prove

$\displaystyle sin^4t=3/8-1/2*cos2t+1/8*cos4t$

get rid of the fraction so $\displaystyle 3-4cos2t+8cos4t$

$\displaystyle 3-4(1-2sin^2t)+8(1-2sin^2t)^2$

$\displaystyle 7+8sin^2t-32sin^2t+32sin^4t$

$\displaystyle 7-24sin^2t+32sin^4t$

this looks nothing like $\displaystyle sin^4$...