# How do I simplify this trig expression?

How do I simplify the expression of $\cos(2x+3x)\cos x+\sin(3x)\cos\left(\frac x2+x\right)$ ?

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I've fixed tex, please check that I didn't modify anything. – Davide Giraudo May 26 '12 at 19:54
Why did you write it as $2x+3x$ instead of $5x$? – Phira May 26 '12 at 20:05
@JoeL. is it possible to use (cos(2x)+cos(3x))⋅cosx ?? i calculated and found (for example,in degrees) cos(15)+cos(30) != cos(45) :( – DrStrangeLove May 26 '12 at 20:06
A simple simplification is $\,\cos (2x+3x)=\cos 5x\,$...is there a typo here? – DonAntonio May 26 '12 at 20:11
hat do you want to simplify it into? It can be done cleanly into a function of $\cos(x/2)$ and $\sin(x/2)$. – ncmathsadist May 26 '12 at 20:12

\begin{align} \cos\alpha\cos\beta - \sin\alpha\sin\beta & = \cos(\alpha+\beta) \\ \cos\alpha\cos\beta + \sin\alpha\sin\beta & = \cos(\alpha-\beta) \end{align} Adding left sides and adding right sides gives $$2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta)$$ so $$\cos\alpha\cos\beta = \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}.$$ A similar thing handles a product of a sine and a cosine.
Later edit: "Dr. Strangelove" doesn't seem to be getting excited about this answer, so I'll add a bit more. The proposed identity includes $\cos(5x)\cos x$. Let $5x$ be $\alpha$ and $x$ be $\beta$ in the identity above. We get $$\cos(2x+3x)\cos x=\cos(5x)\cos x = \frac{\cos(5x+x) + \cos(5x-x)}{2} = \frac{\cos(6x)+\cos(4x)}{2}.$$