Question
Solve the following equations:
$\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$
$\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$
$(\sin4\text{x}+\sin2\text{x})+(\sin3\text{x}+\sin\text{x})=0$
Using, $(\sin\text{A}+\sin\text{B})\text{Formula}\Rightarrow$
$2\sin\Big[\frac{(4\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{4\text{x}-2\text{x}}{2}\Big]+2\sin\Big[\frac{( 3\text{x}+\text{x})}{2}\Big]\cos\Big[\frac{( 3\text{x}-\text{x})}{2}\Big]=0$
$2\sin3\text{x}\cos\text{x}+2\sin2\text{x}\cos\text{x}=0$
$2\cos\text{x}(\sin3\text{x}+\sin2\text{x})=0$
$2\cos\text{x}(2\sin)\Big[\frac{(3\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{(3\text{x}-2\text{x})}{2}\Big]=0$
$4\cos\text{x}\sin\frac{5\text{x}}{2}\cos\frac{\text{x}}{2}=0$
$\cos\text{x}=0;\sin\frac{5\text{x}}{2}=0;\cos\frac{\text{x}}{2}=0$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\frac{5\text{x}}{2}=\text{m}\pi;\frac{\text{x}}{2}=\frac{(2\text{x}+1)\pi}{2}$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\text{x}=\frac{2\text{m}\pi}{5};\text{x}=(2\text{x}+1)\pi,\text{m},\text{r},\text{n}\in\text{Z}$
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Prove that the coefficient of (r + 1)th term in the expansion of $(1+\text{x})^{\text{n+1}}$ is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of $(1+\text{x})^{\text{n}}.$
$\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$