Prove that Cos3A = 4cos^3A - 3cosA ?

ANSWER)

cos (3A)

= 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