Question

Q17. Prove that: sin 10sin 50sin 60sin 70 = .$\frac{ \sqrt{3}}{16}$

6d758920f194723bb97a43c131af5167.docx

Answer 1

we can write it as
sin60sin10sin50sin70=$\sqrt{3} /16$
=$\sqrt{3} /2$(sin10sin50sin70)
=$\sqrt{3} /2$*1/2(2sin10sin50sin70)
=$\sqrt{3} /4$(2sin10sin50.sin70)
since
{2sinAsinB=cos(A-B)-cos(A+B)}
=$\sqrt{3} /4$(cos(50-10)-cos(50+10)*)sin70
=$\sqrt{3} /4$(coc40-cos60)sin70
=$\sqrt{3} /4$(sin70cos40-sn70cos60)
=$\sqrt{3} /8$(2sin70cos40-sin70)
since
{2sinAcosB=sin(A+B)+sin(A-B)}
=$\sqrt{3} /8$(sin(70+40)+sin(70-40)-son70)
=$\sqrt{3} /8$(sin110+sin30-sin70)
=$\sqrt{3} /8$(sin70+1/2-sin70)
=$\sqrt{3} /16$