find the value of Π_(i=1) ^(999) cos (ia) ; where a = ((2π)/(1999))


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 119421 by bemath last updated on 24/Oct/20

$f i n d t h e v a l u e o f \underset{i = 1}{\prod^{999}} \cos (i a); w h e r e a = \frac{2 π}{1999}$
Commented by MJS_new last updated on 24/Oct/20

$study P_{n} = \underset{j = 1}{\prod^{n}} \cos \frac{2 π j}{2 n + 1} to find the answer$ ${it}^{'} s easier than it seems$
	Answered by benjo_mathlover last updated on 24/Oct/20

	$L e t B d e n o t e t h e d e s i r e d p r o d u c t, a n d l e t$ $J = \sin a \sin 2 a \sin 3 a . . . \sin 999 a$ $c o n s i d e r t h a t$ $2^{999} B J = (2 \sin a \cos a) (2 \sin 2 a \cos 2 a) (2 \sin 3 a \cos 3 a) . . . (2 \sin 999 a \cos 999 a)$ $= \sin 2 a \sin 4 a \sin 6 a . . . \sin 1998 a$ $= (\sin 2 a \sin 4 a \sin 6 a . . . \sin 998 a) (- \sin (2 π - 1000 a)) . (- \sin (2 π - 10002 a)) . . . (- \sin (2 π - 1998 a))$ $= \sin 2 a \sin 4 a \sin 6 a . . . \sin 998 a \sin 999 a \sin 997 a . . . \sin a = J$ $s i n c e J \neq 0, h e n c e t h e d e s i r e d p r o d u c t$ $i s B = \frac{1}{2^{999}}$
	Answered by mindispower last updated on 24/Oct/20

	$\underset{i = 1}{\prod^{m}} c o s (i \frac{2 π}{2 m + 1})$ $Z^{2 m + 1} - 1 = 0 \Rightarrow Z = e^{\frac{2 i k π}{2 m + 1}}, k \in [0, 2 m]$ $z^{2 m + 1} - 1 = \prod_{k ⩽ 2 m} (z - e^{\frac{2 i k π}{2 m + 1}})$ $z = - 1 \Rightarrow$ $- 2 = \underset{k = 0}{\prod^{2 m}} (- 1 - e^{\frac{2 i k π}{2 m + 1}}) = - 1 e^{i \frac{π}{2 m + 1} . \frac{2 m (2 m + 1)}{2}} . \prod_{k ⩽ 2 m} (2 c o s (\frac{k π}{2 m + 1})$ $= {(- 1)}^{m + 1} 2^{2 m + 1} \prod_{k ⩽ 2 m} c o s (\frac{k π}{2 m + 1})$ $\prod_{k ⩽ 2 m} c o s (\frac{k π}{2 m + 1}) = \prod_{k ⩽ m} c o s (\frac{k π}{2 m + 1}) . \prod_{m < k ⩽ 2 m} (- c o s (π - \frac{k π}{2 m + 1}))$ $p u t k \to 2 m + 1 - k i n 2 n d \Rightarrow$ $= \prod_{1 ⩽ k ⩽ m} c o s (\frac{k π}{2 m + 1}) . {(- 1)}^{m} . \prod_{k ⩽ m} c o s (\frac{π (2 m + 1) - (2 m + 1 - k) π}{2 m + 1}$ $= {(- 1)}^{m} {(\underset{k = 1}{\prod^{m}} c o s (\frac{k π}{2 m + 1}))}^{2}$ $\Rightarrow - 2 = {(- 1)}^{m + 1} . 2^{2 m + 1} . {(- 1)}^{m} . {(\prod_{k ⩽ m} c o s (\frac{k π}{2 m + 1}))}^{2}$ $\Rightarrow \frac{1}{2^{2 m}} = \prod_{k ⩽ m} {c o s}^{2} (\frac{k π}{2 m + 1})$ $s i n c e 0 < \frac{k π}{2 m + 1} < \frac{m π}{2 m} = \frac{π}{2}$ $\Rightarrow \prod_{k ⩽ m} c o s (\frac{k π}{2 m + 1}) = \frac{1}{2^{m}}$ $p u t m = 999 \Rightarrow \prod_{k ⩽ 999} c o s (\frac{k π}{1999}) = \frac{1}{2^{999}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum