化简:[1-(sin4x-sin2xcos2x+cos4x)]/sin2x+3sin2x.
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 13:05:51
xSKnAJd(S_n_{CUD@رlزFİh;l5(H7`ɇC2M^倍,aMS:y([SN7'ٍގ{jhOG)n'hGnnѰZS]7<^Koً.> Hz"(OBp
Dh,ǬdъRarTJQ0!XpUaa|E\adJEH4q`%
bIteA~Coҹ0{:K_: y-"ս~J/ݴʪpSlY~^0L?M'6fk[
\oV|Msgf{
5˘B.8KD`4&8L"i(|Tp@zs%45:Es^ZTI4IC,K/!
SQHBaTY! Ȑ8ׄUnl`r+oFz^rbyo
化简:[1-(sin4x-sin2xcos2x+cos4x)]/sin2x+3sin2x.
化简:[1-(sin4x-sin2xcos2x+cos4x)]/sin2x+3sin2x.
化简:[1-(sin4x-sin2xcos2x+cos4x)]/sin2x+3sin2x.
因为:1-(sinx)^4-(sinx)^2(cosx)^2+(cosx)^4=1-[(sin²x+cos²x)-3sin²xcos²x]
=3sin²xcos²x,
所以:原式=3sin²xcos²x/sin²x+3sin²x=3cos²x+3sin²x=3(cos²x+sin²x)=3
2xcos2x+cos4x)]/sin2x+