96cos(π)/(33)cos(2π)/(33)cos(4π)/(33)cos(8π)/(33)cos(16π)/(33) is...

Given,

Expression $96 \cdot cos\dfrac{\pi}{33} \cdot cos\dfrac{2\pi}{33} \cdot cos\dfrac{4\pi}{33} . . . . . . . . . cos\dfrac{16\pi}{33}$

Now we know that,

$cosA \cdot \cos 2A \cdot cos{2}^{2}A \cdot cos{2}^{3}A . . . . . \cdot cos{2}^{n - 1}A = \dfrac{sin{2}^{n}A}{{2}^{n}sinA}$

Now using the above formula in given expression we get,

$96 \cdot cos\dfrac{\pi}{33} \cdot cos\dfrac{2\pi}{33} \cdot cos\dfrac{4\pi}{33} . . . . . . . . . cos\dfrac{16\pi}{33}$

$= 96 \times \dfrac{sin\dfrac{32\pi}{33}}{{2}^{5}sin\dfrac{\pi}{33}}$

$= 96 \times \dfrac{\sin \left(\pi - \dfrac{\pi}{33}\right)}{{2}^{5}sin\dfrac{\pi}{33}}$

$= 96 \times \dfrac{\sin \left(\dfrac{\pi}{33}\right)}{{2}^{5}sin\dfrac{\pi}{33}}\left\{\text{as}\sin \left(\pi - \alpha\right) = sin\alpha\right\}$

$= 96 \times \dfrac{1}{32} = 3$