第一步已经知道了,请写出第二步的详细过程已知函数f(x)=(1+1/tanx)sin²X-2 sin(x+∏/4) sin(x-∏/4)① 若tanx=2时,f(x)的值 ②若x属于〔π/12,π/2〕,求f(x)的取值范围f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-
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![第一步已经知道了,请写出第二步的详细过程已知函数f(x)=(1+1/tanx)sin²X-2 sin(x+∏/4) sin(x-∏/4)① 若tanx=2时,f(x)的值 ②若x属于〔π/12,π/2〕,求f(x)的取值范围f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-](/uploads/image/z/13340213-53-3.jpg?t=%E7%AC%AC%E4%B8%80%E6%AD%A5%E5%B7%B2%E7%BB%8F%E7%9F%A5%E9%81%93%E4%BA%86%2C%E8%AF%B7%E5%86%99%E5%87%BA%E7%AC%AC%E4%BA%8C%E6%AD%A5%E7%9A%84%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%EF%BC%881%2B1%2Ftanx%EF%BC%89sin%26sup2%3BX-2+sin%28x%2B%E2%88%8F%2F4%29+sin%28x-%E2%88%8F%2F4%29%E2%91%A0+%E8%8B%A5tanx%3D2%E6%97%B6%2Cf%28x%29%E7%9A%84%E5%80%BC+%E2%91%A1%E8%8B%A5x%E5%B1%9E%E4%BA%8E%E3%80%94%CF%80%2F12%2C%CF%80%2F2%E3%80%95%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4f%28x%29%3D%281%2B1%2Ftanx%29sin%5E2-2sin%28x%2B%CF%80%2F4%29sin%28x-)
第一步已经知道了,请写出第二步的详细过程已知函数f(x)=(1+1/tanx)sin²X-2 sin(x+∏/4) sin(x-∏/4)① 若tanx=2时,f(x)的值 ②若x属于〔π/12,π/2〕,求f(x)的取值范围f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-
第一步已经知道了,请写出第二步的详细过程
已知函数f(x)=(1+1/tanx)sin²X-2 sin(x+∏/4) sin(x-∏/4)
① 若tanx=2时,f(x)的值
②若x属于〔π/12,π/2〕,求f(x)的取值范围
f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).
=sinx(cosx+sinx)+2sin(x+π/4)cos(x+π/4)
=1/2sin2x+sin^2x+sin(2x+π/2)
=1/2sin2x+1/2-1/2cos2x+cos2x
=(sin2x+cos2x+1)/2
tana=sina/cosa=2,所以sin^2a=4cos^2a,即1-cos2a=4*(1+cos2a)得cos2a=-3/5
sin2a=2sinacosa/(sin^2a+cos^2a)=2tana/(tan^2a+1)=4/5
所以f(a) =1/2*(4/5-3/5+1)=3/5
第一步已经知道了,请写出第二步的详细过程已知函数f(x)=(1+1/tanx)sin²X-2 sin(x+∏/4) sin(x-∏/4)① 若tanx=2时,f(x)的值 ②若x属于〔π/12,π/2〕,求f(x)的取值范围f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-
f(x)=(sin2x+cos2x+1)/2
=√2/2sin(2x+π/4)+1/2
x属于〔π/12,π/2〕,2x+π/4属于〔5π/12,5π/4〕,
2x+π/4=π/2,sin(2x+π/4)max=1
2x+π/4=5π/4,sin(2x+π/4)min=-√2/2
f(x)max=√2/2+1/2
f(x)min=0
f(x)的取值范围[0,(√2+1)/2]