Question-57770

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

Answered by Kunal12588 last updated on 11/Apr/19

\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}, \dots

= - \frac{1}{2}, - 1, - \frac{1}{2}, \frac{1}{2}, 1, \frac{1}{2}, - \frac{1}{2}, \dots

\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