如题(1+tan1°)(1+tan2°)(1+tan3°)......(1+tan44°)(1+tan45°)=
括号与括号之间是相乘关系!
来个详细步骤!!。谢谢
如题(1+tan1°)(1+tan2°)(1+tan3°)......(1+tan44°)(1+tan45°)=
括号与括号之间是相乘关系!
来个详细步骤!!。谢谢
(1+tanA)(1+tanB)=1+tanA+tanB+tanA*tanB=(sinAsinB+cosAcosB+sinAcosB+sinBcosA)/(cosAcosB)
sin(A+B) = sinAcosB+cosAsinB ,cos(A-B) = cosAcosB+sinAsinB
sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
化简得,【sin(A+B)+cos(A-B)】/(cosAcosB)=sinA/cosB+sinB/cosA
A+B=45°,(1+tanA)(1+tanB)=2
(1+tan1°)(1+tan2°)(1+tan3°)......(1+tan44°)(1+tan45°)= 2^23
1+tan1°=1+tan(45°-44°)=1+(1-tan44°)/(1+tan44°)=2/(1+tan44°)
1+tan2°=1+tan(45°-43°)=2/(1+tan43°)
1+tan3°=1+tan(45°-42°)=2/(1+tan42°)
..........................
1+tan22°=1+tan(45°-23°)=2/(1+tan23°)
所以:
(1+tan1°)(1+tan2°)(1+tan3°)......(1+tan44°)(1+tan45°)
= 2^22*(1+tan45°)=2^23