已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a(a属于R,a是常数)
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已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a(a属于R,a是常数)
(1)求函数f(x)的最小正周期
(2)若x属于[-π/2,π/2]时,最大值最小值之和为√3,求a的值.
(1)求函数f(x)的最小正周期
(2)若x属于[-π/2,π/2]时,最大值最小值之和为√3,求a的值.
![已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a(a属于R,a是常数)](/uploads/image/z/3946291-43-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x%2B%CF%80%2F6%29%2Bsin%28x-%CF%80%2F6%29%2Bcosx%2Ba%28a%E5%B1%9E%E4%BA%8ER%2Ca%E6%98%AF%E5%B8%B8%E6%95%B0%29)
1、
f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=√[(√3)²+1²]sin(x+z)+a
=2sin(x+z)+a
其中tanz=1/√3
所以T=2π/1=2π
2、
tanz=1/√3
z=π/6
f(x)=2sin(x+π/6)+a
-π/2
f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=√[(√3)²+1²]sin(x+z)+a
=2sin(x+z)+a
其中tanz=1/√3
所以T=2π/1=2π
2、
tanz=1/√3
z=π/6
f(x)=2sin(x+π/6)+a
-π/2
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