**If 2 cos 0 = 3, prove that 3 sin 0 – 4 sin3 0 = 1.**

**Answer :**

2 cos θ = √3

⇒ cos θ = √3/2

sin^{2} θ = 1 – cos^{2} θ

= 1 – (√3/2)^{2}

= 1 – ¾

= ¼

sin θ = √ ¼ = ½

LHS = 3 sin θ – 4 sin^{3} θ

= sinθ (3 – 4sin^{2} θ)

= ½ [3 – (4 × ¼)]

= ½ (3 – 1)

= ½ × 2

= 1

= RHS

Hence proved.

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