5cos(2x-y)+7cosy=0 ,tan(x-y)tanx=?y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
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5cos(2x-y)+7cosy=0,tan(x-y)tanx=?y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域5cos(2x-y)+7cosy=0
5cos(2x-y)+7cosy=0 ,tan(x-y)tanx=?y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
5cos(2x-y)+7cosy=0 ,tan(x-y)tanx=?
y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
5cos(2x-y)+7cosy=0 ,tan(x-y)tanx=?y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
答:
1)
5cos(2x-y)+7cosy=0
5cos[(x-y)+x]+7cos[x-(x-y)]=0
5cos(x-y)cosx-5sin(x-y)sinx +7cosxcos(x-y)+7sinxsin(x-y)=0
所以:
12cos(x-y)cosx+2sin(x-y)sinx=0
所以:sin(x-y)sinx=-6cos(x-y)cosx
所以:tan(x-y)tanx=-6
2)
y=sin²(x+π/4)-sin²(x-π/4),x∈(π/6,π/3)值域
=(1/2)*[1-cos(2x+π/2) ]- (1/2)*[1-cos(2x-π/2)]
=(1/2) * [cos(2x-π/2)-cos(2x+π/2) ]
=(1/2)* (sin2x+sin2x)
=sin2x
π/6
tan(x-y)tanx=13/3
[0,√2]
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