Question

Sin10.sin50.sin60.sin70=√3/16 prove this

Answer 1

Sin(x)*sin(60+x)*sin(60-x)=(sin3x)/4
Sin10sin50sin70=(sin(3*10))/4=1/8
Sin60*1/8=(√3/2)*1/8=√3/16
Proved

Answer 2

Answer and explanation:

To prove : $\sin10\cdot \sin50\cdot \sin60\cdot \sin70=\frac{\sqrt3}{16}$

Proof :

Taking LHS,

$\sin10\cdot \sin50\cdot \sin60\cdot \sin70$

$=\frac{\sqrt 3}{2}(\sin10\cdot \sin50\cdot \sin70)$

Applying identity, $\sin A\sin B=\frac{\cos(A-B)-\cos (A+B)}{2}$

$=\frac{\sqrt 3}{2}(\sin10\cdot\frac{\cos(50-70)-\cos (50+70)}{2})$

$=\frac{\sqrt 3}{2}(\sin10\cdot\frac{\cos(-20)-\cos (120)}{2})$

$=\frac{\sqrt 3}{2}(\frac{\sin10cos(20)-\sin 10\cos (120)}{2})$

$=\frac{\sqrt 3}{2}(\sin 30+\sin(-10)-2\sin 10\cos 120}{4})$

$=\frac{\sqrt 3}{2}(\sin 30+\sin(-10)+\sin 10}{4})$

$=\frac{\sqrt 3}{2}(\frac{1}{2}-\sin(10)+\sin 10}{4})$

$=\frac{\sqrt 3}{2\times 2\times 4}$

$=\frac{\sqrt 3}{16}$

=RHS

Hence proved.