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

f（x）=cos^4x+2sinxcosx-sin^4x高一数学题~

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

---

成华区13210703873： 高一数学题:已知f(x)=cos^2x+sinxcosx g(x)=2sin(x+π/4)sin(x - π/4) - ？
咸湛方德： 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=π2kπ-π/22kπ-3π/4<2x<2kπ+π/4 kπ-3π/8单调增区间:(kπ-3π/8,kπ...

成华区13210703873： 求高中数学题答案f(x)=(cos^2)x+2sinxcosx - (sin^2)x 第一,求减区间…第二,0<x<=(pi/2)求f(x)的最值 - ？
咸湛方德： 先对f(x)化简,用倍角公式,得f(x)=(1+cos2x)/2+sin2x+(1-cos2x)/2 =cos2x+sin2x=√2sin(2x+pi/4)(前面为根号2)1.令a=2x+pi/4,因为sina在(pi/2,3pi/2)内单调递减,反解出x在pi/8+k*pi<x<5pi/8+k*pi,且k属于Z.2.同上令a=2x+pi/4,当0<x<=(pi/2)时pi/4<a<5pi/4,将sina的图像画出来很明显当a=pi/4或5pi/4时f(x)有最小值=1 此时x对应为0和pi/2; 当a=pi/2时,函数有最大值√2(根号2),此时x为pi/8.解答完毕!!!

成华区13210703873： 高一数学题 步骤写明 已知f(x)=(sinx+cosx)平方+2cos平方x; 1、求f(x)的最大值和最小值;2、求f - ？
咸湛方德： f(x)=(sinx+cosx)^2+2(cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx+2(cosx)^2=1+sin(2x)+[cos(2x)+1]=sin(2x)+cos(2x)+2=根号(2)sin(2x+45)+2 最大值和最小值分别为sin(2x+45)=1或者-1时的值希望可以帮到你

成华区13210703873： 已知函数f(x)=cosx^2+ 根号3sinxcosx+1,求函数f(x)的最小正周期和单调增区间 - ？
咸湛方德： 解:f(x)=cosx^2+ 根号3sinxcosx+1 =(1/2)cos2x+(根号3/2)sin2x+3/2 =sin(2x+π/6)+3/2 所以函数的周期为π, 单调增区间为()

成华区13210703873： 高一数学:设0<a≤2,且函数f(x)=cos^2(x) - asinx+b的最大值为0,最小值为 - 4,求a,b的值. - ？
咸湛方德： a=3 b=5

成华区13210703873： 高一数学题 已知函数f(x)=sin wx,g(x)=cos wx, - ？
咸湛方德： ω>0时 f(x),g(x)的周期T=2π/ω 满足|f(x1)-g(x2)|=2的x1,x2 其|x1-x2|min=T/4 得(2π/ω)/4=π/4,ω=2 sin2x(√2)sin(2x-π/4)<0 sin(2x-π/4)<0-π+2kπ<2x-π/4<2kπ,k∈Z-3π/8+kπ<x<π/8+kπ,k∈Z 所以sin2x

成华区13210703873： 求高中数学题答案f(x)=(cos^2)x+2sinxcosx - (sin^2)x 第一,求减区间…第二,0 - ？
咸湛方德：[答案] 先对f(x)化简,用倍角公式,得f(x)=(1+cos2x)/2+sin2x+(1-cos2x)/2 =cos2x+sin2x =√2sin(2x+pi/4)(前面为根号2) 1.令a=2x+pi/4,因为sina在(pi/2,3pi/2)内单调递减,反解出x在pi/8+k*pi

成华区13210703873： 高中数学:已知函数f(x)=cos平方x+sinxcosx(x?R),问题在补充里(1)求f(3派/8)的值;派是园周率的字母/是分数线(2)求f(x)的单调递增区间 - ？
咸湛方德：[答案] f[x]=(1+cos2x)/2+1/2·sin2x =1/2+1/2(cos2x+sin2x) =1/2+√2/2·sin(2x+π/4) (1)f(3π/8)=1/2+√2/2·sin(3π/4+π/4) =1/2. (2)又2kπ-π/2≤2x+π/4≤2kπ+π/2 解得 kπ-3π/8≤x≤kπ+π/8, ∴f(x)的单调递增区间是[kπ-3π/8,kπ+π/8].

成华区13210703873： 高一数学题 已知f(x)=2cos^2(wx/2)+cos(wx+π/3) (w>0)的最小正周期为π - ？
咸湛方德： f(x)=coswx+1+coswxcosπ/3-sinwxsinπ/3=-√3/2*sinwx+3/2*coswx+1=-√[(√3/2)^2+(3/2)^2]sin(wx-z)+1=-√3sin(wx-z)+1 其中tanz=(3/2)/(√3/2)=√3 所以z=π/3 所以f(x)=-√3sin(wx-π/3)+1 T=2π/w=π w=2 f(A)=-√3sin(2A-π/3)+1=-1/2 sin(2A-π/3)=√3/2 ...

成华区13210703873： 一道简单的高一数学题:函数f(x)=cos(派/3)*x (x属于z)的值域为 - ？
咸湛方德： Z表示整数撒 别忘了0那这个函数的值域就是 0.5 -0.5 1 -1