I need help wit a problem....

1-cos= Sin

1+sin

Having trouble after i multipying for common denominator

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- May 9th 2010, 08:23 AMdmidgetproving identities
I need help wit a problem....

1-__cos__= Sin

1+sin

Having trouble after i multipying for common denominator - May 9th 2010, 08:25 AMdmidget
- May 9th 2010, 08:27 AMe^(i*pi)
- May 9th 2010, 08:32 AMdmidget
- May 9th 2010, 08:44 AMe^(i*pi)
I used LaTeX, you can click the image to see how I done it - you can find out more here: http://www.mathhelpforum.com/math-help/latex-help/

I'm not sure how to try it, I've done the difference of two squares but that only gives $\displaystyle \frac{(1-\cos \theta)(1+\cos \theta)}{1+\sin \theta}$

Using the Pythagorean identity gives only $\displaystyle \frac{\sin^2 \theta}{1+\sin \theta}$

Edit: this doesn't appear to be an identity - counterexample $\displaystyle \theta = \frac{\pi}{4}$

$\displaystyle \frac{1-\cos^2 \left(\frac{\pi}{4}\right)}{1+\sin \left(\frac{\pi}{4}\right)} = \frac{1-\frac{1}{2}}{1+\frac{\sqrt2}{2}} = \frac{\frac{1}{2}}{\frac{2+\sqrt2}{2}} = \frac{1}{2+\sqrt2} = \frac{2-\sqrt2}{2} $

$\displaystyle \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2} $ - May 9th 2010, 09:01 AMdmidget
- May 9th 2010, 11:36 AMe^(i*pi)