Question 930463
{{{3-4cos(2x)+cos(4x)/4}}} is = {{{2sin^4(x)}}}....

here is proof:

{{{2sin^2(x)sin^2(x )}}}...use {{{sin^2(x)=1-cos^2(x)}}}


{{{2(1-cos^2(x))(1-cos^2(x)) }}}

{{{2(1-cos^2(x))^2 }}}

{{{2(1-2cos^2(x)+cos^4(x))}}}


{{{(2-4cos^2(x)+2cos^4(x))}}}.........since {{{2cos^4(x)=(1/4) (4cos(2x)+cos(4x)+3)=(cos(2x)+cos(4x)/4+3/4)}}}


{{{2-4cos^2(x)+cos(2x)+cos(4x)/4+3/4}}}........and {{{4cos^2(x)=2 (cos(2x)+1)}}}


{{{2-2(cos(2x)+1)+cos(2x)+cos(4x)/4+3/4}}}


{{{cross(2)-2cos(2x)-cross(2)+cos(2x)+cos(4x)/4+3/4}}}


{{{-cos(2x)+cos(4x)/4+3/4}}}


{{{(1/4)(-4cos(2x)+cos(4x)+3)}}}


{{{(3-4cos(2x)+cos(4x))/4}}}

and

{{{(3-4cos(2x)+cos(4x))/4=2sin^4(x)}}} but it's NOT simplified form...:-)