Hi Leandro,
Multiply both sides by sin x and then repeatedly use the formula for sin 2x.
Chris
Leandro wrote back
I understand... I saw in one site this ident of trig:
sin (mx)(cos nx) = 1/2(sen (m+n)x + sen (m  n)x)
and A = cos x . cos 2x . cos 4x . cos 8x
Making sin x on the both sides:
sin x . A = sin x . cos x . cos 2x . cos 4x . cos 8x
If sin 2x = 2 sin x cos x, so sin x cos x = sinx 2x / 2. So
sin x . A = 1/2(sin 2x . cos 2x . cos 4x . cos 8x)
So, sin 2x . cos 2x = 1/2(sen 4x + sen 0x)
sin x . A = 1/4(sen 4x . cos 4x . cos 8x)
sin x . A = 1/8(sen 8x . cos 8x)
sin x . A = 1/16(sin 16x)
A = 1/16(sin 16x/ sin x)
But I don't know what to do anymore :(
Leandro,
Now you can use properties of the logarithm functions, in particular the log base 2 function.
First for any logarithm
log(a b) = log(a) + log(b) and log(a/b) = log(a)  log(b)
second, for the log base 2
log_{2}(2^{p}) = p
Thus
log_{2}(A)
= log_{2}(2^{4}(sin 16x / sin x )
= log_{2}(2^{4}) + log_{2}(sin 16x / sin x )
= 4 +
log_{2}(sin 16x )  log_{2}(sin x )
Penny
