Quandaries and Queries Who is asking: Student Level: Secondary Question: A = cos x * cos 2x * cos 4x * cos 8x What's the value of log A at base 2? 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 log2(2p) = p Thus log2(A) = log2(2-4(sin 16x / sin x ) = log2(2-4) + log2(sin 16x / sin x ) = -4 + log2(sin 16x ) - log2(sin x ) Penny Go to Math Central