Question

高一数学题：已知f(x)=cos^2x+sinxcosx g(x)=2sin(x+π/4)sin(x-π/4)

1、求f（x）的最小正周期及单调增区间
2、若f（a）+g（a）=5/6，且a属于[3π/8,5π/8]
求sin2a的值
尽量这两天，快交作业了

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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