Question
Prove the following: $\sin(\text{n}+1)\text{x}\sin(\text{n}+2)\text{x}+2\cos(\text{n}+1)\text{x}\cos(\text{n}+2)\text{x}=\cos\text{x}$

Answer

$ \text{L.H.S.}=$ $=\frac{1}{2}[2\sin(\text{n}+1)\text{x}\sin(\text{n}+2)\text{x}+2\cos(\text{n}+1)\text{x}\cos(\text{n}+2)\text{x}]$$=\frac{1}{2}\begin{bmatrix}\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}-\cos\{(\text{n}+1)\text{x}+(\text{n}+2)\text{x}\}\\+\cos\{(\text{n}+1)\text{x}+(\text{n}+2)\text{x}\}+\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}\end{bmatrix}$
$\begin{bmatrix}\therefore-2\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})-\cos(\text{A}-\text{B})\\2\cos\text{A}\cos\text{B}=\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})\end{bmatrix}$
$=\frac{1}{2}\times2\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}$
$=\cos(-\text{x})=\cos\text{x}$
$= \text{R.H.S.}$

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