Question

$sin12^{\circ }sin24^{\circ }sin48^{\circ }sin84^{\circ }=$

• A. $cos20^{\circ }cos40^{\circ }cos60^{\circ }cos80^{\circ }$
• B. $sin20^{\circ }sin40^{\circ }sin60^{\circ }sin80^{\circ }$
• C. $\frac{3}{15}$
• D. None of these

Answer 1

Answer: (A)

$sin12^{\circ }sin24^{\circ }sin48^{\circ }sin84^{\circ }$
$=\frac{1}{4}\left(2sin12^{\circ }sin48^{\circ }\right)\left(2sin24^{\circ }sin84^{\circ }\right)$
$=\frac{1}{2}\left(cos36^{\circ }-cos60^{\circ }\right)\left(cos60^{\circ }-cos108^{\circ }\right)$
$=\frac{1}{4}\left(cos36^{\circ }-\frac{1}{2}\right)\left(\frac{1}{2}+sin18^{\circ }\right)$
$=\frac{1}{4}\left\{\frac{1}{4}\left(\sqrt{5}+1\right)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}\left(\sqrt{5}-1\right)\right\}=\frac{1}{16}$
and $cos20^{\circ }cos40^{\circ }cos60^{\circ }cos80^{\circ }$
$=\frac{1}{2}\left[cos\left(60^{\circ }-20^{\circ }\right)cos20^{\circ }cos\left(60^{\circ }+20^{\circ }\right)\right]$
$=\frac{1}{2}\left[\frac{1}{4}cos3\left(20^{\circ }\right)\right]=\frac{1}{8}cos60^{\circ }=\frac{1}{2}\times \frac{1}{8}=\frac{1}{16}.$