f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8)的值 (2)求这个函数的单调递增区间
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![f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8)的值 (2)求这个函数的单调递增区间](/uploads/image/z/5410035-27-5.jpg?t=f%28x%29%3Dcos%E5%B9%B3%E6%96%B9x%2Bsinxcosx%2B3%2F2+x%E2%88%88R+%281%29f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%B0%8F%E5%91%A8%E6%9C%9F%E5%8F%8Af%EF%BC%88%CF%80%2F8f%28x%29%3Dcos%E5%B9%B3%E6%96%B9x%2Bsinxcosx%2B3%2F2+x%E2%88%88R+%281%29f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%B0%8F%E5%91%A8%E6%9C%9F%E5%8F%8Af%EF%BC%88%CF%80%2F8%EF%BC%89%E7%9A%84%E5%80%BC+%EF%BC%882%EF%BC%89%E6%B1%82%E8%BF%99%E4%B8%AA%E5%87%BD%E6%95%B0%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8)的值 (2)求这个函数的单调递增区间
f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8
f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8)的值 (2)求这个函数的单调递增区间
f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8f(x)=cos平方x+sinxcosx+3/2 x∈R (1)f(x)的小周期及f(π/8)的值 (2)求这个函数的单调递增区间
f(x)=(1+cos2x)/2+(1/2)sin2x+3/2
=(1/2)sin2x+(1/2)cos2x+2
=(√2/2)sin(2x+π/4)+2
(1)T=2π/2=π,f(π/8)=(√2/2)+2
(2)2kπ-π/2≤2x+π/4≤2kπ+π/2
得:kπ-3π/8≤x≤kπ+π/8
∴递增区间为:[kπ-3π/8,kπ+π/8],k∈Z
f(x)=cos²x+sinxcosx+ (3/2)
=(1/2)(1+cos2x)+(1/2)sin2x + (3/2)
=(1/2)(sin2x+cos2x)+2
=(√2/2)sin[2x+(π/4)] + 2
(1)
T = 2π/2 = π
f(π/8) = (√2/2)sin[(π/4)+...
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f(x)=cos²x+sinxcosx+ (3/2)
=(1/2)(1+cos2x)+(1/2)sin2x + (3/2)
=(1/2)(sin2x+cos2x)+2
=(√2/2)sin[2x+(π/4)] + 2
(1)
T = 2π/2 = π
f(π/8) = (√2/2)sin[(π/4)+(π/4)] + 2
=(4+√2)/2
(2)
考察y=sinx可知,该函数在:[2kπ-(π/2),2kπ+(π/2)] (k∈Z)为增函数,因此:
2kπ-(π/2) ≤ 2x+(π/4) ≤ 2kπ+(π/2),时f(x)为增函数,即:
x∈[kπ-(3π/8),kπ+(π/8) ],f(x)是增函数
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