# proving identities

• May 9th 2010, 08:23 AM
dmidget
proving 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 AM
dmidget
Quote:

Originally Posted by dmidget
I need help wit a problem....

1- cos(squared) = Sin
' 1+sin

Having trouble after i multipying for common denominator

revised
• May 9th 2010, 08:27 AM
e^(i*pi)
Quote:

Originally Posted by dmidget
revised

Very confusing...

Do you mean $\frac{1-\cos^2 \theta}{1+\sin \theta} = \sin \theta$
• May 9th 2010, 08:32 AM
dmidget
Quote:

Originally Posted by e^(i*pi)
Very confusing...

Do you mean $\frac{1-\cos^2 \theta}{1+\sin \theta} = \sin \theta$

yeah how did you get it too show like that??
• May 9th 2010, 08:44 AM
e^(i*pi)
Quote:

Originally Posted by dmidget
yeah how did you get it too show like that??

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

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

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

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

$\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2}$
• May 9th 2010, 09:01 AM
dmidget
Quote:

Originally Posted by e^(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 $\frac{(1-\cos \theta)(1+\cos \theta)}{1+\sin \theta}$

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

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

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

$\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2}$

i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin
• May 9th 2010, 11:36 AM
e^(i*pi)
Quote:

Originally Posted by dmidget
i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin

$1-\frac{\cos^2 \theta}{1+\sin \theta}$

Use the Pythagorean identity and then the difference of two squares