# Showing $\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=e^{ax}\cos(bx).$

I've got:

$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}.$$

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

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
$$\frac{e^{ax}(b^2\cos(bx)+a^2\cos(bx))}{a^2+b^2}=\frac{e^{ax}((b^2+a^2)\cos(bx))}{(a^2+b^2)}=\frac{e^{ax}((a^2+b^2)\cos(bx))}{(a^2+b^2)}=e^{ax}\cos(bx).$$