Question

cos3A=4cos^3A-3cosA prove that

Answer 1

LHS= cos(2A + A)

= cos (2A) cos (A) - sin(2A) sin(A)

= [ 2cos^2(A) - 1 ] cos (A) - (2 sin A cos A )sin A

= 2cos^3(A) - cos A - 2sin^2(A) cos A

= 2cos^3(A) - cos A - 2( 1 - cos^2(A)) cos A

= 2cos^3(A) - cos A - 2cos A + 2cos^3(A)

= 4cos^3(A) - 3cos A=RHS.

this answer may help u

Answer 2

Solution :

L.H.S.

$= cos3A \\ \\ = cos(2A + A) \\ \\ = cos2A \: cosA - sin2A \: sinaA \\ \\ =(2 {cos}^{2} A - 1)cosA - 2sinA \: cosA \: sinA \\ \\ = 2 {cos}^{3} A - cosA - 2 {sin}^{2} A \: cosA \\ \\ = 2 {cos}^{3} A - cosA - 2(1 - {cos}^{2} A)cosA \\ \\ = 2 {cos}^{3} A - cosA - 2cosA + 2 {cos}^{3} A \\ \\ = 4 { cos}^{3} A - 3cosA$

= R.H.S. [Proved]

Rules :

$cos2A = 2 {cos}^{2} A - 1 \\ \\ sin2A = 2sinA \: cosA \\ \\ {sin}^{2} A + {cos}^{2} A = 1$