Question

Prove sin3A= 3sinA-4sin³A and cos3A= 4cos³A-3cosA​

Answer 1

$\huge \: sin3A \\ \\ = sin(2A + A) \\ \\ = sin2AcosA + cos2AsinA \\ \\ = 2sinAcosA.cosA + (1 - 2 {sin}^{2} A)sinA \\ \\ = 2sinA {cos}^{2} A + sinA - 2 {sin}^{3} A \\ \\ = 2sinA(1 - {sin}^{2} A) + sinA - 2 {sin}^{3} A \\ \\ = 2sinA - 2 {sin}^{3} A + sinA - 2 {sin}^{3} A \\ \\ :\implies 3sinA - 4 {sin}^{3} A \\ \\ \\ \huge \: cos3A \\ \\ = cos(2A + A) \\ \\ = cos2AcosA - sin2AsinA \\ \\ = (2 {cos}^{2} A - 1)cosA - 2sinAcosA.sinA \\ \\ = 2 {cos}^{3} A - cosA - 2 {sin}^{2} AcosA \\ \\ = 2 {cos}^{3} A - cosA - 2(1 - {cos}^{2} A)cosA \\ \\ = 2 {cos}^{3} A - cosA - 2cosA + 2 {cos}^{3} A \\ \\ : \implies 4 {cos}^{3} A - 3cosA$