How do I simplify the expression of $\cos(2x+3x)\cos x+\sin(3x)\cos\left(\frac x2+x\right)$ ?
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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 $$ \cos(2x+3x)\cos x=\cos(5x)\cos x = \frac{\cos(5x+x) + \cos(5x-x)}{2} = \frac{\cos(6x)+\cos(4x)}{2}. $$ |
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