sin^2 (π/11)+sin^2 (2π/11)+...+sin^2 (5π/11)=?


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

$\sin^{2} (π / 11) + \sin^{2} (2 π / 11) + . . . + \sin^{2} (5 π / 11) = ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jun/18

	${s i n}^{2} a + {s i n}^{2} 2 a + {s i n}^{2} 3 a + {s i n}^{2} 4 a + {s i n}^{2} 5 a$ $a = \frac{Π}{11}$ $= \frac{1 - c o s 2 a}{2} + \frac{1 - c o s 4 a}{2} + \frac{1 - c o s 6 a}{2} + \frac{1 - c o s 8 a}{2} +$ $\frac{1 - c o s 10 a}{2}$ $= 5 \times \frac{1}{2} - \frac{1}{2} (c 2 a + c 4 a + c 6 a + c 8 a + c 10 a)$ $w h e n c 2 a = c o s 2 a e t c$ $s = c 2 a + c 4 a + c 6 a + c 8 a + c 10 a$ $s \times 2 s i n a = 2 s i n a . (c 2 a + c 4 a + c 6 a + c 8 a + c 10 a)$ $s \times 2 s i n a = 2 s i n a . c o s 2 a + 2 s i n a . c o s 4 a +$ $2 s i n a c o s 6 a + 2 s i n a . c o s 8 a + 2 s i n a . c o s 10 a$ $n o w l o o k$ $2 s i n a . c o s 2 a = s i n 3 a - s i n a$ $2 s i n a . c o s 4 a = s i n 5 a - s i n 3 a$ $2 s i n a . c o s 6 a = s i n 7 a - s i n 5 a$ $2 s i n a . c o s 8 a = s i n 9 a - s i n 7 a$ $2 s i n a . c o s 10 a = s i n 11 a - s i n 9 a$ $n o w a d d t h e m . . . a d d i t i o n o f r i g h t s i d e i s$ $= s i n 11 a - s i n a$ $= 2 c o s 6 a . s i n 5 a$ $a d d i t i o n o f l e f t i s s \times 2 s i n a$ $s o$ $s \times 2 s i n a = 2 c o s 6 a . s i n 5 a$ $s = \frac{c o s 6 a . s i n 5 a}{s i n a}$ $\frac{5}{2} - \frac{1}{2} (\frac{c o s 6 a . s i n 5 a}{s i n a})$ $= \frac{5}{2} - \frac{1}{4} (\frac{2 c o s 6 a . s i n 5 a}{s i n a})$ $= \frac{5}{2} - \frac{1}{4} (\frac{s i n 11 a - s i n a}{s i n a})$ $a = \frac{Π}{11} s o 11 a = Π$ $s i n 11 a = s i n Π = 0$ $= \frac{5}{2} - \frac{1}{4} (\frac{0 - s i n a}{s i n a})$ $= \frac{5}{2} + \frac{1}{4}$ $= \frac{10 + 1}{4} = \frac{11}{4}$
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