Question

Prove that sin 20(degree) sin 40(degree ) sin 80(degree) = √3/8​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :

$\sin 20^\circ \sin 40^\circ \sin 80 ^\circ = \frac{\sqrt{3}}{8}$

To Find :

$\text{Prove LHS = RHS}$

Solution :

$\text{LHS} =\sin 20^\circ \sin 40^\circ \sin 80 ^\circ$

$\implies \sin 20^\circ \sin (60^ \circ - \sin 20^ \circ)\sin (60^\circ + \sin 20^\circ)$

$( \because (a-b)(a+b)=a^2-b^2)$

$\implies \sin 20^\circ [ \sin^2 60^ \circ - \sin ^2 20^\circ]$

$\implies \sin 20^\circ [(\frac{\sqrt{3}}{2} )^2 - \sin^2 20^\circ] (\because \sin 60^\circ = \frac{\sqrt{3}}{2})$

$\implies \sin 20^\circ[\frac{3}{4}-\sin^2 20 ^\circ ]$

$\implies \sin 20^ \circ [ \frac{3-4 \sin ^2 20 ^\circ}{4}]$

$\implies \frac{3 \sin 20^\circ - 4 \sin^3 20^\circ }{4}$ $(\because \sin 3A = 3 \sin A - 4 \sin ^3 A)$

$\implies \frac {\sin 3(20^\circ)}{4}$

$\implies \frac{\sin 60^\circ}{4} \implies \frac{\frac{\sqrt{3}}{2}}{4} \implies \frac{\sqrt{3}}{2 \times 4}$

$\boxed {\frac{\sqrt{3}}{8}} = \text{RHS}$

Hence Proved.