# proving identities

#### dmidget

I need help wit a problem....

1 - cos = Sin
1+sin

Having trouble after i multipying for common denominator

#### dmidget

I need help wit a problem....

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

Having trouble after i multipying for common denominator
revised

#### e^(i*pi)

MHF Hall of Honor
Very confusing...

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

• dmidget

#### dmidget

Very confusing...

Do you mean $$\displaystyle \frac{1-\cos^2 \theta}{1+\sin \theta} = \sin \theta$$
yeah how did you get it too show like that??

#### e^(i*pi)

MHF Hall of Honor
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 $$\displaystyle \frac{(1-\cos \theta)(1+\cos \theta)}{1+\sin \theta}$$

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

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

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

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

#### dmidget

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 $$\displaystyle \theta = \frac{\pi}{4}$$

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

$$\displaystyle \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} \neq \frac{2-\sqrt2}{2}$$
i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin

#### e^(i*pi)

MHF Hall of Honor
i got the problem wrong., its one minus cos(squared) over 1 plus sin= sin
$$\displaystyle 1-\frac{\cos^2 \theta}{1+\sin \theta}$$

Use the Pythagorean identity and then the difference of two squares