Q. 43 B3.7( 3 Votes )

If , prove that: 3sinθ – 4sin3θ = 1.

Answer :


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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