三角函数求解!难题我采纳!10.化简.（1）【sin(2π-α)cos([π/2]+α)cos([π/2]-α)】/【sin(3π-α)sin(-π-α)sin([π/2]+α)】（2）【an(2π-θ)sin(2π-θ)cos(6π-θ)】/【(-cosθ)sin(5π+θ)】11.已知tanα=-2,求sin^2α-4sinαcosα+5cos^2α的值；
问题描述：
三角函数求解!难题我采纳!
10.化简.
（1）【sin(2π-α)cos([π/2]+α)cos([π/2]-α)】/【sin(3π-α)sin(-π-α)sin([π/2]+α)】
（2）【an(2π-θ)sin(2π-θ)cos(6π-θ)】/【(-cosθ)sin(5π+θ)】
11.已知tanα=-2,求sin^2α-4sinαcosα+5cos^2α的值；
答
1. = -sinA*sinA*sinA/(sinA*sinA*(-cosA)) = tanA
(2) ...
sin^2 - 4sincos + 4cos^2 = (sin-2cos)^2 = cos^2 (tan-2)^2 = 16 cos^2 A
原式=17 cos^2 A
tanA = sinA/cosA = sqrt(1-cos^2)/cos
tan^2 A = (1-cos^2) / cos^2 = 2
cos^2 = 1/3
17 cos^2 A = 17/3
答
10.化简.（1）【sin(2π-α)cos([π/2]+α)cos([π/2]-α)】/【sin(3π-α)sin(-π-α)sin([π/2]+α)】=[-sinα(-sinα)sinα]/(sinα*sinα*cosα)=tanα（2）【tan(2π-θ)sin(2π-θ)cos(6π-θ)】/【(-cosθ)si...