Question

prove that
sin3A=3sinA-4sin^3

Answer 1

heya

sin(2a+a)

=sin2a.cosa+cos2a.sina

=2sina.cosa.cosa+(cos^2 a-sin^2a)sina

=2sina.cos^2 a+sina-2sin^3a

=2sina(1-sin^2a)+sina-2sin^3 a

=2sina-2sin^3a+sina-2sin^3a

=3sin-4sin^3 a

hope it help you

@rajukumar.☺

Answer 2

$sin3A \\ \\ = sin(2A + A) \\ \\ = sin2AcosA + cos2AsinA \\ \\ = 2sinAcosA.cosA + (1 - 2 {sin}^{2} A)sinA \\ \\ = 2sinA {cos}^{2} A + sinA - 2 {sin}^{3} A \\ \\ = 2sinA(1 - {sin}^{2} A) + sinA - 2 {sin}^{3} A \\ \\ = 2sinA - 2 {sin}^{3} A + sinA - 2 {sin}^{3} A \\ \\ = 3sinA - 4 {sin}^{3} A$