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y sinx+Cosx的单调区间

y=sinx+cosxy'=cosx-sinx=√2[√2/2cosx-√2/2sinx]=√2cos(x+π/4)下面求单调递增区间y'>=0,2kπ-π/2 ≤x+π/4 ≤ 2kπ+π/2 2kπ-3π/4 ≤x≤ 2kπ+π/4 单调递增区间[2kπ-3π/4 , 2kπ+π/4] 下面求单调递减区间 y' 评论0 0 0

y=sinx cosx=√2[√2sinx/2 √2cosx/2]=√2sin(x π/4)下面求单调递增区间2kπ-π/2 ≤x π/4 ≤ 2kπ π/22kπ-3π/4 ≤x≤ 2kπ π/4单调递增区间[2kπ-3π/4 , 2kπ π/4]下面求单调递减区间2kπ π/2 ≤x π/4 ≤ 2kπ 3π/22kπ π/4 ≤x≤ 2kπ 4π/3单调递减区间[2kπ π/4 , 2kπ 4π/3]

y=sinx+cosx=√2[√2sinx/2+√2cosx/2]=√2sin(x+π/4) 下面求单调递增区间2kπ-π/2 ≤x+π/4 ≤ 2kπ+π/22kπ-3π/4 ≤x≤ 2kπ+π/4 单调递增区间[2kπ-3π/4 , 2kπ+π/4] 下面求单调递减区间2kπ+π/2 ≤x+π/4 ≤ 2kπ+3π/22kπ+π/4 ≤x≤ 2kπ+4π/3 单调递减区间[2kπ+π/4 , 2kπ+4π/3]

y=|sinx|+|cosx|=√(1+|sin2x|)故只要求|sin2x|的单调区间由于|sin2x|的周期为π/2显然它在〔0,π/4〕为增,在〔π/4,π/2〕为减故函数的增区间 和减区间只要在上面两个区间的端点处分别加上kπ/2,就可以了.

y=sinx-sinxcosx=(1-cos2x)/2-1/2*sin2x=-1/2(sin2x+cos2x)+1/2=-√2/2*sin(2x+π/4)+1/2 sin(2x+π/4)增则-√2/2*sin(2x+π/4)减 所以y和sin(2x+π/4)单调性相反 sinx增则2kπ-π/2<x<2kπ+π/2 所以这里即2kπ-π/2<2x+π/4<2kπ+π/22kπ-3π/4<2x<2kπ+π/4 kπ-3

y=sinx+cosx=sinx+sin(π/2-x)=2sinπ/4cos(x-π/4)=√2cos(x-π/4) 极值点.-√2<=y<=√2 单调区间,递增2kπ-π/4<=x<=2kπ+3π/4 递减2kπ+3π/4<=x<=2kπ+7π/4

因为函数y=|sinx|+|cosx|的值总为正数,所以给该函数整体平方后,函数的单调区间不会发生变化. 设f(x)=y 则f(x)=(|sinx|+|cosx|) =1+2|sinxcosx|=1+|sin2x| 因为y=|sin2x|的增区间为[kπ/2,kπ/2+π/4](k∈z), 所以原函数的增区间为[kπ/2,kπ/2+π/4](k∈z).

dy=cosx+sinx=[(√2)/2]sin(x+π/4) 第一步:求导 第二步:使用辅助角公式 第三步:求单调区间(自己做吧)

y=sinx+cosx=sinx+sin(π/2-x)=2sinπ/4cos(x-π/4)=√2cos(x-π/4)极值点.-√2<=y<=√2单调区间,递增2kπ-π/4<=x<=2kπ+3π/4递减2kπ+3π/4<=x<=2kπ+7π/4

设f(x)=|sinx|+|cosx|,f(x+π/2)=|sinx|+|cosx|,知f(x)=f(x+π/2),于是周期为π/2故只需画出【0,π/2】图像.1、当x∈(0,π/4),sinx和cosx均大于0;此时f(x)=|sinx|+|cosx|=sinx+cosx=√2sin(x+π/4)于是f(x)在x∈(0,π/4)可以画出来,是单调递增的.2、当x∈(π/4

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