Prove that

Solution:

To prove: $3 \sin 40^{\circ}-\sin ^{3} 40^{\circ}=\frac{\sqrt{3}}{2}$

Taking LHS,

$=3 \sin 40^{\circ}-\sin ^{3} 40^{\circ} \ldots$ (i)

We know that

$\sin 3 x=3 \sin x-\sin ^{3} x$

Here, x = 40°

So, eq. (i) becomes

$=\sin 3\left(40^{\circ}\right)$

$=\sin 120^{\circ}$

$=\sin \left(180^{\circ}-60^{\circ}\right)$

$=\sin 60^{\circ}\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta\right]$

$=\frac{\sqrt{3}}{2}\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

= RHS

∴ LHS = RHS

Hence Proved