已知sina+cosa=2/3,a∈(0,π),求sina与cosa及sin^3a+cos^3a的值
1个回答

sina+cosa=2/3 (0)

(sina+cosa)^2=4/9

(sina)^2+(cosa)^2+2sinacosa=4/9

1+2sinacosa=4/9

2sinacosa=4/9-1

2sinacosa=-5/9 (1)

sinacosa=-5/18 (2)

(sina-cosa)^2

=(sina)^2+(cosa)^2-2sinacosa

=1+5/9

=14/9

a∈(0,π),

sina>0

由(2)知:cosa0

sina-cosa=(根号14)/3 (3)

由(0),(3),解方程:

sina=(2+根号14)/6

cosa=(2-根号14)/6

sin^3a+cos^3a

=(sina+coaa)(sin^2a+coa^2a+sinacosa)

=(-5/9)*(1-5/18)

=(-5/9)*13/18

=-65/162