三角方程

敬畏数学

$  2\sin (20 \du +\alpha )\sin 10\du =\sin \alpha $，$ \alpha =$     

战巡

\[2[\sin(20^{\du})\cos(\alpha)+\cos(20^{\du})\sin(\alpha)]\sin(10^{\du})=\sin(\alpha)\]
\[\sin(20^{\du})\cot(\alpha)\sin(10^{\du})+\cos(20^{\du})\sin(10^{\du})=\frac{1}{2}=\sin(30^{\du})\]
\[\sin(20^{\du})\cot(\alpha)\sin(10^{\du})+\cos(20^{\du})\sin(10^{\du})=\sin(20^{\du})\cos(10^{\du})+\cos(20^{\du})\sin(10^{\du})\]
\[\sin(20^{\du})\cot(\alpha)\sin(10^{\du})=\sin(20^{\du})\cos(10^{\du})\]
\[\cot(\alpha)=\cot(10^{\du})\]
.................

isee

\begin{gather*}
\sin (20 \du +\alpha )\sin 10\du =\sin30\du\sin \alpha\\
\cos(30^\circ+\alpha)-\cos(10^\circ+\alpha)=\cos(30^\circ+\alpha)-\cos(30^\circ-\alpha)\\
\cos(30^\circ-\alpha)-\cos(10^\circ+\alpha)=0\\
-2\sin20^\circ\sin(10^\circ-\alpha)=0\\
10^\circ-\alpha=k\cdot 180^\circ,\,k\in\mathbb  Z
\end{gather*}

敬畏数学

战巡 发表于 2025-3-26 11:17
\[2[\sin(20^{\du})\cos(\alpha)+\cos(20^{\du})\sin(\alpha)]\sin(10^{\du})=\sin(\alpha)\]
\[\sin(20^{\ ...

敬畏数学

isee 发表于 2025-3-26 11:18
\begin{gather*}
\sin (20 \du +\alpha )\sin 10\du =\sin30\du\sin \alpha\\
\cos(30^\circ+\alpha)-\cos( ...