Question 927300
{{{cos^4(theta/2) =cos^2(theta/2) *cos^2(theta/2)}}} .....since {{{cos^2(theta/2) =(1/2) (cos(theta)+1)}}}, we have


{{{cos^4(theta/2) =(1/2) (cos(theta)+1)*(1/2) (cos(theta)+1)}}}


{{{cos^4(theta/2) =(1/4) (cos(theta)+1)* (cos(theta)+1)}}}


{{{cos^4(theta/2) =(1/4)(cos^2(theta)+cos(theta)+cos(theta)+1)}}}


{{{cos^4(theta/2) =(1/4)(cos^2(theta)+2cos(theta)+1)}}} .....since {{{cos^2(theta)=(1/2)(cos(2 theta)+1)}}}, we have


{{{cos^4(theta/2) =(1/4)((1/2)(cos(2theta)+1)+2cos(theta)+1) }}}


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


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


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


{{{cos^4(theta/2) =(1/8) (4cos(theta)+cos(2theta)+3))}}}





{{{sin^4 (2x) =16 sin^4(x) cos^4(x)}}} ...............=>{{{sin^4(x)= (1/8) (-4 cos(2 x)+cos(4 x)+3)}}} and {{{cos^4(x)=(1/8) (4 cos(2x)+cos(4x)+3)}}}


{{{sin^4(2x)=16((1/8)(-4cos(2x)+cos(4x)+3)) ((1/8)(4cos(2x)+cos(4x)+3))}}}


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