已知sinα +cosα=(1-√3)⼀2,且0<α<π,则tanα=

2025-02-23 17:16:04
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回答1:

(sinα +cosα)²=sin²α+cos²α+2sinα *cosα=1+2sinα *cosα=(1-√3)²/4=1-(√3)/2
sinα *cosα=-(√3)/4
sinα[(1-√3)/2-sinα]=-(√3)/4
sin²α-(1/2-√3/2)sinα-√3/4=0
(sinα+√3/2)(sinα-1/2)=0
sinα=1/2
或sinα=-√3/2(不在0<α<π上,舍去)
cosα=-√3/2
tan=-√3/3