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 March 11th, 2011, 03:20 PM #1 Member   Joined: Apr 2010 Posts: 46 Thanks: 1 Simple Trig Simplify the following as much as possible: $\dfrac{cos^3x-3cosxsin^2x}{3sinx-4sin^3x}$ I get that the top simplifies to $sin3x$, but the bottom isn't pretty and I'm not sure what to do with it. Could anyone help me out?
March 11th, 2011, 03:52 PM   #2
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Re: Simple Trig

Hello, clandarkfire!

Quote:
 $\text{Simplify the following as much as possible: }\:\frac{\cos^3\!x\,-\,3\,\!\cos x\,\!\sin^2\!x}{3\,\!\sin x\,-\,4\,\!\sin^3\!x}$

It helps if you know some multiple-angle identities:

[color=beige]. . / / / [/color]$\begin{array}{cccccccccc}\sin\,\!3\!A=&3\,\!\sin A \,-\,4\,\!\sin^3\!A \\ \\ \\ \\ \\ \cos\,\!3\!A=&4\,\!\cos^3\!A\,-\,3\,\!\cos A \end{array}=$

$\text{The numerator is:}$

[color=beige]. . [/color]$\cos^3\!x\,-\,3\,\!\cos x\,\!\sin^2\!x \;=\;\cos^3\!x \,-\,3\,\!\cos x(1\,-\,\cos^2\!x) \;=\;\cos^3\!x\,-\,3\,\!\cos x\,+\,3\,\!\cos^3\!x$

[color=beige]. . . . . . . . . . . . . . . . .[/color]$=\;4\,\!\cos^3\!x\,-\,3\,\!\cos x \;=\;\cos3x$

$\text{The denominator is: }\:3\,\!\sin x \,-\,4\,\!\sin^3\!x \;=\;\sin 3x$

$\text{The problem becomes: }\:\frac{\cos3x}{\sin3x} \;=\;\cot 3x$

 March 11th, 2011, 05:32 PM #3 Member   Joined: Apr 2010 Posts: 46 Thanks: 1 Re: Simple Trig Thankyou! I got the numerator, but my method of finding cos3x gave me $cos3x=cos^3x-cos x sin^2x,$ and somehow I didn't get that that was equal to $3sin x-4sin^3x.$
 March 12th, 2011, 02:51 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 If you can't find your mistakes, we can identify them if you post your work here.
 March 12th, 2011, 05:14 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 By using the right trigonometric identities, you should be able to arrive at the correct answer.

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