1/[(cotX)^2-1]

[(sin^x)^2-(cosx)^2]/[(sinx)^2-sinxcosx]

I've spent the last 3 hours trying to simplify these, and havn't had any luck.

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- Jun 8th 2010, 05:44 PMYungsoonTrig expression simplification
1/[(cotX)^2-1]

[(sin^x)^2-(cosx)^2]/[(sinx)^2-sinxcosx]

I've spent the last 3 hours trying to simplify these, and havn't had any luck. - Jun 8th 2010, 05:57 PMslider142
The denominator should remind you of $\displaystyle \sin^2 x + \cos^2 x = 1$, or $\displaystyle \sin^2 x = 1 - \cos^2 x$. Divide both sides of this equation by $\displaystyle \sin^2 x$ and remember the definitions of the complementary trigonometric functions. Most applications of identities will be like this. You can usually manipulate the closest identity into an applicable form.

Quote:

[(sin^x)^2-(cosx)^2]/[(sinx)^2-sinxcosx]

$\displaystyle \frac{x^2 - y^2}{x^2 - xy} = \cdots$

Can you simplify this fraction? - Jun 8th 2010, 07:30 PMYungsoon
Ok, I see how to get the 2nd one now, I think I was mostly having issues with it due to frustration over the first one.

I'm still not seeing how to solve the first problem

(and thank you for your help so far, it's been a rough day) - Jun 9th 2010, 09:10 AMslider142
Try to find $\displaystyle \cot^2 x$ in trigonometric identities you already know. For example, we know the common identity $\displaystyle \sin^2 x + \cos^2 x = 1$. We also know that $\displaystyle \cot x = \frac{\cos x}{\sin x}$. We can thus get an identity involving $\displaystyle \cot^2 x$ by dividing both sides of $\displaystyle \sin^2 x + \cos^2 x = 1$ by $\displaystyle \sin^2 x$. We can then use this identity to simplify our expression.