# Help Simplifying

• December 14th 2009, 01:29 PM
FaRmBoX
Help Simplifying
• December 14th 2009, 01:35 PM
e^(i*pi)
Quote:

Originally Posted by FaRmBoX

What are you trying to ask?

A. $sin^3(x)+cos^3(x)$

Use the sum of two cubes $a^3+b^3 = (a+b)(a^2-ab+b^2)$ and the Pythagorean identity

B. $sin^2(x)-cos^2(x)$

You could use the difference of two squares but it'd get you nowhere really. I would rewrite as $-(cos^2(x)-sin^2(x))$ and now you can use the double angle identity for cos(2x)
• December 14th 2009, 01:44 PM
FaRmBoX
That was my bad, it the first value divided by the 2nd value. I put them one above the other as I typed it and I didnt realize the entire post ended being one line. ( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) )
• December 14th 2009, 01:54 PM
e^(i*pi)
Quote:

Originally Posted by FaRmBoX
That was my bad, it the first value divided by the 2nd value. I put them one above the other as I typed it and I didnt realize the entire post ended being one line. ( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) )

$\frac{sin^3(x)+cos^3(x)}{sin^2(x)-cos^2(x)}$

Slightly different this time. Again use the sum of two cubes on the top

$sin^3(x)+cos^3(x) = [sin(x)+cos(x)][sin^2(x)-sin(x)cos(x)+cos^2(x)]$

As $sin^2(x)+cos^2(x) = 1$ we get $[sin(x)+cos(x)][1-sin(x)cos(x)]$

As far as I know once $sin(x)+cos(x)$ 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

$sin^2(x)-cos^2(x) = [sin(x)+cos(x)][sin(x)-cos(x)]$

Bring both parts together

$\frac{[sin(x)+cos(x)][1-cos(x)sin(x)]}{[sin(x)+cos(x)][sin(x)-cos(x)]}$