作业帮函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围
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![函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围](/uploads/image/z/3009801-57-1.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3D%281%2B1%2Ftanx%29sin%5E2x-2sin%28x%2B%CF%80%2F4%29sin%28x-%CF%80%2F4%29.%E8%8B%A5tana%3D2%2C%E6%B1%82f%28a%29.%E8%8B%A5x%E5%B1%9E%E4%BA%8E%5B%CF%80%2F12%2C%CF%80%2F2%5D%2C%E6%B1%82%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E5%87%BD%E6%95%B0f%28x%29%3D%281%2B1%2Ftanx%29sin%5E2x-2sin%28x%2B%CF%80%2F4%29sin%28x-%CF%80%2F4%29.%E8%8B%A5tana%3D2%2C%E6%B1%82f%28a%29.%E8%8B%A5x%E5%B1%9E%E4%BA%8E%5B%CF%80%2F12%2C%CF%80%2F2%5D%E6%B1%82f%28x%29%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围
函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围
函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围
函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2],求取值范围函数f(x)=(1+1/tanx)sin^2x-2sin(x+π/4)sin(x-π/4).若tana=2,求f(a).若x属于[π/12,π/2]求f(x)取值范围
f(x)=sin^2x+sinxcosx-sin^2x+cos^2x
=sinxcosx+cos^2x
=sin2x/2+(1+cos2x)/2
=sin2x/2+cos2x/2+1/2
(1)f(a)=sin2a/2+cos2a/2+1/2
根据万能公式得 sin2a=tana/(1+tan^2a)=2/5 cos2a=(1-tan^2a)/(1+tan^2a)=-3/5
f(a)=1/5-3/10+1/2=2/5
(2)f(x)=sin2x/2+cos2x/2+1/2=√2sin(2x+π/4)/2+1/2
π/12