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Mathematics Question on Trigonometric Functions

The value of sin⁡10∘⋅sin⁡30∘⋅sin⁡50∘−sin⁡70∘\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ}-\sin 70^{\circ} is

A

18\frac{1}{8}

B

316\frac{3}{16}

C

316\frac{\sqrt{3}}{16}

D

116\frac{1}{16}

Answer

116\frac{1}{16}

Explanation

Solution

sin⁡10∘⋅sin⁡30∘⋅sin⁡50∘⋅sin⁡70∘\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}
=12⋅sin⁡10∘⋅12(2sin⁡70∘⋅sin⁡50∘)=\frac{1}{2} \cdot \sin 10^{\circ} \cdot \frac{1}{2}\left(2 \sin 70^{\circ} \cdot \sin 50^{\circ}\right)
=\frac{1}{2} \sin 10^{\circ} \cdot \frac{1}{2}\left\\{\cos \left(70^{\circ}, 50^{\circ}\right)\right.\\\ \left.-\cos \left(70^{\circ}+50^{\circ}\right)\right\\}
=\frac{1}{2} \sin 10^{\circ} \cdot \frac{1}{2}\left\\{\cos 20^{\circ}-\cos 120^{\circ}\right\\}
=12sin⁡10∘⋅12(cos⁡10∘+12)=\frac{1}{2} \sin 10^{\circ} \cdot \frac{1}{2}\left(\cos 10^{\circ}+\frac{1}{2}\right)
=14sin⁡10∘⋅cos⁡20∘+18sin⁡10∘=\frac{1}{4} \sin 10^{\circ} \cdot \cos 20^{\circ}+\frac{1}{8} \sin 10^{\circ}
=14⋅12(sin⁡30∘−sin⁡10∘)+18⋅sin⁡10∘=\frac{1}{4} \cdot \frac{1}{2}\left(\sin 30^{\circ}-\sin 10^{\circ}\right)+\frac{1}{8} \cdot \sin 10^{\circ}
=18⋅sin⁡30∘−18sin⁡10∘+1/8sin⁡10∘=\frac{1}{8} \cdot \sin 30^{\circ}-\frac{1}{8} \sin 10^{\circ}+1 / 8 \sin 10^{\circ}
=18⋅12−0=116=\frac{1}{8} \cdot \frac{1}{2}-0=\frac{1}{16}