函数y=sin(3x+π/4)+2cos(3x+π/4)的最小正周期是
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解决时间 2021-01-26 16:11
- 提问者网友：聂風
- 2021-01-25 20:58
函数y=sin(3x+π/4)+2cos(3x+π/4)的最小正周期是
最佳答案
- 五星知识达人网友：执傲
- 2021-01-25 22:02
令cosa=√5/5,sina=2√5/5
y=sin(3x+π/4)+2cos(3x+π/4)
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[sin(3x+π/4)cosa+cos(3x+π/4)sina]
=√5sin(3x+π/4+a)
T=2π/3
y=sin(3x+π/4)+2cos(3x+π/4)
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[sin(3x+π/4)cosa+cos(3x+π/4)sina]
=√5sin(3x+π/4+a)
T=2π/3
全部回答
- 1楼网友：像个废品
- 2021-01-26 00:26
y=sin(3x+π/4)+2cos(3x+π/4)
=√5[(1/√5)sin(3x+π/4)+(2/√5)cos(3x+π/4)]
=√5sin(3x+π/4+α)
其中sinα=2/√5
故原函数的最小正周期为2π/3
- 2楼网友：千杯敬自由
- 2021-01-25 23:16
y=sin3xcosπ/4+cos3xsinπ/4+2cos3xcosπ/4-2sin3xsinπ/4
=√2/2*(sin3x+cos3x+2cos3x-2sin3x)
=-√2/2*(sin3x-3cos3x)
=-√2/2*√(1+9)sin(3x-z)
=-√10sin(3x-z)
其中tanz=3/1=3
所以T=2π/3
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