已知-π⼀2<x<π⼀2,sinx+cosx=1⼀5,求tanx

2025-06-28 09:20:54
推荐回答(2个)
回答1:

-π/20
sinx+cosx=1/5
sinx=1/5-cosx
sin²x+cos²x=1
(1/5-cosx)²+cos²x=1
25cos²x-5cosx-12=0
(5cosx+3)(5cosx-4)=0
cosx=-3/5(<0,舍去)或cosx=4/5
sinx=1/5-cosx=1/5-4/5=-3/5
tanx=sinx/cosx=(-3/5)/(4/5)=-3/4

回答2:

(sinx+cosx)^2=1/25 sinx^2+2sinxcosx+cosx^2=1/25 2sinxcosx/(cosx^2+sinx^2)=-24/25
2tanx/(1+tanx^2)=-24/25 2tanx+24tanx^2/25+24/25=0 tanx=-3/4或tanx=-4/3 有根据定义域
tanx∈[-1,1] tanx=-3/4