Check out some similar questions!

K_T.T. · Nov 20, 2011, 03:59 PM

Please help me solve this, I have tried several ways and still doesn't come out corectly.

Prove the following:

cos^3x-sin^3x/ cosx-sinx=1 sin^2x.

K_T.T. · Nov 20, 2011, 04:02 PM

Please help me solve this, I have tried several ways and still doesn't come out correctly.

Prove the following:

cos^3x-sin^3x/ cosx-sinx=1 sin^2x.

Fr_Chuck · Nov 20, 2011, 04:04 PM

Show us what you tired, we do not "do" the work for you, but will suggest where you are making your errors.

K_T.T. · Nov 20, 2011, 04:10 PM

this is how far I got to,

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

since the top and bottom cancel out, I got

cos^2x+sinxcosx+sin^2x

I reduced further to

1+sinxcosx

this is where I got stuck at because on the other side of the equation equals to 1+sin^2x

Aurora2000 · Nov 21, 2011, 03:10 AM

What you are doing is correct, but are you sure the request is

$\dfrac{\cos^3x-\sin^3x}{\cos x-\sin x}=1+\sin^2x$ ?

Try $x=\pi/6$ : $\cos \pi/6=\sqrt{3}/2$ , $\sin \pi/6=1/2$
then the left side reads
$\dfrac{(\sqrt{3}/2)^3 -1/8 }{\sqrt{3}/2-1/2}$ which is an irrational number
while right side is 5/4.

Aurora2000 · Nov 21, 2011, 03:13 AM

Oops, some problem with mimeTex (it does not recognize the command \ dfrac as LaTeX does)

What you are doing is correct, but are you sure the request is

$\frac{cos^3 x-\sin^3 x}{\cos x-\sin x}=1+\sin^2 x$ ?

Try $x=\pi/6$ , you will get an irrational value on the left side, and 5/4 on the right side.