来自高妍方的问题
求证:(1)(sin2α-cos2α)的平方=1-sin4α(2)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
求证:(1)(sin2α-cos2α)的平方=1-sin4α(2)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
1回答
2020-06-24 13:09
求证:(1)(sin2α-cos2α)的平方=1-sin4α(2)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
求证:(1)(sin2α-cos2α)的平方=1-sin4α(2)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
1)左边=(sin2α-cos2α)^2==(sin2α)^2-2sin2αcos2α+(cos2α)^2=1--2sin2αcos2α=1-sin4α=右边(证毕)
(2)左边=tan(x/2+π/4)+tan(x/2-π/4)=(tanx/2+tanπ/4)/(1-tanx/2+tanπ/4)+(tanx/2-tanπ/4)/(1+tanx/2tanπ/4)=(tanx/2+1)/(1-tanx/2)+(tanx/2-1)/(1+tanx/2)={(tanx/2+1)^2-(1-tanx/2)^2}/{1-(tanx/2)^2}=4tanx/2/{1-(tanx/2)^2}=2tanx=右边