fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)求最小正周期 、单调区间

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fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)求最小正周期 、单调区间

fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)求最小正周期 、单调区间
fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
求最小正周期 、单调区间

fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)求最小正周期 、单调区间
1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
=1-sin^2(x+π/4)-sin^2(x+π/4)+sin(2x+π/4)
=cos^2(x+π/4)-sin^2(x+π/4)+sin(2x+π/4)
=cos(2x+π/2)+sin(2x+π/4)
=sin(2x+π/4)-sin2x
所以最小正周期T=2π/2=π

fx)=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
=cos(2x+π/2)+sin(2x+π/4)
=(2-√2)/2sin2x+√2/2*cos2x
[(2-√2)/2]^2+(√2/2)^2=2-√2
假设sint=√2/[2√(2-√2)]
f(x)=√(2-√2)sin(2x+t)
最小正周期∏,
单调增区间-π/2+2kπ<=2x+t<=π/2+2kπ
-π/4-t/2+kπ<=x<=-t/2+π/4+kπ