已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,
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已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,
(1)求f(x)在区间[π/8,3π/4]上的取值范围
(2)当tanα=2时,f(α)=3/5,求m的值.
(1)求f(x)在区间[π/8,3π/4]上的取值范围
(2)当tanα=2时,f(α)=3/5,求m的值.
1、将m=0带入f(x),f(x)=(1+1/tanx)sin^2x利用半角公式化简整理得f(x)=1+√2*sin(2x-π/4)/2
x属于[π/8,3π/4],2x-π/4属于[0,5π/4],sin(2x-π/4)∈[-√2/2,1],带入可得f(x)的取值范围[0,1+√2/2]
2、tana=sina/cosa=2,sin^2a+cos^2a=1,sina=2√5/5,cos=√5/5,
带入求得m=-2
x属于[π/8,3π/4],2x-π/4属于[0,5π/4],sin(2x-π/4)∈[-√2/2,1],带入可得f(x)的取值范围[0,1+√2/2]
2、tana=sina/cosa=2,sin^2a+cos^2a=1,sina=2√5/5,cos=√5/5,
带入求得m=-2
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