题目

1.已知0

提问时间：2020-10-24

答案

1、解: 由tan(a/2)+1/tan(a/2)=tan(x/2)+cot(x/2)=5/2 则: sin(x/2)/cos(x/2)+cos(x/2)/sin(x/2)=5/2 [sin^2(x/2)+cos^2(x/2)]/[sin(x/2)cos(x/2)]=5/2 2/[2sin(x/2)cos(x/2)=5/2 2/sin(x)=5/2 5sin(x)=4 sin(x)=4/5 由于0 0 则: cos(x)=根号[1-sin^2(x)] =3/5 则:sin(x-Л/3) =sinxcospi/3-sinpi/3cosx =4/5*1/2-根号3/2*3/5 =(4-3根号3)/10 2、因为cosa+cos^2a=1 又因为 sin^2a+cos^2a=1 所以得:sin^2a=cosa 则: sin^2a+sin^6a+sin^8a =sin^2a+(sin^2a)^3+(sin^2a)^4 =cosa+cos^3a+cos^4a =cosa+cos^2a(cosa+cos^2a) =cosa+cos^2a(sin^2a+cos^2a) =cosa+cos^2a =sin^2a+cos^2a =1 3、