2006-09-24 03:02:03 · 7 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
yes it i true based upon the identity
a^3+b^3=(a+b)(a^2-ab+b^2)
so sin^3x+cos^3x
=(sinx+cosx)(sin^2x-sinxcosx+cos^2x)
but sin^2x +cos^2x=1
therefore substituting
sin^3x+cos^3x=(sinx+cosx)(1-sinxcosx)
2006-09-24 03:07:20 · answer #1 · answered by raj 7 · 1⤊ 0⤋
a^3+b^3=(a+b)(a^2+b^2-ab)
substituting sinx as a and cosx as b we satisfy the relation
(sinx)^3+(cosx)^3=(sinx+cosx)(sin^2x+cos^2x-sinxcosx)
sin^2x+cos^2x=1 so,
RHS=(sinx+cosx)(1-sinxcosx)
HENCE THE PROOF
2006-09-24 10:20:43 · answer #2 · answered by KSA 3 · 1⤊ 0⤋
(sin(x))^3 + (cos(x))^3 = (sin(x) + cos(x))(sin(x)^2 - (sin(x)cos(x) + cos(x)^2)
this can also be written as
sin(x)^3 + cos(x)^3 = (sin(x) + cos(x))((sin(x)^2 + cos(x)^2) - (sin(x)cos(x))
sin(x)^2 + cos(x)^2 = 1, so
sin(x)^3 + cos(x)^3 = (sin(x) + cos(x))(1 - (sin(x)cos(x))
So yes it is TRUE.
2006-09-24 10:55:32 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
Ah, I forgot all my trigonometry when I left school and was having a go at it with pencil and paper when rfamilymember answered first.
2006-09-24 10:12:17 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Nerds lol, jk. I guess, anythings possible to an extent.
2006-09-24 10:09:51 · answer #5 · answered by callaway126 1 · 0⤊ 1⤋
Yes it is true.
2006-09-24 10:11:08 · answer #6 · answered by Anonymous · 0⤊ 0⤋
possible
2006-09-24 10:04:28 · answer #7 · answered by http://hogshead.pokerknave.com/ 6 · 0⤊ 0⤋