Q. 43 B3.7( 3 Votes )

# 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

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