已知α∈﹙0 π/4﹚ β∈﹙0 π﹚ 且tan﹙α-β﹚=1/2 tanβ=﹣1/7 求2α-β

问题描述:

已知α∈﹙0 π/4﹚ β∈﹙0 π﹚ 且tan﹙α-β﹚=1/2 tanβ=﹣1/7 求2α-β

tan(a-b)=[ tan(a) - tan(b) ]/[ 1 + tan(a)tan(b) ],由tan﹙α-β﹚=1/2 tanβ=﹣1/7可以解出tanα=1/3,而tan(2α-β)=tan( α+α-β )=【tanα+tan(α-β)】/(1-tanαtan(α-β))=1,根据已知条件α∈﹙0 π/4﹚ β∈﹙0 π﹚,可知2α-β=-135°