已知 -π/6≤x≤π/4 ,求函数y=log2(1+sinx)+log2(1-sinx)的最大值和最小值
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已知 -π/6≤x≤π/4 ,求函数y=log2(1+sinx)+log2(1-sinx)的最大值和最小值
已知 -π/6≤x≤π/4 ,求函数y=log2(1+sinx)+log2(1-sinx)的最大值和最小值
已知 -π/6≤x≤π/4 ,求函数y=log2(1+sinx)+log2(1-sinx)的最大值和最小值
y=log2(1+sinx)+log2(1-sinx)
y=log2(1-sin平方x )
y=log2(cos平方x )
y=2log2(|cosx| )
因为-π/6≤x≤π/4 ,所以 二分之根号2
函数y=log2(1+sinx)+log2(1-sinx)=
=log2[(1+sinx)(1-sinx)]=
=log2[1-(sinx)^2]=
=log2[(cosx)^2],
当x=-π/6,0,π/4 时,
cosx=√3/2,1,√2/2,
(cosx)^2=3/4,1,1/2,
y=log2(cosx)^2=log2(3/4)...
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函数y=log2(1+sinx)+log2(1-sinx)=
=log2[(1+sinx)(1-sinx)]=
=log2[1-(sinx)^2]=
=log2[(cosx)^2],
当x=-π/6,0,π/4 时,
cosx=√3/2,1,√2/2,
(cosx)^2=3/4,1,1/2,
y=log2(cosx)^2=log2(3/4)=log2(3)-2log2(2)=log2(3)-2,log2(1)=0,log2(1/2)=-log2(2)=-1,
当-π/6≤x≤0时,在此区域(cosx)是增函数,√3/2≤cosx≤1,3/4≤(cosx)^2≤1,
log2(3)-2≤log2(cosx)^2≤0,
当0≤x≤π/4时,在此区域(cosx)是减函数,√2/2≤cosx≤1,1/2≤(cosx)^2≤1,-1≤log2(cosx)^2≤0,
其值域为:log2(3)-2≤log2(cosx)^2≤0,和 -1≤log2(cosx)^2≤0, 因为-1
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