calculate trigonometric values..-1

Geometry Level 3

Evaluate $96.\sqrt{3}.sin\dfrac{\pi}{48}.cos\dfrac{\pi}{48}.cos\dfrac{\pi}{24}.cos\dfrac{\pi}{12}.cos\dfrac{\pi}{6}$

1 solution

Sujoy Roy
Dec 2, 2014

$96\sqrt{3}\sin\frac{\pi}{48}\cos\frac{\pi}{48}\cos\frac{\pi}{24}\cos\frac{\pi}{12}\cos\frac{\pi}{6}$

$=48\sqrt{3}(2\sin\frac{\pi}{48}\cos\frac{\pi}{48})\cos\frac{\pi}{24}\cos\frac{\pi}{12}\cos\frac{\pi}{6}$

$=24\sqrt{3}(2\sin\frac{\pi}{24}\cos\frac{\pi}{24})\cos\frac{\pi}{12}\cos\frac{\pi}{6}$

$=12\sqrt{3}(2\sin\frac{\pi}{12}\cos\frac{\pi}{12})\cos\frac{\pi}{6}$

$=6\sqrt{3}(2\sin\frac{\pi}{6}\cos\frac{\pi}{6})$

$=6\sqrt{3}(\sin\frac{\pi}{3})$

$=6\sqrt{3}(\frac{\sqrt{3}}{2})$

$=\boxed{9}$

Once you start solving it goes on automatically ! Btw Nice Solution ! :)

Keshav Tiwari - 6 years, 6 months ago