Question

已知sinx+cosx=1/5，且0＜x＜π。求sin^3*x -cos^3*x的值

Answer 1

因为sinx+cosx=1/5，所以（sinx+cosx）^2=1/25，所以sinx*cosx=-12/25，所以x》90°，sin^3*x -cos^3*x=（sinx-cosx）*（sinx^2+sinx*cosx+cosx^2）=根号（sinx-cosx）^2*（sinx^2+sinx*cosx+cosx^2）=根号（1-2*sinx*cosx）*（sinx^2+sinx*cosx+cosx^2）=7/5*13/25=91/125

Answer 2

(sinx+cosx)^2=1+2sinxcosx=1/25 2sinxcosx=-24/25 π/2<x<π
(sinx-cosx)^2=1-2sinxcosx=1+24/25=49/25 sinx-cosx=7/5
sin^3*x -cos^3*x=(sinx-cosx)（sin^2*x+sinxcosx+cos^2*x）=(7/5)*(1-12/25)=91/125

Answer 3

sinx+cosx=1/5
(sinx+cosx)^2 =1/25
sin^2*x + 2 sinx cosx + cos^2*x =1/25
1 + 2 sinx cosx =1/25
sinx cosx = -12/25

sin^3*x -cos^3*x
= (sinx - cosx) (sin^2*x + sinx*cosx + cos^2*x)
= (sinx - cosx) (1 + sinx*cosx )
= (sinx - cosx) (13/25)
= (13/25)√(sinx - cosx)^2
= (13/25)√{sin^2(x) - 2 sinx cosx + cos^2(x) }
= (13/25)√(1 - 2 sinx cosx )
= (13/25)√(49/25)
= 91/125

Answer 4

sin3x+cos3x=1/5
sin^3(x)+cos^3(x)=91/125

Answer 5

-53/125