Question 466502
First we should rewrite cos 84 as sin 6 (these two are equal). Denote our desired answer A (which should turn out to be 1/16)


*[tex \LARGE (\sin 6)(\cos 12)(\cos 24)(\cos 48) = A] (remember sin 6 is the cos 84)


Suppose we multiply both sides by 16 cos 6:


*[tex \LARGE 16 (\sin 6 )(\cos 6)(\cos 12)( \cos 24 )(\cos 48) = 16 (\cos 6)A] 


Applying the double angle identity for sine, we have 2 sin 6 cos 6 = sin 12, and we replace sin 6 cos 6 with (1/2) sin 12 (here I will divide coefficient by 2)


*[tex \LARGE 8(\sin 12)(\cos 12)(\cos 24)(\cos 48) = 16 (\cos 6)A]


We apply it again and again:


*[tex \LARGE 4(\sin 24)(\cos 24)(\cos 48) = 16 (\cos 6)A]


*[tex \LARGE 2(\sin 48)(\cos 48) = 16(\cos 6)A]


*[tex \LARGE \sin 96 = 16 (\cos 6)A]


Here, sin 96 is equal to cos 6 (you can verify this either using the sum identity or the unit circle), so we can cancel them out, leaving


*[tex \LARGE 1 = 16A]


*[tex \LARGE A = \frac{1}{16}], as desired.