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Mathematics Question on Trigonometric Functions

The value of sin⁡ 20∘ sin⁡ 40∘sin⁡ 60∘ sin⁡ 80∘\sin \, 20^{\circ} \, \sin \, 40^{\circ} \sin \, 60^{\circ} \, \sin \, 80^{\circ} is

A

−316\frac{-3}{16}

B

516\frac{5}{16}

C

316\frac{3}{16}

D

116\frac{1}{16}

Answer

316\frac{3}{16}

Explanation

Solution

Indeed sin⁡ 20∘ sin⁡ 40∘ sin⁡ 60∘ sin⁡ 80∘\sin \, 20^{\circ} \, \sin \, 40^{\circ} \, \sin \, 60^{\circ} \, \sin \, 80^{\circ}. =32sin⁡ 20∘sin⁡(60∘−20∘)sin⁡(60∘+20∘) = \frac{\sqrt{3}}{2} \sin \, 20^{\circ} \sin (60^{\circ} - 20^{\circ} ) \sin (60^{\circ} + 20^{\circ} ) (since sin⁡60∘=32\sin 60^{\circ} = \frac{\sqrt{3}}{2} ) =32sin⁡20∘[sin⁡260∘−sin⁡220∘] = \frac{\sqrt{3}}{2} \sin20^{\circ} \left[\sin^{2} 60^{\circ} - \sin^{2}20^{\circ}\right] =32sin⁡20∘[34−sin⁡220∘]= \frac{\sqrt{3}}{2} \sin 20^{\circ} \left[\frac{3}{4} - \sin^{2} 20^{\circ}\right] =32×14[3sin⁡20∘−4sin⁡320∘]= \frac{\sqrt{3}}{2} \times\frac{1}{4} \left[3 \sin20^{\circ} - 4 \sin^{3} 20^{\circ}\right] =32×14(sin⁡60∘)= \frac{\sqrt{3}}{2} \times\frac{1}{4} \left(\sin60^{\circ}\right) =32×14×32=316= \frac{\sqrt{3}}{2} \times \frac{1}{4} \times\frac{\sqrt{3}}{2} = \frac{3}{16}