设函数f(x)=cos(2x+π/3)+sin^2x-1/2当x属于[0,π]时,求f(x)的递减区间当f(a-π/8)=√3/3时 求f(2a)的值
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![设函数f(x)=cos(2x+π/3)+sin^2x-1/2当x属于[0,π]时,求f(x)的递减区间当f(a-π/8)=√3/3时 求f(2a)的值](/uploads/image/z/1729882-10-2.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2x-1%2F2%E5%BD%93x%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%5D%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E5%BD%93f%28a-%CF%80%2F8%29%3D%E2%88%9A3%2F3%E6%97%B6+%E6%B1%82f%282a%29%E7%9A%84%E5%80%BC)
设函数f(x)=cos(2x+π/3)+sin^2x-1/2当x属于[0,π]时,求f(x)的递减区间当f(a-π/8)=√3/3时 求f(2a)的值
设函数f(x)=cos(2x+π/3)+sin^2x-1/2
当x属于[0,π]时,求f(x)的递减区间
当f(a-π/8)=√3/3时 求f(2a)的值
设函数f(x)=cos(2x+π/3)+sin^2x-1/2当x属于[0,π]时,求f(x)的递减区间当f(a-π/8)=√3/3时 求f(2a)的值
f(x)=cos(2x+π/3)+sin^2x-1/2
=cos(2x+π/3)+(1-cos2x)/2-1/2
=cos2xcos(π/3)-sin2xsin(π/3)-cos2x*1/2
=-√3/2*sin2x
当x∈[0,π]时,2x∈[0,2π],f(x)的递减区间即sin2x的递增区间,显然需:
0≤2x≤π/2,或3π/2≤2x≤2π
得0≤x≤π/4,或3π/4≤x≤π
f(a-π/8)=√3/3=-√3/2*sin[2(a-π/8)]
得sin(2a-π/4)=-2/3
f(2a)=-√3/2*sin4a=-√3/2*cos(4a-π/2)=-√3/2*cos[2(2a-π/4)]
=-√3/2*[1-2sin^2 (2a-π/4)]
=-√3/2*[1-2*(-2/3)^2]
=-√3/18