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求详细解释这题,谢谢.函数F(x)=Cos(3x+π/4)Co...

f(x)=2 cos 2 (x- π 4 )-[sin(x+ π 4 )+cos(x+ π 4 )] 2 = 1+cos(2x- π 2 )-[ 2 sin(x+ π 2 ) ] 2 =1+sin2x-2cos 2 x=sin2x-cos2x = 2 sin(2x- π 4 ) (1)T= 2π 2 =π(2)当 f(x)取最大值时, sin(2x- π 4 )=1 ,即 2x- π 4 = π 2 +2kπ ?{x|...

(1) f(x)= 3 sinωxcosωx-co s 2 ωx= 3 2 sin2ωx- 1 2 (1+cos2ωx) = 3 2 sin2ωx- 1 2 cos2ωx- 1 2 =sin(2ωx- π 6 )- 1 2 …(3分)由已知, 2π 2ω = π 2 ∴ω=2 …(1分)(2)写出函数f(x)图象的对称轴 f(x)=sin(4x- π 6 )- 1 2 由 4x- π 6 =kπ...

(Ⅰ)f(x)=cos(2x- 4π 3 )+2cos 2 x=(cos2xcos 4π 3 +sin2xsin 4π 3 )+(1+cos2x)= 1 2 cos2x- 3 2 sin2x+1=cos(2x+ π 3 )+1,(3分)∵-1≤cos(2x+ π 3 )≤1,即cos(2x+ π 3 )最大值为1,∴f(x)的最大值为2,(4分)要使f(x)取最...

解:(1) ∴ ;(2) ∵ ∴ 。

(1)由 f(x)=co s 2 ωx- 3 sinωx?cosωx ,得 f(x)= 1+cos2ωx 2 - 3 2 sin2ωx = cos(2ωx+ π 3 )+ 1 2 .由 T= 2π 2ω =π ,得ω=1,所以 f(x)=cos(2x+ π 3 )+ 1 2 .由 -π+2kπ≤2x+ π 3 ≤2kπ,k∈Z ,解得 - 2π 3 +kπ≤x≤- π 6 +kπ,k∈Z .所以函数f...

(1)f(x)= sin π 4 xcos π 6 -cos π 4 xsin π 6 -cos π 4 x = 3 2 sin π 4 x- 3 2 cos π 4 x = 3 sin( π 4 x- π 3 ) 故f(x)的最小正周期为T= 2π π 4 =8(2)在y=g(x)的图象上任取一点(x,g(x)),它关于x=1的对称点(2-x,g(x))....

函数 f(x)=cos(2x+ π 3 )+2co s 2 x =cos2xcos π 3 -sin2xsin π 3 +cos2x+1= 3 2 cos2x- 3 2 sin2x+1 =1- 3 sin(2x- π 3 ).(Ⅰ)函数f(x)的周期T=π,由∵ π 2 +2kπ≤2x- π 3 ≤ 3π 2 +2kπ,k∈Z∴ 5π 12 +kπ≤x≤ 11π 12 +kπ,k∈Z所以y=1- 3 sin...

解:(1) f ( x )=3(1+cos2 x )- sin2 x -3 =2 ( ) =2 cos(2 x + )……………… ……………………………3分 f ( x )的值域为[-2 ,2 ],周期为π; ……………………4分(2)由 f ( A )=2 cos(2 A + )=-2 得cos(2 A + )=-1,∵0< A < , < 2 A + < ,∴2 A...

(1) f(x)=5 3 co s 2 x+ 3 si n 2 x-4sinxcosx = 5 3 (cos2x+1) 2 + 3 (1-cos2x) 2 -2sin2x=3 3 +2 3 cos2x-2sin2x = 3 3 +4cos(2x+ π 6 ) 当x∈R时,f(x)的最小值为3 3 -4.(2)∵ π 4 ≤x≤ 7π 24 ∴ π 2 ≤2x≤ 7π 12 ,∴ 2π 3 ≤2x+ π 6 ≤ 3π ...

(Ⅰ) 因为 f(x)= 3 2 sin2ωx- 1 2 (1+cos2ωx) = sin(2ωx- π 6 )- 1 2 ,所以 T= 2π 2ω = π 2 ,∴,ω=2.(Ⅱ) 因为x是△ABC的最小内角,所以 x∈(0, π 3 ] ,又 f(x)=sin(4x- π 6 )- 1 2 ,所以 f(x)∈[-1, 1 2 ] .

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