Prove that

Solution:

To Prove: $8 \cos ^{3} 20^{\circ}-6 \cos 20^{\circ}=1$

Taking LHS,

$=8 \cos ^{3} 20^{\circ}-6 \cos 20^{\circ}$

Taking 2 common, we get

$=2\left(4 \cos ^{3} 20^{\circ}-3 \cos 20^{\circ}\right) \ldots$ (i)

We know that,

$\cos 3 x=4 \cos ^{3} x-3 \cos x$

Here, x = 20°

So, eq. (i) becomes

$=2\left[\cos 3\left(20^{\circ}\right)\right]$

$=2\left[\cos 60^{\circ}\right]$

$=2 \times \frac{1}{2}\left[\because \cos \left(60^{\circ}\right)=\frac{1}{2}\right]$

= 1

= RHS

∴ LHS = RHS

Hence Proved