If $\text{2 sin x} = {\sqrt3}$, evaluate.

(i) 4 sin³ x - 3 sin x.

(ii) 3 cos x - 4 cos³ x.

(i) $\text{2 sin x} = {\sqrt3}$

⇒ sin x = $\dfrac{\sqrt3}{2}$

Taking cube on both sides, we get

$⇒ (\text {sin x})^3 = \Big(\dfrac{\sqrt3}{2}\Big)^3\\[1em] ⇒ \text {sin}^3 \text{x} = \dfrac{3\sqrt3}{8}\\[1em]$

Now, 4 sin³ x - 3 sin x

$= 4 \times \dfrac{3\sqrt3}{8} - 3 \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{12\sqrt3}{8} - \dfrac{3\sqrt3}{2}\\[1em] = \dfrac{12\sqrt3}{8} - \dfrac{3\sqrt3 \times 4}{2 \times 4}\\[1em] = \dfrac{12\sqrt3}{8} - \dfrac{12\sqrt3}{8}\\[1em] = 0$

Hence, 4 sin³ x - 3 sin x = 0.

(ii) sin x = $\dfrac{\sqrt3}{2}$

sin x = $\dfrac{Perpendicular}{Hypotenuse} = \dfrac{\sqrt3}{2}$

∴ If length of AB = $\sqrt{3}$ a unit, length of AC = 2a unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AO is hypotenuse)

⇒ (2a)² = $(\sqrt{3}\text{a})^2$ + BC²

⇒ 4a² = 3a² + BC²

⇒ BC² = 4a² - 3a²

⇒ BC² = a²

⇒ BC = $\sqrt{\text{a}^2}$

⇒ BC = a

cos x = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{a}{2a} = \dfrac{1}{2}$

Taking cube on both sides, we get

$⇒ (\text {cos x})^3 = \Big(\dfrac{1}{2}\Big)^3\\[1em] ⇒ \text {cos}^3 \text{x} = \dfrac{1}{8}\\[1em]$

Now, 3 cos x - 4 cos³ x

$= 3 \times \dfrac{1}{2} - 4 \times \dfrac{1}{8}\\[1em] = \dfrac{3}{2} - \dfrac{1}{2}\\[1em] = \dfrac{3 - 1}{2}\\[1em] = \dfrac{2}{2}\\[1em] = 1$

Hence, 3 cos x - 4 cos³ x = 1.