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I've got:


Could someone show me how it simplifies to:

$e^{ax} \cos(bx)$?

It looks like the denominator is canceled by the terms that are being added, but then how do I get rid of one of the cosines?

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Notice b^(2)cos(bx)+a^(2)cos(bx)=(a^2+b^2)(cos(bx)) – yunone Feb 8 '11 at 7:54
Someone should edit this title! – user2468 Feb 8 '11 at 8:14
up vote 4 down vote accepted

You use the distributive law, which says that $(X+Y)\cdot Z=(X\cdot Z)+(Y\cdot Z)$ for any $X$, $Y$, and $Z$. In your case, we have $X=b^2$, $Y=a^2$, and $Z=\cos(bx)$, and so


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