已知函数f(x)=sin(x+θ) +acos(x+2θ) π 2 ) (1)当a= 2 ,
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已知函数f(x)=sin(x+θ) +acos(x+2θ) π 2 ) (1)当a= 2 ,
已知函数f(x)=sin(x+θ)
+acos(x+2θ)
π 2 )
(1)当a= 2 ,θ= π 4 时,求f(x)在区间
[0,π]上的最大值与最小值;
(2)若f( π 2 )=0,f(π)=1,求a,θ的
值.
已知函数f(x)=sin(x+θ)
+acos(x+2θ)
π 2 )
(1)当a= 2 ,θ= π 4 时,求f(x)在区间
[0,π]上的最大值与最小值;
(2)若f( π 2 )=0,f(π)=1,求a,θ的
值.
第一问:
f(x) = sin(x+θ) + acos(x+2θ)
a=√2,θ=π/4时:
f(x) = sin(x+π/4)+√2cos(x+π/2)
= √2/2sinx+√2/2cosx-√2cosx
= √2/2sinx -√2/2cosx
= sin(x-π/4)
x∈【0,π】
x-π/4∈【-π/4,3π/4】
x-π/4=π/2时,f(x)取最大值,f(x)max = sin(π/2)=1
x-π/4=-π/4时,f(x)取最小值,f(x)min = sin(-π/4)=-√2/2
第二问:
f(π/2)=0,即:
sin(π/2+θ) + acos(π/2+2θ) = cosθ-asin(2θ) = cosθ-2asinθcosθ = cosθ(1-2asinθ)=0
θ∈(-π/2,π/2),∴cosθ>0
∴1-2asinθ=0
∴sinθ=1/(2a)
f(π)=1,即:
sin(π+θ) + acos(π+2θ) = -sinθ-acos2θ = -sinθ - a(1-2sin²θ) = 1
∴-1/(2a) - a [1-2/(4a²)] = 1
∴-1/(2a) - a + 1/(2a)] = -a = 1
∴a = -1
sinθ = 1/(2a) = -1/2
θ∈(-π/2,π/2)
∴θ = -π/6
f(x) = sin(x+θ) + acos(x+2θ)
a=√2,θ=π/4时:
f(x) = sin(x+π/4)+√2cos(x+π/2)
= √2/2sinx+√2/2cosx-√2cosx
= √2/2sinx -√2/2cosx
= sin(x-π/4)
x∈【0,π】
x-π/4∈【-π/4,3π/4】
x-π/4=π/2时,f(x)取最大值,f(x)max = sin(π/2)=1
x-π/4=-π/4时,f(x)取最小值,f(x)min = sin(-π/4)=-√2/2
第二问:
f(π/2)=0,即:
sin(π/2+θ) + acos(π/2+2θ) = cosθ-asin(2θ) = cosθ-2asinθcosθ = cosθ(1-2asinθ)=0
θ∈(-π/2,π/2),∴cosθ>0
∴1-2asinθ=0
∴sinθ=1/(2a)
f(π)=1,即:
sin(π+θ) + acos(π+2θ) = -sinθ-acos2θ = -sinθ - a(1-2sin²θ) = 1
∴-1/(2a) - a [1-2/(4a²)] = 1
∴-1/(2a) - a + 1/(2a)] = -a = 1
∴a = -1
sinθ = 1/(2a) = -1/2
θ∈(-π/2,π/2)
∴θ = -π/6
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