I want to simplify the expression to the expression given by Wolfram. It does not give the steps. Can I have a hint?
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Originally Posted by Stuck Man I want to simplify the expression to the expression given by Wolfram. It does not give the steps. Can I have a hint? Posting both the original expression and Wolfram's attempt would be a good start
simplify[((2cosx+cos2x)(2c osx+2cos2x)-(2sinx+sin2x)(-2sinx-2sin2x))/((2cosx+cos2x) ;^2+(2sinx+sin2x)^2)&# 93; - Wolfram|Alpha
Are you having a laugh? It's this: $\displaystyle \frac{(2cosx+cos2x)(2cosx+2cos2x)-(2sinx+sin2x)(-2sinx-2sin2x)}{(2cosx+cos2x)^2+(2sinx+sin2x)^2}$? I can't see a fast approach, but I assume there must be one.
The question is "Express the derivative of arctan((2sinx+sin2x)/(2cosx+cos2x)) in terms of cos x."
Honestly, I'm struggling to find a shortcut. Expanding and simplifying the brackets reduces the number of terms, but it's a complete mess.
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