If $\text{cos α} = \dfrac{1}{2}$, find the value of 4 sin³ α - 3 sin α.

Given:

$\text{cos α} = \dfrac{1}{2}\\[1em] ⇒ \text{cos α} = \text{cos 60°}\\[1em]$

Thus, α = 60°.

Now,

$4 \text{sin}^3 α - 3 \text{sin} α\\[1em] = 4 \text{sin}^3 60° - 3 \text{sin} 60°\\[1em] = 4 \times \Big(\dfrac{\sqrt3}{2}\Big)^3 - 3 \times \dfrac{\sqrt3}{2}\\[1em] = 4 \times \dfrac{3\sqrt3}{8} - 3 \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{3\sqrt3}{2} - \dfrac{3\sqrt3}{2}\\[1em] = 0$

Hence, 4 sin³ α - 3 sin α = 0.