已知函数f(x)=(cosx/2)²-(sinx/2)²+sinx (1)求函数的最小正周期
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f(x)=(cosx/2)²-(sinx/2)²+sinx
=cosx+sinx
=√2sin(x+π/4)
函数的最小正周期=2π/1=2π
2)当x0∈(0,π/4)
x0+π/4∈(π/4,π/2)
f(x0)=5分之4倍的根号2
sin(x0+π/4)=4/5
cos(x0+π/4)=3/5
f(x0+π/6)=√2sin(x0+π/6+π/4)
=√2{sin(x0+π/4)cosπ/6+cos(x0+π/4)sinπ/6}
=√2[4/5*√3/2+3/5*1/2]
=(4√6+3/2)/10
=cosx+sinx
=√2sin(x+π/4)
函数的最小正周期=2π/1=2π
2)当x0∈(0,π/4)
x0+π/4∈(π/4,π/2)
f(x0)=5分之4倍的根号2
sin(x0+π/4)=4/5
cos(x0+π/4)=3/5
f(x0+π/6)=√2sin(x0+π/6+π/4)
=√2{sin(x0+π/4)cosπ/6+cos(x0+π/4)sinπ/6}
=√2[4/5*√3/2+3/5*1/2]
=(4√6+3/2)/10
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