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Question

Question: $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $...

sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘=\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ =

A

32\frac{3}{2}

B

34\frac{3}{4}

C

316\frac{3}{16}

D

38\frac{3}{8}

Answer

316\frac{3}{16}

Explanation

Solution

We have:

sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘=(sin⁡20∘sin⁡40∘sin⁡80∘)sin⁡60∘\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \left(\sin20^\circ \sin40^\circ \sin80^\circ\right)\sin60^\circ

It is a known trigonometric identity that:

sin⁡20∘sin⁡40∘sin⁡80∘=38\sin20^\circ \sin40^\circ \sin80^\circ = \frac{\sqrt{3}}{8}

And since:

sin⁡60∘=32\sin60^\circ = \frac{\sqrt{3}}{2},

we obtain:

sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘=38×32=316\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2} = \frac{3}{16}.