(1)(sin2α-cos2α)^2=(sin2α)^2+(cos2α)^2-2sin2αcos2α=1-sin4α ;
(2)因为tanα/2=(1-cosα)/sinα,也有tanα/2=sinα/(1+cosα),于是1/(tanα/2)=(1+cosα)/sinα,因此有结论,即tanα/2-1/(tanα/2)=(1-cosα)/sinα-(1+cosα)/sinα=-2/tanα.
(1)(sin2α-cos2α)^2=(sin2α)^2+(cos2α)^2-2sin2αcos2α=1-sin4α ;
(2)因为tanα/2=(1-cosα)/sinα,也有tanα/2=sinα/(1+cosα),于是1/(tanα/2)=(1+cosα)/sinα,因此有结论,即tanα/2-1/(tanα/2)=(1-cosα)/sinα-(1+cosα)/sinα=-2/tanα.