函数y=sinx+cosx,x属于R的单调递增区间为?
1个回答
2013-09-04
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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/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
2kπ π/2 ≤x π/4 ≤ 2kπ 3π/2
2kπ π/4 ≤x≤ 2kπ 4π/3
单调递减区间[2kπ π/4 , 2kπ 4π/3]
=√2sin(x π/4)
下面求单调递增区间
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
2kπ π/2 ≤x π/4 ≤ 2kπ 3π/2
2kπ π/4 ≤x≤ 2kπ 4π/3
单调递减区间[2kπ π/4 , 2kπ 4π/3]
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