Prove that 4sinαsin(α+π/3)sin(α+2π/3) = sin3α
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LHS = 4sina sin( a + π/3) sin(a +2π/3)
= 2{ 2sina.sin(a +π/3).sin(a +2π/3)
=2{cos(a -a -π/3) - cos( 2a +π/3) }sin(a +2π/3)
=2×1/2sin(a +2π/3) - 2sin(a +2π/3).cos(2a +π/3)
=sin(a + 2π/3) - sin(3a + π) -sin( a +2π/3)
=sin3a =RHS
= 2{ 2sina.sin(a +π/3).sin(a +2π/3)
=2{cos(a -a -π/3) - cos( 2a +π/3) }sin(a +2π/3)
=2×1/2sin(a +2π/3) - 2sin(a +2π/3).cos(2a +π/3)
=sin(a + 2π/3) - sin(3a + π) -sin( a +2π/3)
=sin3a =RHS
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