sin-2-pi-11-sin-2-2pi-11-sin-2-5pi-11-

Question Number 37244 by subhendubera201314@gmail.com last updated on 11/Jun/18

\sin^{2} (π / 11) + \sin^{2} (2 π / 11) + \dots + \sin^{2} (5 π / 11) = ?

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jun/18

{s i n}^{2} a + {s i n}^{2} 2 a + {s i n}^{2} 3 a + {s i n}^{2} 4 a + {s i n}^{2} 5 a

a = \frac{Π}{11}

= \frac{1 - c o s 2 a}{2} + \frac{1 - c o s 4 a}{2} + \frac{1 - c o s 6 a}{2} + \frac{1 - c o s 8 a}{2} +

\frac{1 - c o s 10 a}{2}

= 5 \times \frac{1}{2} - \frac{1}{2} (c 2 a + c 4 a + c 6 a + c 8 a + c 10 a)

w h e n c 2 a = c o s 2 a e t c

s = c 2 a + c 4 a + c 6 a + c 8 a + c 10 a

s \times 2 s i n a = 2 s i n a . (c 2 a + c 4 a + c 6 a + c 8 a + c 10 a)

s \times 2 s i n a = 2 s i n a . c o s 2 a + 2 s i n a . c o s 4 a +

2 s i n a c o s 6 a + 2 s i n a . c o s 8 a + 2 s i n a . c o s 10 a

n o w l o o k

2 s i n a . c o s 2 a = s i n 3 a - s i n a

2 s i n a . c o s 4 a = s i n 5 a - s i n 3 a

2 s i n a . c o s 6 a = s i n 7 a - s i n 5 a

2 s i n a . c o s 8 a = s i n 9 a - s i n 7 a

2 s i n a . c o s 10 a = s i n 11 a - s i n 9 a

n o w a d d t h e m \dots a d d i t i o n o f r i g h t s i d e i s

= s i n 11 a - s i n a

= 2 c o s 6 a . s i n 5 a

a d d i t i o n o f l e f t i s s \times 2 s i n a

s o

s \times 2 s i n a = 2 c o s 6 a . s i n 5 a

s = \frac{c o s 6 a . s i n 5 a}{s i n a}

\frac{5}{2} - \frac{1}{2} (\frac{c o s 6 a . s i n 5 a}{s i n a})

= \frac{5}{2} - \frac{1}{4} (\frac{2 c o s 6 a . s i n 5 a}{s i n a})

= \frac{5}{2} - \frac{1}{4} (\frac{s i n 11 a - s i n a}{s i n a})

a = \frac{Π}{11} s o 11 a = Π

s i n 11 a = s i n Π = 0

= \frac{5}{2} - \frac{1}{4} (\frac{0 - s i n a}{s i n a})

= \frac{5}{2} + \frac{1}{4}

= \frac{10 + 1}{4} = \frac{11}{4}