MHB Mangoqueen54's question at Yahoo Answers involving a trigonometric identity

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The discussion revolves around proving the trigonometric identity (Sin2x - sinx) / (cos2x + cosx) = (1 - cosx) / sinx. The proof begins by transforming the left side using double-angle identities, resulting in a fraction that can be simplified through factoring. Common factors are canceled, leading to a form that allows the application of Pythagorean identities. The final simplification confirms the identity as desired. The conversation highlights the challenges of responding to questions on platforms where users may delete their inquiries.
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Here is the question:

Trigonometry proof SAVE ME!?

(Sin2x - sinx) / (cos2x + cosx) = (1- cosx) / sinx

show your work

Thanks so much!

Here is a link to the question:

http://answers.yahoo.com/question/index?qid=20130130130636AAOqgvz

I have posted a link there so the OP can find my response.
 
Last edited:
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Re: mangoqueen54's question at Yahoo! Answers involging a trigonometric identity

Hello mangoqueen54,

We are given to prove:

$\displaystyle \frac{\sin(2x)-\sin(x)}{\cos(2x)+\cos(x)}=\frac{1-\cos(x)}{\sin(x)}$

Traditionally we begin on the left side, and try to obtain the right side.

Applying the double-angle identities for sine and cosine, the left side becomes:

$\displaystyle \frac{2\sin(x)\cos(x)-\sin(x)}{2\cos^2(x)+\cos(x)-1}$

We may factor as follows:

$\displaystyle \frac{\sin(x)(2\cos(x)-1)}{(2\cos(x)-1)(\cos(x)+1)}$

Divide out common factors and multiply by $\displaystyle 1=\frac{\cos(x)-1}{\cos(x)-1}$

$\displaystyle \frac{\sin(x)}{\cos(x)+1}\cdot\frac{\cos(x)-1}{\cos(x)-1}$

$\displaystyle \frac{\sin(x)(\cos(x)-1)}{\cos^2(x)-1}$

Use a Pythagorean identity in the denominator and multiply by $\displaystyle 1=\frac{-1}{-1}$ :

$\displaystyle \frac{\sin(x)(1-\cos(x))}{\sin^2(x)}$

Simplify:

$\displaystyle \frac{1-\cos(x)}{\sin(x)}$

Shown as desired.
 
Re: mangoqueen54's question at Yahoo! Answers involging a trigonometric identity

It's very frustrating at Yahoo as so many people there delete their question while you are working on a reply. (Headbang)(Headbang)(Headbang)
 
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