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Mathematics Question on Inverse Trigonometric Functions

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

A

cos 20∘ cos 40∘ cos 60∘ cos 80∘cos \,20^{\circ} \,cos \,40^{\circ} \,cos\,60^{\circ}\, cos \,80^{\circ}

B

sin 20∘ sin 40∘ sin 60∘ sin 80∘sin \,20^{\circ} \,sin \,40^{\circ} \,sin \,60^{\circ}\, sin \,80^{\circ}

C

315\frac{3}{15}

D

None of these

Answer

cos 20∘ cos 40∘ cos 60∘ cos 80∘cos \,20^{\circ} \,cos \,40^{\circ} \,cos\,60^{\circ}\, cos \,80^{\circ}

Explanation

Solution

sin 12∘ sin 24∘ sin 48∘ sin 84∘sin \,12^{\circ} \,sin \,24^{\circ} \,sin \,48^{\circ}\, sin \,84^{\circ} =14(2 sin 12∘ sin 48∘) (2 sin 24∘ sin 84∘)= \frac{1}{4}\left(2\,sin \,12^{\circ} \,sin \,48^{\circ}\right)\,\left(2\,sin \,24^{\circ } \,sin \,84^{\circ }\right) =12(cos 36∘−cos 60∘) (cos 60∘−cos 108∘)= \frac{1}{2}\left(cos \,36^{\circ }- cos \,60^{\circ }\right)\,\left(cos \,60^{\circ }- cos \,108^{\circ }\right) =14(cos 36∘−12)(12+sin 18∘)= \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∘ cos 40∘ cos 60∘ cos 80∘cos \,20^{\circ} \,cos \,40^{\circ} \,cos\,60^{\circ}\, cos \,80^{\circ} =12[cos(60∘−20∘)cos 20∘ cos(60∘+20∘)]= \frac{1}{2}\left[cos\left(60^{\circ}-20^{\circ}\right)cos \,20^{\circ } \,cos \left(60^{\circ }+20^{\circ }\right)\right] =12[14cos 3(20∘)]=18 cos 60∘=12×18=116.= \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}.