现将各项都乘开,然后将sin^2x=1-cos^2x带入
得8cos^4x过程能不能大略写一下-cos^2x(2cos^2x+3)-sin^2x(2cos^2x+3)-4sin^2xcos^2x+3=2cos^4x+3cos^2x-2cos^2xsin^2x-3sin^2x-4sin^2xcos^2x+3=2cos^4x+3cos^2x-6cos^2xsin^2x-3sin^2x+3=2cos^4x+3cos^2x-6cos^2x(1-cos^2x)-3(1-cos^2x)+3=2cos^4x+3cos^2x-6cos^2x+6cos^4x-3+3cos^2x+3=8cos^4x