函数fx=2sinxcosx+sin(π/2-2x) 求 f(π/4)的值 求fx的最小正周期和最小值 求fx的单调递增区间
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f(x)=2sinxcosx+sin(π/2-2x)
=sin2x+cos2x
=√2(√2/2*sin2x+√2/2*cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
f(π/4)=√2sin(2*π/4+π/4)
=√2sin(3π/4)
=√2*√2/2
=1
T=2π/2=π
-1<=sin(2x+π/4)<=1
-√2<=√2sin(2x+π/4)<=√2
所以f(x)的最小值为：-√2
f(x)的单调递增区间:
2kπ-π/2<2x+π/4<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8<x<kπ+π/8
=sin2x+cos2x
=√2(√2/2*sin2x+√2/2*cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
f(π/4)=√2sin(2*π/4+π/4)
=√2sin(3π/4)
=√2*√2/2
=1
T=2π/2=π
-1<=sin(2x+π/4)<=1
-√2<=√2sin(2x+π/4)<=√2
所以f(x)的最小值为：-√2
f(x)的单调递增区间:
2kπ-π/2<2x+π/4<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8<x<kπ+π/8
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谢谢哦 好厉害
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