已知函數f(x)=√3sinwx*coswx-cos^2(wx)(w>0)的周期爲π/2 求w的值.
設三角形ABC的三邊爲a,b,c滿足b^2=ac,且邊b所對的角爲x,求此時函數f(x)的值域 網友廻答:
- 已知函數f(x)=√3sinωxcosωx-cos²ωx,(ω>0)的周期爲π/2.
(1)求ω的值及f(x)的表達式.
(2)設△ABC的三邊b²=ac,且b所對角爲x,求此時函數f(x)的值域.
(1)f(x) = √3sinωxcosωx-cos²ωx
= (√3/2)sin(2ωx)-(1/2)cos(2ωx)-(1/2)
= sin(2ωx-π/6) - 1/2
T = 2π/2ω = π/2 ---> ω = 2
--->f(x) = sin(4x-π/6)-1/2
(2)cosx=(a² c²-b²)/(2ac)≥(2ac-ac)/(2ac)=1/2
--->0<x≤π/3--->-π/6<4x-π/6≤7π/6
--->-1/2≤sin(4x-π/6)≤1
--->-1≤f(x)≤1/2
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