高一数学题（三角函数的诱导公式）若[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2,则tanα=?
问题描述：
高一数学题（三角函数的诱导公式）
若[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2,则tanα=?
答
[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2
[-cosa*sin^2(a)]/[-sina*cos^2(a)]=1/2
sina/cosa=1/2
tana=1/2
答
[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)
=[-cos α·sin^2(α)]/[-sin α·cos^2(α)]
=tan α
=1/2
答
这里用a代替α （打字方便）
[cos(π+a)*sin^2(-a)]/[sin(π+a)*cos^2(-a)=1/2
[-cos(a)*sin^2a]/[-sin(a)*cos^2(a)]=1/2
sin(a)/cos(a)=-tan(a)=1/2
tan(a)=1/2