f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
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f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
1.求f(-11π/12)的值
2.当X∈[0,π/4)时 求g(x)=f(x)/2+sin2x的最大值和最小值
1.求f(-11π/12)的值
2.当X∈[0,π/4)时 求g(x)=f(x)/2+sin2x的最大值和最小值
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
=(cos2x)^2/((sinπ/4cosx+cosπ/4sinx)(sinπ/4cosx-cosπ/4sinx))
=(cos2x)^2/(cosx^2/2-sinx^2/2)
=2cos2x
1.f(-11π/12)=2cos(-11π/6)=√3
2.g(x)=cos2x+sin2x=√2sin(2x+π/4)
当X∈[0,π/4)时,2x+π/4∈[π/4,3π/4).所以最大值为√2,最小值为1
=(cos2x)^2/((sinπ/4cosx+cosπ/4sinx)(sinπ/4cosx-cosπ/4sinx))
=(cos2x)^2/(cosx^2/2-sinx^2/2)
=2cos2x
1.f(-11π/12)=2cos(-11π/6)=√3
2.g(x)=cos2x+sin2x=√2sin(2x+π/4)
当X∈[0,π/4)时,2x+π/4∈[π/4,3π/4).所以最大值为√2,最小值为1
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
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