# "This can be done by twice using the double angle formula for cos." #
# "We have:" #
# \qquad \qquad cos2x \ = \ cos^2x - sin^2 x; \qquad \quad \ color{blue}{"double angle formula for cos"} #
# \qquad \qquad \qquad \qquad \quad \ \ = \ cos^2x - ( 1 - cos^2x ) #
# \qquad \qquad \qquad \qquad \quad \ \ = \ cos^2x - 1 + cos^2x #
# \qquad \qquad \qquad \qquad \quad \ \ = \ 2 cos^2x - 1. #
# "Thus:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad cos2x \ = \ 2 cos^2x - 1. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (I) #
# "So now:" #
# \qquad \qquad cos4x \ = \ cos( 2 cdot (2 x ) ); \qquad \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{"now let:" \qquad A = 2x"} #
# \qquad \qquad \qquad= \ cos( 2 A ); #
# \qquad \qquad \qquad= \ 2 cos^2A - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ color{blue}{"by (I)"} #
# \qquad \qquad \qquad= \ 2 ( cosA )^2 - 1; #
# \qquad \qquad \qquad= \ 2 ( cos 2 x )^2 - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{"since:" \qquad A = 2x"} #
# \qquad \qquad \qquad= \ 2 ( 2 cos^2x - 1 )^2 - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ color{blue}{"by (I)"} #
# \qquad \qquad \qquad= \ 2 ( [ 2 cos^2x ]^2 - 2 [ 2 cos^2x ] + 1 ) - 1; #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{ "as:" qquad ( a + b )^2 = a^2 + 2 a b + b^2 }#
# \qquad \qquad \qquad= \ 2 ( 4 cos^4x - 4 cos^2x + 1 ) - 1; #
# \qquad \qquad \qquad= \ 8 cos^4x - 8 cos^2x + 2 - 1; #
# \qquad \qquad \qquad= \ 8 cos^4x - 8 cos^2x + 1. #
# "This is what we wanted to show !!" #
# "Summarizing, we have shown:" #
# \qquad \qquad \qquad \qquad \qquad \qquad cos4x \ = \ 8 cos^4x - 8 cos^2x + 1. #
