Question

Another Proving of Identities:
sin^3x+cos^3x/(sinx+cosx) = 1-sinxcosx ?
How ?

Answer 1

a^3 + b^3 =?

Answer 2

um? sir? sorry can't follow.

Answer 3

lol, you and your `sir`'s :p

Answer 4

???

Answer 5

zepdrix will answer. hehehe.

Answer 6

ok.

Answer 7

There are formulas for the sum and difference of `cubes`.
They don't come up very often but it's still worth remembering them I think. :)\[\large a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 8

oh yes yes I remember that.

Answer 9

\[\large \sin^3x+\cos^3x=?\]

Answer 10

1

Answer 11

No no no silly :3 that's \(\large \sin^2x+\cos^2x\)

Answer 12

oh sorry. (_ _)

Answer 13

(sinx+cosx)(sin^2x+2sinxcosx+cos^2x) ?

Answer 14

wait the degrees are wrong I think

Answer 15

Follow the formula buster! :O

Answer 16

The powers looks correct, its a couple other minor mistakes i see though :U

Answer 17

(sinx+cosx)(sin^2x-2sinxcosx+cos^2x) ? it's minus

Answer 18

Yah it's minus :)
Also, that 2 shouldn't be there in the middle, on the -2sinxcosx.

Answer 19

so it's sinxcosx only?

Answer 20

yup

Answer 21

Ya,
(sinx+cosx)(sin^2x-sinxcosx+cos^2x)

Answer 22

oh right.:)
then how can I prove it?

Answer 23

So using that identity, it allows us to rewrite our numerator like this, correct?
\[\large \frac{\sin^3x+\cos^3x}{(\sin x+\cos x)} \qquad=\qquad \frac{\color{orangered}{(\sin x+\cos x)}(\sin^2x-\sin x \cos x+\cos^2x)}{\color{orangered}{(\sin x+ \cos x)}}\]

See any nice cancellations from here? :3
Maybe I made it way too obvious lol

Answer 24

I get it now

Answer 25

oh ok c: too quick for me hehe

Answer 26

thanks sir. :)

Answer 27

gr* stuff man @zepdrix

Answer 28

i mean great :)