发布网友 发布时间:2022-04-28 10:32
共1个回答
热心网友 时间:2023-09-26 10:48
sinx + cosx =√2 = √2·sin[x + (π/4)],∴sin[x + (π/4)] = 1,∴x + (π/4) = 2kπ + (π/2),k∈Z
∴x/2 = kπ + (π/8),而tan(x/2) = tan[kπ + (π/8)] = tan(π/8)>0
∵1 = tan(π/4) = 2tan(π/8)/{1 - [tan(π/8)]^2},解得tan(π/8) = √2 - 1,负值舍
∴当x满足sinx+cosx =√2时,tan(x/2)值为 √2 - 1