Монгол Бодлогын Сан

Заавар: Нийлбэр ба ялгавар өнцгийн томьёонууд ашигла.

Бодолт: $\sin^2\alpha+\cos^2\alpha=1$ тул $\cos^2\alpha=1-\sin^2\alpha$. $\cos\alpha>0$ тул $$\cos\alpha=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}$$ $\sin^2\beta+\cos^2\beta=1$ ба $\sin\beta< 0$ тул $$\sin\beta=-\sqrt{1-\left(-\dfrac35\right)^2}=-\dfrac45$$ \begin{align*} \sin(\alpha-\beta)&=\dfrac{5}{13}\cdot\dfrac{-3}{5}-\dfrac{12}{13}\cdot\dfrac{-4}{5}=\dfrac{33}{65}\\ \sin(\alpha+\beta)&=\dfrac{5}{13}\cdot\dfrac{-3}{5}+\dfrac{12}{13}\cdot\dfrac{-4}{5}=-\dfrac{63}{65} \end{align*}