If 4 cos² A - 3 = 0, show that :

cos 3A = 4 cos³ A - 3 cos A

Given,

⇒ 4 cos² A - 3 = 0

⇒ 4 cos² A = 3

⇒ cos² A = $\dfrac{3}{4}$

⇒ cos A = $\sqrt{\dfrac{3}{4}}$

⇒ cos A = $\dfrac{\sqrt{3}}{2}$.

⇒ cos A = cos 30°

⇒ A = 30°.

L.H.S. = cos 3A = cos 3(30°) = cos 90° = 0.

R.H.S. = 4 cos³ A - 3 cos A

= 4 cos³ 30° - 3 cos 30°

= $4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 - 3 \times \dfrac{\sqrt{3}}{2}$

= $4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2}$

= $\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2}$

= 0.

Since, L.H.S. = R.H.S.

Hence, proved that cos 3A = 4 cos³ A - 3 cos A.