If 4 cos² A - 3 = 0 and 0° ≤ A ≤ 90°; then prove that :

(i) sin 3A = 3 sin A - 4 sin³ A

(ii) 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 = $\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}$.

⇒ cos A = cos 30°

⇒ A = 30°.

(i) To prove:

sin 3A = 3 sin A - 4 sin³ A

Solving L.H.S. of the equation :

⇒ sin 3A = sin 3(30°)

= sin 90° = 1.

Solving R.H.S. of the equation :

⇒ 3 sin A - 4 sin³ A

⇒ 3 sin 30° - 4 sin³ 30°

⇒ 3 $\times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3$

⇒ $\dfrac{3}{2} - 4 \times \dfrac{1}{8}$

⇒ $\dfrac{3}{2} - \dfrac{1}{2}$

⇒ $\dfrac{2}{2}$

⇒ 1.

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

Hence, proved that sin 3A = 3 sin A - 4 sin³ A.

(ii) To prove:

cos 3A = 4 cos³ A - 3 cos A

Solving L.H.S.

⇒ cos 3A = cos 3(30°) = cos 90° = 0.

Solving 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