Question

Trig Identities: Prove. (sin^3x + cos^3x)/sinx +cosx= 1 -sinxcosx

Answer 1

is it like this?
\[
\frac{\sin^3x+\cos^3x}{\sin x}+\cos x
\]

Answer 2

no the sinx + cosx are both a the bottom

Answer 3

\[
\frac{\sin^3x+\cos^3x}{\sin x+\cos x}\\
\quad=\frac{\sin x(1-\cos^2x)+\cos x(1-\sin^2x)}{\sin x+\cos x}\\
\quad=\frac{\sin x-\cos^2x\sin x+\cos x-\sin^2x\cos x}{\sin x+\cos x}\\
\quad=\frac{(\sin x+\cos x)-\sin x\cos x(\sin x+\cos x)}{\sin x+\cos x}\\
\quad=\frac{\cancel{(\sin x+\cos x)}(1-\sin x\cos x)}{\cancel{\sin x+\cos x}}
\]

Answer 4

okay i see... so in the last step when you cross them out that is the one? @electrokid

Answer 5

yes

Answer 6

that is the one the entire thing should equal to right?

so, QED.

Answer 7

Thank You!