Question

若sina-cosa=根号二\3,0<a<π\2,则sina+cosa=

救命啊。。。

Answer 1

sina-cosa=√2/3
两边平方：
sin^2a+cos^2a-2sinacosa=2/9
1-2sinacosa = 2/9
2sinacosa = 7/9

(sina+cosa)^2 = sin^2a+cos^2a+2sinacosa = 1+2sinacosa = 1+7/9 = 16/9
0<a<π/2，sina＞0，cosa＞0，sina+cosa＞0
sina+cosa = √(16/9) = 4/3

Answer 2

sina-cosa=√2/3
又(sina-cosa)^2=1-2sinacosa=2/9
所以2sinacosa=sin2a=7/9
又(sina+cosa)^2=1+sin2a=16/9
0<a<π/2,所以sina+cosa>0
所以sina+cosa=4/3

Answer 3

sina-cosa=根号二\3,0<a<π\2
1-2sinacosa=2/9
则（sina+cosa）²=1+2sinacosa=1+7/9=16/9
从而sina+cosa=4/3