解:
4/5/(3/5+1/2)*2
=4/5/(6/10+5/10)*2
=4/5/(11/10)*2
=4/5*10/11*2
=40/55*2
=80/55
=16/11
f(x)=A[1-cos(2ωx+2φ)]/2
=A[-1/2*cos(2ωx+2φ)+1/2]
最大=A(1/2+1/2)=A=2
所以f(x)=-cos(2ωx+2φ)+1
相邻两对称轴间距是半个周期
所以T/2=2
T=4
T=2π/2ω=4
ω=π/4
所以f(x)=-cos(πx/2+2φ)+1
把点代入
2=-cos(π/2+2φ)+1
cos(π/2+2φ)=-1
-sin(2φ)=-1
sin(2φ)=1
2φ=π/2
所以f(x)=-cos(πx/2+π/2)+1
即f(x)=sin(πx/2)+1
只能用脱式计算
=(4/5)/(3/5*2+1/2*2)
=(4/5)/(6/5-1)
=(4/5)/(1/5)
=4