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

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    @JoeL. : _How_ did you conclude that $\cos(2x+3x)\cos x$ is the same as $(\cos(2x)+\cos(3x))\cos x$ The two functions don't have the same period.2012-05-26

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$ \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}. $