Prove that Sin 10 degree sin 30 degree sin 50 degree sin 70 degree = 1/16?

Jun 2, 2018

$L H S = \sin 10 \sin 30 \sin 50 \sin 70$

$= \cos \left(90 - 10\right) \sin 30 \cos \left(90 - 50\right) \cos \left(90 - 70\right)$

$= \cos \left(80\right) \cdot \frac{1}{2} \cdot \cos \left(40\right) \cos \left(20\right)$
$= \frac{1}{4 \sin 20} \cos \left(80\right) \cos \left(40\right) \cdot 2 \sin 20 \cos \left(20\right)$

$= \frac{1}{8 \sin 20} \cos \left(80\right) \cdot 2 \cos \left(40\right) \sin \left(40\right)$

$= \frac{1}{16 \sin 20} \cdot 2 \cos \left(80\right) \sin \left(80\right)$

$= \frac{1}{16 \sin 20} \cdot \sin \left(160\right)$

$= \frac{1}{16 \sin 20} \cdot \sin \left(180 - 20\right)$

$= \frac{1}{16 \sin 20} \cdot \sin \left(20\right) = \frac{1}{16} = R H S$

Jun 2, 2018

As, $\sin x \sin \left(60 + x\right) \sin \left(60 - x\right) = \frac{1}{4} \sin 3 x$

$\rightarrow \sin 10 \sin 30 \sin 50 \sin 70$

$= \frac{1}{2} \left[\sin 10 \sin \left(60 + 10\right) \sin \left(60 - 10\right)\right]$

$= \frac{1}{2} \left[\frac{1}{4} \sin \left(3 \times 10\right)\right] = \frac{1}{8} \sin 30 = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}$