Question

If cos a . cos 2a . cos 3a ............... cos999a = $\frac{1}{2^x}$.Where a = $\frac{2\pi }{1999}$.Find the value of x.

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Answer 1

Given series cosa .cos 2a . cos 3a . . . . . . cos 999 is

Let p = cos a cos 2a cos 3a . . . . . . cos 999a

To proceed any cosine product series, we should multiple and divide by same angle sine function and use the formula sin 2a = 2 sin a . cos a

Let Q = sina .sin2a . sin 3a . . . . . . sin 999a

P $\times$ Q = (sin a .cos a) . (sin 2a . cos 2a) . - - - - - . (sin 999a. cos 999a)multiply both side by $2^{999}$

$2^{999}$ P $\times$ Q = (2 sina .cos a) . (2 sin 2a .cos 2a) . - - - - - - (2 sin 999a .cos 999a)

$2^{999}$ PQ = sin 2a .sin 4a . sin 8a . - - - - - - sin 1998a - - - - - - (1)

Let's take RHS here.

sin 2a . sin 4a . sin 8a . - - - - - - sin 998 .sin 1000a . 1002a - - - - - - sin 1998a

After sin 998a,

sin 1000a = sin 1000 $\times \frac{2\pi }{1999}$ = -sin$(2\pi -1000\times \frac{2\pi }{1999})$ $\{a=\frac{2\pi }{1999}\}$

= -sin$\left(2\pi (1-\frac{1000}{1999})\right)$

= -sin$\left(2\pi \frac{999}{1999}\right)$

= -sin$\left(999.\frac{2\pi }{1999}\right)$ = -sin 999a

Similarly sin 10002a = sin 1002 $\times$ -sin$(2\pi -1002\times \frac{2\pi }{1999})$

$=-sin\left\{2\pi (1-\frac{1002}{1999})\right\}$

= -sin2a $\times$ $\frac{997}{1999}$ = -sin 997a

sin 1004a = -sin 995a

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sin 1998a = -sin a

When we multiply the terms after sin 998a we will get all the odd angles terms sin a $\times$ sin 3a $\times$ sin 5a . . . . . sin 999a

RHS

= sin 2a .sin 4a .sin 8a . . . . . . sin 998a {(-sin 999a) (-sin 997a) (-sin 995a) . . . . . . (-sin a)}

(-) sign will be 998 times after multiplication 998 times it becomes positive (+ 1)

= sin a .sin 2a . sin 3a . sin 4a . sin 5a . . . . . sin 999a

= Q

From equation 1

$2^{999}$ PQ = Q

p = $\frac{1}{2^{999}}$

p = $\frac{1}{2^x}$

x = 999