原式=[sin(2π+x+π)+cos(π+x)]/[-sinx-cos(π+x)]
=[sin(π+x)+cos(π+x)]/[-sinx-cos(π+x)]
=(-sinx-cosx)/(-sinx+cosx) (应用诱导公式)
=(sinx+cosx)/(sinx-cosx)
=(sinx/cosx+1)/(sinx/cosx-1)
=(tanx+1)/(tanx-1)
=(m+1)/(m-1)
tanx=m,则{sin(x+3π）+cos(π+x)}/{sin(-x)-cos(π+x）}=?
1个回答
相关问题
-
已tanx=4/3,求cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x)
-
已知(1+tanx)/(1-tanx)=2 求cos^2(π+x)+3sin(π-x)cos(-x)+2sin^2(2π
-
若sin(x-π)=2cos(2π-x),求(sin(π-x)+5cos(2π-x))/(3cos(π-x)-sin(-
-
已知 f(x)= sin(x-3π)•cos(2π-x)•sin(-x+ 3π 2 ) cos(-x-π)•cos( π
-
已知函数f(x)= [sin(2π-x)sin(π+x)cos(-x-π)] ／[2cos(π-x)sin(3π-x)]
-
已知f(x)＝sin(x−3π)•cos(2π−x)•sin(−x+3π2)cos(−x−π)•cos(π2−x)
-
已知tanx=2,求2sin（π-x）cos（x-π/2）+sinxsin（3π/2-x）+cos²（x-π）
-
f(x)sinx·sin(π-x)+√3sin(π/2+x)cos(π/2+x)+2cos(π+x)cos(π-x) 求
-
若2cos(3π-x)=cos(π/2+x),则sin(π+2x)=
-
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π