Question

Express cos 3α in terms of cos α and tan 3α in terms of tan α

Answer 1

Cos3x = cos (2x+ x)
=cos2x*cosx - sin2x*sinx
=(2*cos^2 (x) - 1) *cosx - 2sinx*cosx*sinx { since cos2x = 2cos^2 (X) -1 , sin2x = 2*sinx*cosx }
=2 cos^3 (x) - cos x - 2 cosx *sin^2 (x)
= 2 cos^3 (x) - cos x (1 + 2 sin^2 (x) )
= 2 cos^3 (x) - cos x (1 + 2 ( 1 - cos^2 (x) ))
= 2 cos^3 (x) - cos x ( 1+ 2 - 2*cos^2(x) )
=2 cos^3 (x) - cos x (3 -2*cos^2(x) )
=2 cos^3 (x) - 3cos x + 2*cos^3(x)
=4 cos^3 (x) - 3cos x

tan3x=tan(2x+x)        
         =tan2x+tanx / 1-tan2x*tanx       
         =2 tanx/1-t 2 x +tanx /1-2tanx*tanx/1-tan 2 x        
         =2tanx+tanx-tan 3 x/tan 2 x-2tan 2 x       
          =3tanx-tan 3 x/1-3tan 2 x