Question

prove (sin^3x+cos^3x)/(sinx+cosx)=1-sinxcosx

Answer 1

\[\frac{b^3+a^3}{b+a}=\frac{(b+a)(b^2-ab+a^2)}{b+a}=b^2-ab+a^2\] is a start

Answer 2

this has nothing to do with trig it is just algebra

Answer 3

now back to trig
since \(a^2+b^2=\cos^2(x)+\sin^2(x)\) in this case , that part is 1

Answer 4

Wait, what?

Answer 5

ok i will wait

Answer 6

the first part of this has nothing whatsoever to do with trig
it is just algebra, but it is hard to see with all the sines and cosines

Answer 7

it is always true that
\[x^3+y^3=(x+y)(x^2-xy+y^2)\] and so it is always true that
\[\frac{x^3+y^3}{x+y}=x^2-xy+y^2\]

Answer 8

if this means that
\[\frac{\cos^3(x)+\sin^3(x)}{\cos(x)+\sin(x)}=\cos^2(x)-\cos(x)\sin(x)+\sin^2(x)\]

Answer 9

Sorry for sounding stupid, but I don't remember learning that b^3+a^3=b^2-ba+a^2

Answer 10

it is how you factor the "sum of two cubes"

Answer 11

Okay, I will keep that in mind.

Answer 12

it is not what you wrote, it is not
\[x^3+y^3=x^2-xy+y^2\] but rather
\[x^3+y^3=(x+y)(x^2-xy+y^2)\]

Answer 13

Okay, thanks.

Answer 14

you can check by multiplying out and see that it works

Answer 15

also
\[x^3-y^3=(x-y)(x^2+xy+y^2)\]

Answer 16

yw