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mangoqueen54's question at Yahoo! Answers involving a trigonometric identity

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MarkFL

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Feb 24, 2012
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MarkFL

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Feb 24, 2012
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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.
 
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MarkFL

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Feb 24, 2012
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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)