If , prove that: 3sinθ – 4sin³θ = 1.

We know that,

Let AB =k√3 and BC = 2k

In right angled ∆ABC, we have

B² + P² = H²

⇒ (k√3)² + P² = (2k)²

⇒ P² + 3k² = 4k²

⇒ P² = 4k² – 3k²

⇒ P² = k²

⇒ P =√k²

⇒ P = ±k

⇒ P = k [taking positive square root since, side cannot be negative]

Now,

We know that,

Or

Now, LHS = 3sin θ – 4sin³ θ

⇒ 1 = RHS

Hence Proved