Question

y=sinx/2-cosx/2的单调递增区间

y=sinx/2+cosx/2在(-2π,2π)内的递增区间为多少?答案是-3/2π<=x<=π/2
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Answer 1

asinx+bcosx=√(a^2+b^2)*sin(x+z)
其中tanz=b/a
所以y=√2sin(x/2+π/4)
-2π<x<2π
所以-π<x 2<π -3π/4<x 4<br="" 4 sin增区间是[2kπ-π/2,2kπ+π/2]
-3π/4<x 2<="x/2+π/4<=π/2<br/" 在此范围内的是-π="" 4="" 4 -π/2<=x/2+π/4<=π/2
-3π/4<=x/2<=π/4
-3π/2<=x<=π/2 </x<2π