( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) ) Please help me simplify this one. Giving me a headache.
What are you trying to ask?
A. $\displaystyle sin^3(x)+cos^3(x)$
Use the sum of two cubes $\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$ and the Pythagorean identity
B. $\displaystyle sin^2(x)-cos^2(x)$
You could use the difference of two squares but it'd get you nowhere really. I would rewrite as $\displaystyle -(cos^2(x)-sin^2(x))$ and now you can use the double angle identity for cos(2x)
$\displaystyle \frac{sin^3(x)+cos^3(x)}{sin^2(x)-cos^2(x)}$
Slightly different this time. Again use the sum of two cubes on the top
$\displaystyle sin^3(x)+cos^3(x) = [sin(x)+cos(x)][sin^2(x)-sin(x)cos(x)+cos^2(x)]$
As $\displaystyle sin^2(x)+cos^2(x) = 1$ we get $\displaystyle [sin(x)+cos(x)][1-sin(x)cos(x)]$
As far as I know once $\displaystyle sin(x)+cos(x)$ goes it cannot be reduced further
This time use the difference of two squares on the bottom to ensure something will cancel in the step after this
$\displaystyle sin^2(x)-cos^2(x) = [sin(x)+cos(x)][sin(x)-cos(x)]$
Bring both parts together
$\displaystyle \frac{[sin(x)+cos(x)][1-cos(x)sin(x)]}{[sin(x)+cos(x)][sin(x)-cos(x)]}$