高一数学题:已知f(x)=cos^2x+sinxcosx g(x)=2sin(x+π/4)sin(x-π/4)
解f(x)=cos^4x+2sinxcosx-sin^4x
=cos^4x-sin^4x+2sinxcosx
=(cos^2x-sin^2x)(cos^2x+sin^2x)+2sinxcosx
=(cos^2x-sin^2x)+2sinxcosx
=cos2x+sin2x
=√2sin(2x+π/4)
函数的周期T=2π/2=π
由0≤x≤π/2
即0≤2x≤π
即π/4≤2x+π/4≤5π/4
即当2x+π/4=π/2时,即x=π/8时,f(x)有最大值√2
此时x值的集合为{x/x=π/8}。
(1)y=cos2x+sinxcosx=1+cos2x2+12sin2x=22 sin(2x+π4)+12∴T=2π2=π,由 2kπ?π2≤2x+π4≤π2+2kπ k∈Z,即 kπ?3π8≤x≤π8+kπ k∈Z,所以函数的单调增区间为:[?38π+kπ,π8+kπ] (k∈Z).(2)g(x)=2sin(x+π4)sin(x?π4)=-sin(2x+π2)=-cos2x,因为f(x)+g(x)=1+cos2x2+12sin2x-cos2x=12+12sin2x?12cos2x=12+22 sin(2x-π4)f(α)+g(α)=56,<table cellpadding="-1" cellspacing="-1" style="margin-right:1px"
f(x)=cos^2x+sinxcosx=(1+cos2x)/2+1/2*sin2x
=1/2+1/2(cos2x+sin2x)
=√2/2*(√2/2*cos2x+√2/2sin2x)+1/2
=√2/2*(sinπ/4cos2x+cosπ/4sin2x)+1/2
=√2/2*sin(π/4+2x)+1/2
T=2π/2=π
2kπ-π/2<π/4+2x<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8<x<kπ+π/8
单调增区间:(kπ-3π/8,kπ+π/8)
g(x)=2sin(x+π/4)sin(x-π/4)
=2sin(x+π/2-π/4)sin(x-π/4)
=2cos(π/4-x)sin(x-π/4)
=2cos(x-π/4)sin(x-π/4)
=sin(2x-π/2)
=-sin(π/2-2x)
=-cos2x
f(a)+g(a)=5/6
cos^2a+sinacosa-cos2a=5/6
(1+cos2a)/2+1/2*sin2a-cos2a=5/6
1/2*cos2a+1/2*sin2a-cos2a=1/3
1/2*sin2a-1/2*cos2a=1/3
sin2a-cos2a=2/3
sin2a-cos2a=2/3
√2(√2/2*sin2a-√2/2cos2a)=2/3
√2*(sin2acosπ/4-cos2asinπ/4)=2/3
√2*sin(2a-π/4)=2/3
sin(2a-π/4)=√2/3
a∈[3π/8,5π/8]
2a∈[3π/4,5π/4]
2a-π/4∈[π/2,π]
cos(2a-π/4)=-√7/3
sin2a=sin(2a-π/4+π/4)
=sin(2a-π/4)cosπ/4+cos(2a-π/4)sinπ/4
=√2/3*√2/2+(-√7/3)*√2/2
=1/3-√14/6
f(x)=cos^2x+sinxcosx
=(1+xos2x)/2+1/2sin2x
=1/2(sin2x+cos2x)+1/2
=根号2/2sin(2x+π/4)+1/2
T=2π/2=π
单调增区间
2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<=kπ+π/8
单调增区间[kπ-3π/8,kπ+π/8] k∈Z
g(x)=2sin(x+π/4)sin(x-π/4)
=-2sin(x+π/4)cos(x+π/4)
=-sin(2x+π/2)
=-cos2x
f(a)+g(a)==1/2(sin2a+cos2a)+1/2-cos2a=5/6
1/2sin2a-1/2cos2a=1/3
sin2a-cos2a=2/3
2sin2a*cos2a=5/9
a属于[3π/8,5π/8]
2a属于[3π/4,5π/4]
cos2a<0 sin2a<0
sin2a=(1-根号6)/2
1)先化简:f(x)=cos^2x+sinxcosx=cos2x/2+1/2+sin2x/2=√2sin(2x+π/4)/2+1/2
所以T=2π/2=π
单调增区间求法:2kπ-π/2<=2x+π/4<=2kπ+π/2
得:[kπ-3π/8,kπ+π/8](k为整数)为单调增区间
2)g(x)=2sin(x+π/4)sin(x-π/4)=cos(π/2)-cos(2x)(三角恒等变换中的积化和差公式)
f(a)+g(a)=sin(2a+π/4)+1/2-cos(2a)=sin2a /2-cos2a /2+1/2=5/6
所以sin2a-cos2a=2/3
又a属于[3π/8,5π/8]
2a属于[3π/4,5π/4]
cos2a<0 sin2a<0
根据(sinx)2+(cosx)^2=1
可解出sin2a=(1-√6)/2
解:1、f(x)=cos^2x+sinxcosx
=1/2(1+cos2x)+1/2sin2x
=1/2[cos2x+sin2x]+1/2
=根号2/4*sin(2x+π/4)+1/2 所以f(x)的最小正周期为π
单调增区间为: 2kπ-π/2<2x+π/4<2kπ+π/2 => 2kπ-3/4π<2x<2kπ+π/4
=>kπ-3/8π<x<kπ+π/8 ( k属于整数)
单调递减区间:2kπ+π/2<2x+π/4<2kπ+3/2π => 2kπ+π/4<2x<2kπ+5/4π
=> kπ+π/8<x<kπ+5/8π (k属于整数)
2、f(a)+g(a)=5/6 => cos^2(a)+1/2sin2a+2sin(a+π/4)sin(a-π/4) =5/6
=> 1/2+1/2cos2a+1/2sin2a+2[根号2*sina+根号2*cosa]*[根号2×sina-根号2*cosa]=5/6
=> 1/2-cos2a+1/2sin2a -cos2a=5/6
=> 1/2sin2a-1/2cos2a=5/6-3/6=1/3
=> sin2a-cos2a=2/3>0 因为a属于[3π/8,5π/8] 所以2a属于[3π/4,5π/4]
-√2/2《sin2a《√2/2 -1《cos2a《-√2/2 √2/2《-cos2a《1
0《sin2a-cos2a《1+√2/2 所以 sin2a<0
=> sin2a-√1+cos^2(2a)=2/3 解次一元二次方程有:
sin2a=(1-√6)/2
1)先化简:f(x)=cos^2x+sinxcosx=cos2x/2+1/2+sin2x/2=√2sin(2x+π/4)/2+1/2
所以T=2π/2=π
单调增区间求法:2kπ-π/2<=2x+π/4<=2kπ+π/2
得:[kπ-3π/8,kπ+π/8](k为整数)为单调增区间
2)g(x)=2sin(x+π/4)sin(x-π/4)=cos(π/2)-cos(2x)(三角恒等变换中的积化和差公式)
f(a)+g(a)=sin(2a+π/4)+1/2-cos(2a)=sin2a /2-cos2a /2+1/2=5/6
所以sin2a-cos2a=2/3
又a属于[3π/8,5π/8]
2a属于[3π/4,5π/4]
cos2a<0 sin2a<0
根据(sinx)2+(cosx)^2=1
可解出sin2a=(1-√6)/2
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