求证:(1)(sin2α - cos2α)的平方=1 - sin4α (2) tan(x/2 + π/4)+tan(x/
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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 -2sin2α cos2α +(cos2α)^2=1--2sin2α cos2α =1 - sin4α =右边(证毕)
(2)左边= tan(x/2 + π/4)+tan(x/2 - π/4)=(tanx/2+tan π/4)/(1-tanx/2+tan π/4)+(tanx/2-tan π/4)/(1+tanx/2tan π/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
(2)左边= tan(x/2 + π/4)+tan(x/2 - π/4)=(tanx/2+tan π/4)/(1-tanx/2+tan π/4)+(tanx/2-tan π/4)/(1+tanx/2tan π/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
求证:(1)(sin2α - cos2α)的平方=1 - sin4α (2) tan(x/2 + π/4)+tan(x/
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