Question

Prove that
sin10sin30sin50sin70 = 1/16

Answer 1

Sin10*sin30*sin50*sin70
= sin10* $\frac{1}{2}$ *sin50*sin70
= $\frac{1}{2}$ * (sin10*sin50) * sin70
= $\frac{1}{4}$ * (2 sin10*sin50) * sin70
= $\frac{1}{4}$ * (cos(50-10)-cos(50+10)) * sin70

( using Cos(A-B) - Cos(A+B) = 2SinA SinB )

= $\frac{1}{4}$ * (cos40 - cos60) * sin70
= $\frac{1}{4}$ * (cos40-1/2) * sin70
= $\frac{1}{4}$ cos40 sin70 - $\frac{1}{8}$ sin70
= $\frac{1}{8}$ * (2 cos40 sin70) - $\frac{1}{8}$ sin70
= $\frac{1}{8}$ * (sin(70+40) + sin(70-40)) - $\frac{1}{8}$ sin70

( using Sin(A+B) + Sin(A-B) = 2CosA SinB )

= $\frac{1}{8}$ * (sin110 + sin30) - $\frac{1}{8}$ sin70
= $\frac{1}{8}$ * (sin110 + 1/2) - $\frac{1}{8}$ sin(180-70)
= $\frac{1}{8}$ * sin110 + $\frac{1}{16}$ - $\frac{1}{8}$ sin110
= $\frac{1}{16}$