If 2 cos θ = $\sqrt{3}$, prove that 3 sin θ - 4 sin³ θ = 1.

Given,

⇒ 2 cos θ = $\sqrt{3}$

⇒ cos θ = $\dfrac{\sqrt{3}}{2}$

Squaring both sides we get :

⇒ cos² θ = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2$ = $\dfrac{3}{4}$.

⇒ 1 - sin² θ = $\dfrac{3}{4}$.

⇒ sin² θ = $1 - \dfrac{3}{4}$.

⇒ sin² θ = $\dfrac{4 - 3}{4}$.

⇒ sin² θ = $\dfrac{1}{4}$.

⇒ sin θ = $\sqrt{\dfrac{1}{4}}$

⇒ sin θ = $\dfrac{1}{2}$.

Substituting values in L.H.S. of equation 3 sin θ - 4 sin³ θ = 1 we get :

$\Rightarrow 3\text{sin θ} - 4\text{ sin}^3 θ \\[1em] \Rightarrow \text{sin θ }(3 - 4\text{ sin}^2 θ) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 4 \times \dfrac{1}{4}) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 1) \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \\[1em] \Rightarrow 1.$

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

Hence, proved that 3sin θ - 4 sin³ θ = 1.