sina+sinβ=1,求cosa+cosβ的取值范围
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sina+sinβ=1,求cosa+cosβ的取值范围
sina+sinβ=1,求cosa+cosβ的取值范围
sina+sinβ=1,求cosa+cosβ的取值范围
用b
sin²a+sin²b+2sinacosa=1²
令cosa+cosb=k
cos²a+cos²b+2cosacosb=k²
相加,sin²+cos²=1
所以2+2(cosacosb+sinasinb)=k²+1
cos(a-b)=(k²-1)/2
则-1<=(k²-1)/2<=1
-1<=k²<=3
即0<=k²<=3
所以-√3<=cosa+cosb<=√3
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