一道高中数学三角函数化简题。大家帮帮忙

解：144/sin^2(x)*[（sinxcos30+cosxsin30）^2+(sinxcos30-cosxsin30)^2]
=144/sin^2(x)*[(根号3/2*sinx+1/2*cosx)^2+(根号3/2*sinx-1/2*cosx)^2]
=144/sin^2(x)*[3/4*sin^2(x)+根号3/2*sinxcosx+1/4*cos^2(x)+3/4*sin^2(x)-根号3/2*sinxcosx+1/4*cos^2(x)]
=144/sin^2(x)*(3/2*sin^2(x)+1/2*cos^2(x))
=144/sin^2(x)*3/2*sin^2(x)+144/sin^2(x)*1/2*cos^2(x)
=72(3+cos^2(x)/sin^2(x)*)
=72[3+cot^2(x)]
那么辛苦希望多加分

sin^2(x+30度) + sin^2(x-30度)
=1-cos^2(x+30度) + sin^2(x-30度)
=1-sin^2[90度-(x+30度)] + sin^2(x-30度)
=1-[sin^2(-x+60度) - sin^2(x-30度)]
=1-[sin(-x+60度) - sin(x-30度)][sin(-x+60度) + sin(x-30度)]
=1-[2cos15度sin(45度-x)][2sin15度cos(45度-x)]
=1-(2cos15度sin15度)[2sin(45度-x)cos(45度-x)]
=1-sin30度sin(90度-2x)
=1-cos2x/2
=1-[2(cosx)^2-1]/2
=3/2-(cosx)^2
所以原式=[144/(sinx)^2][3/2-(cosx)^2]
=[216-144(cosx)^2]/(sinx)^2
=216(cscx)^2-144(cotx)^2
=216[(cotx)^2+1]-144(cotx)^2
=72(cotx)^2+216
=72[(cotx)^2+3