Sin theta - 2 sin ^3 theta / 2 cos^3 - cos theta = tan theta.......prove this
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(Sinθ - 2Sin³θ) / ( 2Cos³θ - Cosθ) = Tanθ
Step-by-step explanation:
(Sinθ - 2Sin³θ) / ( 2Cos³θ - Cosθ) = Tanθ
LHS = (Sinθ - 2Sin³θ) / ( 2Cos³θ - Cosθ)
= Sinθ( 1 - 2Sin²θ) /Cosθ( 2Cos²θ - 1)
1 = Cos²θ + Sin²θ
= Sinθ( Cos²θ + Sin²θ - 2Sin²θ) /Cosθ( 2Cos²θ - ( Cos²θ + Sin²θ))
= Sinθ( Cos²θ - Sin²θ) /Cosθ( 2Cos²θ - Cos²θ - Sin²θ)
= Sinθ( Cos²θ - Sin²θ) /Cosθ( Cos²θ - Sin²θ)
Cancelling Cos²θ - Sin²θ from numerator & denominator
= Sinθ /Cosθ
= Tanθ
= RHS
QED
proved
(Sinθ - 2Sin³θ) / ( 2Cos³θ - Cosθ) = Tanθ
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