Question 1160082


{{{cos(4theta)-4cos(2theta)+3 = 8sin^4(theta)}}}

manipulate left side

{{{cos(4theta)-4cos(2theta)+3}}}.........since {{{cos(4theta)=cos(2theta+2theta)=cos(2theta) cos(2theta) - sin(2theta) sin(2theta) =cos^2(2theta)-sin^2(2theta)=cos^2(2theta)-(1-cos^2(2theta))= cos^2(2theta)-1+cos^2(2theta)) =2cos^2(2theta)-1}}}


then we have

{{{2cos^2(2theta)-1-4cos(2theta)+3}}}

{{{(2cos^2(2theta)-4cos(2theta))+2}}}

{{{2cos(2theta)(cos(2theta)-2+1/cos(2theta))}}}

{{{2cos(2theta)((cos^2(2theta)-2cos(2theta)+1)/cos(2theta))}}}..........simplify

{{{2(cos^2(2theta)-2cos(2theta)+1) }}}

{{{2(cos(2theta)-1) ^2}}}...........since {{{cos(2theta)=cos(theta+theta)=cos^2(theta) - sin^2(theta)}}}

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

{{{2(1-sin^2(theta) - sin^2(theta)-1) ^2}}}

{{{2(-2sin^2(theta) ) ^2}}}

{{{2(4sin^4(theta) )}}} 

{{{8sin^4(theta)}}}-> proven