Let $A = | ({log}_{2} | cos 24^{o} |) + ({log}_{2} | cos 48^{o} |) + ({log}_{2} | cos 96^{o} |) + ({log}_{2} | cos 192^{o} |) |$ ?

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Solution

$A = l o g_{2} (cos 24. cos 48. cos 96. cos 192)$

$\Rightarrow$ Multiply numerator & denominator by $2 sin 24$

$= \frac{(2. sin 24. cos 24)}{2 sin 24} . cos 48 . cos 96 . cos 192$

$= \frac{(2 . sin 48 . cos 48)}{4 sin 24} cos 96. cos 192$

$= \frac{2 . sin 96 . cos 96}{8 sin 24} cos 192$

$= \frac{2. sin 192. cos 192}{16 . sin 24} = \frac{sin 384}{16 sin 24} = \frac{1}{16}$

$A = l o g_{2}^{\frac{1}{16}} l o g_{2}^{2 - 4} = - 4 l o g_{2}^{2}$

$A = - 4$

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