Question

$sin \,12^{\circ} \,sin \,24^{\circ} \,sin \,48^{\circ}\, sin \,84^{\circ} =$

• A. $cos \,20^{\circ} \,cos \,40^{\circ} \,cos\,60^{\circ}\, cos \,80^{\circ} $
• B. $sin \,20^{\circ} \,sin \,40^{\circ} \,sin \,60^{\circ}\, sin \,80^{\circ} $
• C. $\frac{3}{15}$
• D. None of these

Answer 1

Answer: (A)

$sin \,12^{\circ} \,sin \,24^{\circ} \,sin \,48^{\circ}\, sin \,84^{\circ} $
$= \frac{1}{4}\left(2\,sin \,12^{\circ} \,sin \,48^{\circ}\right)\,\left(2\,sin \,24^{\circ } \,sin \,84^{\circ }\right)$
$= \frac{1}{2}\left(cos \,36^{\circ }- cos \,60^{\circ }\right)\,\left(cos \,60^{\circ }- cos \,108^{\circ }\right)$
$= \frac{1}{4}\left(cos \,36^{\circ }-\frac{1}{2}\right)\left(\frac{1}{2}+sin \,18^{\circ }\right)$
$= \frac{1}{4}\left\{\frac{1}{4}\left(\sqrt{5}+1\right)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}\left(\sqrt{5}-1\right)\right\} = \frac{1}{16}$
and $cos \,20^{\circ} \,cos \,40^{\circ} \,cos\,60^{\circ}\, cos \,80^{\circ} $
$= \frac{1}{2}\left[cos\left(60^{\circ}-20^{\circ}\right)cos \,20^{\circ } \,cos \left(60^{\circ }+20^{\circ }\right)\right] $
$= \frac{1}{2}\left[\frac{1}{4}cos\,3\left(20^{\circ}\right)\right] = \frac{1}{8} \,cos\,60^{\circ } = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}.$