How do you get $\displaystyle \frac{x-xcosx}{2sinx-2xcosx}$

to be....

$\displaystyle \frac{1+xsinx-cosx}{2xsinx}$

any ideas?

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- Apr 28th 2008, 01:14 PMMr. EngineerHelp with an explanation
How do you get $\displaystyle \frac{x-xcosx}{2sinx-2xcosx}$

to be....

$\displaystyle \frac{1+xsinx-cosx}{2xsinx}$

any ideas? - Apr 28th 2008, 01:56 PMPlato
I do not understand what you are asking.

Those two expressions are not equal.

A simple graphing utility will verify that fact.

What are you asking? - Apr 28th 2008, 02:04 PMMr. EngineerThanks
Thats all I needed to know, because I thought through a process of trig formulas you could attain that 2nd form, but thank you fot telling me they are not equivalent

- Apr 28th 2008, 02:20 PMPlato
- May 1st 2008, 02:00 AMravali reddy