sinA+cosB=二分之根号二,求tanA的值在△ABC中

问题描述:

sinA+cosB=二分之根号二,求tanA的值
在△ABC中

sinA+cosA=√2/2,求tanA的值
√2/2 sinA+√2/2 cosA=1/2
sin(A+π/4)=1/2
A=2kπ+π/6-π/4=2kπ-π/12
tan(-A)=tanπ/12
= sinπ/12 /cosπ/12
= √¯[(1 - cosπ/6 )/(1 + cosπ/6 )]
= √¯[(2 - √3)/(2 + √3)]
= 2 - √3
所以tanA的值为 √3 - 2