三角函数:已知f(x)=5cos²x+sin²x-(4根号3)sinxcosx(1)化简f(x)的解析式,并求f(x)的最小正周期(2) 当x∈[-π/6,π/4]时,求f(x)的值域

问题描述:

三角函数:已知f(x)=5cos²x+sin²x-(4根号3)sinxcosx
(1)化简f(x)的解析式,并求f(x)的最小正周期
(2) 当x∈[-π/6,π/4]时,求f(x)的值域

f(x)=5cos²x+sin²x-4√3sinxcosx
=4cos²x+1-2√3sin2x
=2+2cos2x+1-2√3sin2x
=3+2cos2x-2√3sin2x
=3+4(1/2cos2x-√3/2sin2x)
=3+4cos(2x+π/3)
T=2π/2=π
(2)x∈[-π/6,π/4]
2x∈[-π/3,π/2]
2x+π/3∈[0,5π/6]
f(x)max=7 f(x)min=3-2√3

f(x)=1+4cosx*cosx-2sqrt(3)sin2x=1+2+2cos2x-2sqrt(3)sin2x,下面和别人的相同!

(1)f(x)=4cos²x+(cos²x+sin²x)-2√3sin2x
=2(cos2x+1)+1-2√3sin2x (cos²x+sin²x=1 2cos²x=cos2x+1)
=-2√3sin2x+2cos2x+3 (括号打开,合并)
=-4sin(2x-π/6)+3 (辅助角公式)
T=2π/2=π (T=2π/w)
(2)x=-π/6时 2x-π/6=-π/2
x=π/4时 2x-π/6=π/3 (整体带入)
由y=-4sinu+3的图像可知在[-π/2,π/3]内,y∈[-2√3+3,7]

1,f(x)=5cos²x+sin²x-4√3sinxcosx
=4cos²x+1-2√3sin2x
=2(1+cos2x)+1-2√3sin2x
=2cos2x-2√3sin2x+3
=4(1/2*cos2x-√3/2*sinx)+3
=4cos(2x+π/3)+3
那么最小正周期T=2π/2=π
2,x∈[-π/6,π/4]
2x+π/3∈[0,5π/6]
cos(2x+π/3)∈[-√3/2,1]
4cos(2x+π/3)+3∈[3-2√3,7]
那么值域为[3-2√3,7]

f(x)=5*(1+cos2x)/2+(1-cos2x)/2-2√3sin2x=2cos2x-2√3sin2x+3=4[cos(2x)*(1/2)-sin(2x)*(√3/2)]+3=4[cos(2x)cos(π/3)-sin(2x)sin(π/3)]+3=4cos(2x+π/3)+3(1) T=2π/2=π(2) x∈[-π/6,π/4]2x+π/3∈[0,5π/6]...

f(x)=5cos²x+sin²x-4√3sinxcosx
=4cos²x+1-2√3sin2x
=2cos2x-2√3sin2x+3
=-4sin(2x-π/6)+3
f(x)的最小正周期=π


当x∈[-π/6,π/4]时,
2x-π/6∈[-π/2,π/3]
sin(2x-π/6)∈[-1,√3/2]
-4sin(2x-π/6)+3)∈[-2√3+3,7]

f(x)的值域[-2√3+3,7]