已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)求函数在区间[0,π/2]上的值域
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已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)求函数在区间[0,π/2]上的值域
求函数单调区间
打错了...f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
求函数单调区间
打错了...f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)=cos(2x-π/3)+sin(2x-π/2)
=cos(2x-π/3)+cos2x=2cos(2x-π/6)cosπ/6=√3cos(2x-π/6)
∵x∈[0,π/2] ∴2x-π/6∈[﹣π/6,5π/6] ∴f(x)∈[﹣3/2,√3]
∴当2x-π/6∈[2kπ,2kπ+π] 即x∈[kπ+π/12,kπ+7π/12] 时,单调递增
当2x-π/6∈[2kπ+π,2kπ+2π] 即x∈[kπ+7π/12,kπ+13π/12] 时,单调递减
=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)=cos(2x-π/3)+sin(2x-π/2)
=cos(2x-π/3)+cos2x=2cos(2x-π/6)cosπ/6=√3cos(2x-π/6)
∵x∈[0,π/2] ∴2x-π/6∈[﹣π/6,5π/6] ∴f(x)∈[﹣3/2,√3]
∴当2x-π/6∈[2kπ,2kπ+π] 即x∈[kπ+π/12,kπ+7π/12] 时,单调递增
当2x-π/6∈[2kπ+π,2kπ+2π] 即x∈[kπ+7π/12,kπ+13π/12] 时,单调递减
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