find-the-value-of-i-1-999-cos-ia-where-a-2pi-1999-

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 \dots \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) \dots (2 \sin 999 a \cos 999 a)

= \sin 2 a \sin 4 a \sin 6 a \dots \sin 1998 a

= (\sin 2 a \sin 4 a \sin 6 a \dots \sin 998 a) (- \sin (2 π - 1000 a)) . (- \sin (2 π - 10002 a)) \dots (- \sin (2 π - 1998 a))

= \sin 2 a \sin 4 a \sin 6 a \dots \sin 998 a \sin 999 a \sin 997 a \dots \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}}