230问答网 > 在三角形ABC,sinA+cosA的取值范围是

在三角形ABC,sinA+cosA的取值范围是

请说明理由！
2025-03-13 12:18:17
推荐回答（2个）
回答1：

解：y=sina+cosa=(√2)sin[a+(π/4)].因角a为⊿的内角，故0π/4-(√2)/2-1<(√2)sin[a+(π/4)]≤√2.===>-1

回答2：

（sinA+sinB）2＝sinA2+sinB2+2sinAsinB
∵sinA2+sinB2＝1
sinAsinB＞0
∴（sinA+sinB）2＞1
sinA+sinB＞1

注:后面的2是平方

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