I've been trying to figure this out for hrs now, somebody please assist me.

sin^3 alpha + cos^3 aplha/ sin alpha + cos alpha = 1- sin alpha cos alpha

Printable View

- Apr 17th 2009, 04:10 AMdmak263verifying trigonometric identities
I've been trying to figure this out for hrs now, somebody please assist me.

sin^3 alpha + cos^3 aplha/ sin alpha + cos alpha = 1- sin alpha cos alpha - Apr 17th 2009, 04:47 AMADARSH
$\displaystyle \frac{(sin^3(\alpha) + cos^3(\alpha))}{(sin(\alpha)+cos(\alpha))}$

Remember that

a^3 +b^3 = (a+b)(a^2+b^2-ab)

Put a =sin(alpha) , b = cos(alpha)

Remembering that

sin^2(alpha)+cos^2(alpha) = 1

-----------------------------------------

So the fraction becomes

$\displaystyle

\frac{(sin(\alpha) + cos(\alpha))(1-sin(\alpha)cos(\alpha))}{(sin(\alpha)+cos(\alpha)) }$

Now this = $\displaystyle (1-sin(\alpha)cos(\alpha))$

Adarsh - Apr 18th 2009, 03:18 AMdmak263
thank you.