已知函数f(x)=(1+1/tanx)sin^2x+msin(x+π/4)sin(x-π4),当m=0时,求f(x)在区
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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,求f(x)在区间[π/8,3π/4]上的取值范围
(步骤,O(∩_∩)O谢谢)不要复制网上的,我没看懂
(步骤,O(∩_∩)O谢谢)不要复制网上的,我没看懂
主要还是化简,先看前面一部分(1+1/tanx)sin^2x
其中
1+1/tanx
=1+cosx/sinx
=(sinx+cosx)/sinx
(1+1/tanx)sin^2x
=sinx(sinx+cosx)
=sin^2x+sinxcosx
=1/2[1-cos2x]+1/2sin2x
=1/2+1/2[sin2x-cos2x]
=1/2+√2/2sin(2x-π/4) 这个注意下,是个技巧,以后经常用的要掌握一下
对于本题,m=0
f(x)=1/2+ √2/2sin(2x-π/4)
当x在区间[π/8,3π/4]
0
其中
1+1/tanx
=1+cosx/sinx
=(sinx+cosx)/sinx
(1+1/tanx)sin^2x
=sinx(sinx+cosx)
=sin^2x+sinxcosx
=1/2[1-cos2x]+1/2sin2x
=1/2+1/2[sin2x-cos2x]
=1/2+√2/2sin(2x-π/4) 这个注意下,是个技巧,以后经常用的要掌握一下
对于本题,m=0
f(x)=1/2+ √2/2sin(2x-π/4)
当x在区间[π/8,3π/4]
0
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