Question

2道高中数学三角函数题 ！！！！！急

1.已知函数f(x)=2cos^x+2sinxcosx 记函数g(x)=f(x-π/8)×f(x+π/8),求函数g(x)的值域。 2.已知函数f(x)=cos^x+(2根号3)sinxcosx-sin^x 在△ABC中，A、B、C分别为三边a、b、c所对的角，若a=根号3，f(A)=1,求b+c的最大值。

Answer 1

1

先化简，cos2x=2cos^x-1 sin2x=2sinxcosx
f(x)=cos2x+sin2x+1=√2(sin2x+π/4)+1
g(x)=[sin(2x-π/4+π/4)+1]*[[sin(2x+π/4+π/4)+1]
=[sin2x+1]*[cos2x+1]
=1/2*sin2x+√2sin(x+π/4)+1
sin2x=-cos(2x+π/2)=2sin^(x+π/4)-1
g(x)=sin^(x+π/4)+√2sin(x+π/4)+1/2=(sin(x+π/4)+1/2)^2
值域就是[0,9/4]
2
f(x)=cos2x+√3sin2x=2sin(2x+π/6)=1
2A+π/6=π/6,5π/6,
A=0,π/3
根据几何上面的特性，等腰时最大；
sin^(A/2)=1/2
这时b+c=√3/sin(A/2)=2√3

Answer 2

1.f(x)=1+cos2x+sin2x=(根号2）sin(2x+π/4)+1
g(x)=((根号2sin2x+1)*(根号2cos2x+1)=2sin2xcos2x+根号2(sin2x+cos2x)+1
=（cos2x+sin2x)^-1+根号2(sin2x+cos2x)+1
令cos2x+sin2x=t
=t^+根号2t
又t属于（-根号2，根号2）
所以g(x)属于（0，3/2)
2.f(x)=cos2x+根号3sin2x
=2sin(2x+π/6)
又f(A)=1，所以A=π/3
所以a/sinA=2
b=2sinb,c=2sinc
b+c=2sinb+2sinc
B=3π/2-c
b+c=2sin(c+π/3)
c+π/3属于（π/3,π）
所以b+cmax=1/2*2=1