如图，摩天轮的半径为50m，圆心O点距地面的高度为60m，摩天轮做匀速转动，每3min转一圈，摩天轮上的点P的起始位置在最低点处，已知在时刻t（min）时点P距离地面的高度f（t）=Asin（ωt+φ）+

Question

Answer 1

（1）由题意可知：A=50，h=60，T=3，∴ω＝(2)/(3)π，
即f(t)＝50sin((2)/(3)πt+ϕ)+60．
又∵f(0)＝10，故ϕ＝−(π)/(2)．∴f(t)＝50sin((2)/(3)πt−(π)/(2))+60，
得f（2006）=85．即点P距离地面的高度为85m．
（2）由（1）知f(t)＝50sin((2)/(3)πt−(π)/(2))+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． ω＝(2)/(3)222333π，
即f(t)＝50sin((2)/(3)πt+ϕ)+60．
又∵f(0)＝10，故ϕ＝−(π)/(2)．∴f(t)＝50sin((2)/(3)πt−(π)/(2))+60，
得f（2006）=85．即点P距离地面的高度为85m．
（2）由（1）知f(t)＝50sin((2)/(3)πt−(π)/(2))+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． f(t)＝50sin((2)/(3)222333πt+ϕ)+60．
又∵f(0)＝10，故ϕ＝−(π)/(2)．∴f(t)＝50sin((2)/(3)πt−(π)/(2))+60，
得f（2006）=85．即点P距离地面的高度为85m．
（2）由（1）知f(t)＝50sin((2)/(3)πt−(π)/(2))+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． f(0)＝10，故ϕ＝−(π)/(2)πππ222．∴f(t)＝50sin((2)/(3)πt−(π)/(2))+60，
得f（2006）=85．即点P距离地面的高度为85m．
（2）由（1）知f(t)＝50sin((2)/(3)πt−(π)/(2))+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． f(t)＝50sin((2)/(3)222333πt−(π)/(2)πππ222)+60，
得f（2006）=85．即点P距离地面的高度为85m．
（2）由（1）知f(t)＝50sin((2)/(3)πt−(π)/(2))+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． f(t)＝50sin((2)/(3)222333πt−(π)/(2)πππ222)+60=60−50cos(2)/(3)πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． 60−50cos(2)/(3)222333πt．
∴f（t）+f（t+1）+f（t+2）=60−50cos(2)/(3)πt+60−50cos(2)/(3)π(t+1)+60−50cos(2)/(3)π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． 60−50cos(2)/(3)222333πt+60−50cos(2)/(3)222333π(t+1)+60−50cos(2)/(3)222333π(t+2)
=180−50[cos(2)/(3)πt+cos((2)/(3)πt+(2)/(3)π)+cos((2)/(3)πt+(4)/(3)π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． 180−50[cos(2)/(3)222333πt+cos((2)/(3)222333πt+(2)/(3)222333π)+cos((2)/(3)222333πt+(4)/(3)444333π)]=180−50[cos(2π)/(3)t+(−(1)/(2))cos(2π)/(3)t−(√(3))/(2)sin(2πt)/(3)+(−(1)/(2))cos(2π)/(3)t−(−(√(3))/(2))sin(2π)/(3)t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值． 180−50[cos(2π)/(3)2π2π2π333t+(−(1)/(2)111222)cos(2π)/(3)2π2π2π333t−(√(3))/(2)√(3)√(3)√(3)

3

33222sin(2πt)/(3)2πt2πt2πt333+(−(1)/(2)111222)cos(2π)/(3)2π2π2π333t−(−(√(3))/(2)√(3)√(3)√(3)

3

33222)sin(2π)/(3)2π2π2π333t]
=180
故不论t为何值，f（t）+f（t+1）+f（t+2）是定值．