Prove that sin3A=3sinA-4sin^3A.

 sin3A = sin(2A+A)

= sin 2A cos A+cos 2AsinA

= 2sinAcosA.cos A+(1-2sin²A)sinA

[Here sin2A=2sinAcosA and cos2A=2cos²-1=1-2sin²A]

=2sinAcos²A+(1-2sin²A)sinA

=2sinA(1-sin²A)+(1-2sin²A)sinA

[Here sin²A+cos²A=1 => cos²A=1-sin²A]

=2sinA-2sin^3A+sinA-2sin^3A

=3sinA-4sin^3A