrt{3} + 1 - 2 \sqrt{3} - 4 + \sqrt{3} + 1) + 2 (\sqrt{3} - 1)$ $= 4 (0) + 2 (\sqrt{3} - 1)$ $= 2 (\sqrt{3} - 1)$ $A n s : C$
	Commented by Tawa1 last updated on 11/Apr/19

	$God bless you sir .$
	Commented by abbas-alsadi last updated on 12/Apr/19


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