The expression 3[sin4(3π2−α)+sin4(3π+α)]−2[sin6(π2+α)+sin6(5π+α)] is equal to
1
3[sin4(3π2−α)+sin4(3π+α)]−2[sin6(π2+α)+sin6(5π+α)]
=3[cos4α+sin4α]−2[cos6α+sin6α]
=3cos4α+3sin4α−2cos4α(1−sin2α)−2sin4α(1−cos4α)
=sin4α+cos4α+2cos4αsin2α+2sin4αcos2α
=(sin2α+cos2α)2−2sin2αcos2α+2cos2αsin2α(cos2α+sin2α)
=1−2sin2αcos2α+2cos2αsin2α
=1