一题高中数学,如图,请写出详细解答过程,谢谢! 20
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18、
(1) f(x)=mcosx+sin(x+π/6)
过P(π/6,√3)
√3=mcosπ/6+sin(π/6+π/6)
√3=m·√3/2+sinπ/3
√3=m·√3/2+√3/2
1=m/2+1/2
1/2=m/2
1=m
m=1
f(x)=cosx+sin(x+π/6)
=cosx+sinxcosπ/6+cosxsinπ/6
=cosx+√3/2sinx+1/2cosx
=3/2cosx+√3/2sinx
=√3(√3/2cosx+1/2sinx)
=√3sin(x+π/3)
-π/2+2kπ=<x+π/3<=π/2+2kπ
-5π/6+2kπ=<x<=π/6+2kπ
单调递增区间:[-5π/6+2kπ,π/6+2kπ] (k∈Z)
(2) f(α)=√3/3
√3sin(α+π/3)=√3/3
sin(α+π/3)=1/3
cos(α+π/3)=±√[1-sin^2(α+π/3)]
=±√[1-(1/3)^2]
=±2√2/3
sinα=sin[(α+π/3)-π/3]
=sin(α+π/3)cosπ/3-cos(α+π/3)sinπ/3
=(1/3)(1/2)-(±2√2/3)(√3/2)
=(1±2√2)/6
∵0<α<2π/3
∴sinα=(1+2√2)/6
(1) f(x)=mcosx+sin(x+π/6)
过P(π/6,√3)
√3=mcosπ/6+sin(π/6+π/6)
√3=m·√3/2+sinπ/3
√3=m·√3/2+√3/2
1=m/2+1/2
1/2=m/2
1=m
m=1
f(x)=cosx+sin(x+π/6)
=cosx+sinxcosπ/6+cosxsinπ/6
=cosx+√3/2sinx+1/2cosx
=3/2cosx+√3/2sinx
=√3(√3/2cosx+1/2sinx)
=√3sin(x+π/3)
-π/2+2kπ=<x+π/3<=π/2+2kπ
-5π/6+2kπ=<x<=π/6+2kπ
单调递增区间:[-5π/6+2kπ,π/6+2kπ] (k∈Z)
(2) f(α)=√3/3
√3sin(α+π/3)=√3/3
sin(α+π/3)=1/3
cos(α+π/3)=±√[1-sin^2(α+π/3)]
=±√[1-(1/3)^2]
=±2√2/3
sinα=sin[(α+π/3)-π/3]
=sin(α+π/3)cosπ/3-cos(α+π/3)sinπ/3
=(1/3)(1/2)-(±2√2/3)(√3/2)
=(1±2√2)/6
∵0<α<2π/3
∴sinα=(1+2√2)/6
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