已知sinα是方程5x²-7x-6=0的根，求值过程

显示全部楼层 · 2011-8-26 19:15:00

求[sin(-α-3/2π)sin(3/2π-α)tan2(2π-α)]/[cos(π/2-α)cos(π/2+α)cos2(π-α)]的值
过程!!

千问 · 2011-8-26 19:15:00

由5x2-7x-6=0 得 (5x+3)(x-2)=0sina=-3/5或sina=2（舍去）下面是每一步化简过程sin(-α-3/2π)=-cosa sin(3/2π-α)=cosatan2(2π-α)=tan2α cos(π/2-α)=sinacos(π/2+α)=-sina cos2(π-α)]=cos2α 最后代入得原式=25/16...

千问 · 2011-8-26 19:15:00

5x2-7x-6=0(5x+3)(x-2)=0sina=-3/5或sina=2舍去[sin(-α-3/2π)sin(3/2π-α)tan2(2π-α)]/[cos(π/2-α)cos(π/2+α)cos2(π-α)]=[sin(π/2-a)sin(a-π/2)tan2a]/[-sinasinacos2...