Монгол Бодлогын Сан

Заавар: Нийлбэр, үржвэрийн томьёонууд.

Нийлбэр, ялгаврыг үржвэрт хувиргах: \begin{align*} \sin\alpha+\sin\beta&=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}\\ \sin\alpha-\sin\beta&=2\sin\dfrac{\alpha-\beta}{2}\cos\dfrac{\alpha+\beta}{2}\\ \cos\alpha+\cos\beta&=2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}\\ \cos\alpha-\cos\beta&=-2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2} \end{align*}

Үржвэрийг нийлбэрт хувиргах томьёонууд: $$\sin\alpha\cdot\sin\beta=\dfrac12\{\cos(\alpha-\beta)-\cos(\alpha+\beta)\}$$ $$\sin\alpha\cdot\cos\beta=\dfrac12\{\sin(\alpha-\beta)+\sin(\alpha+\beta)\}$$ $$\cos\alpha\cdot\cos\beta=\dfrac12\{\cos(\alpha-\beta)+\cos(\alpha+\beta)\}$$

Бодолт:

1. $\sin 75^{\circ}\cdot \cos 15^{\circ}=\dfrac{1}{2}\{\sin(75^\circ-15^\circ)+\sin(75^\circ+15^\circ)\}=\dfrac{1}{2}\Big(\dfrac{\sqrt{3}}{2}+1\Big)=\dfrac{2+\sqrt{3}}{4}$
2. $\cos 45^{\circ}\cdot \sin 75^{\circ}=\dfrac{1}{2}\{\sin(75^\circ-45^\circ)+\sin(75^\circ+45^\circ)\}=\dfrac{1}{2}\Big(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\Big)=\dfrac{1+\sqrt{3}}{4}$
3. $\sin 105^{\circ}\cdot \sin 45^{\circ}=\dfrac{1}{2}\{\cos(105^\circ-45^\circ)-\cos(105^\circ+45^\circ)\}=\dfrac{1}{2}\Big\{\dfrac{1}{2}-\Big(-\dfrac{\sqrt3}{2}\Big)\Big\}=\dfrac{1+\sqrt{3}}{4}$
4. $\cos45^{\circ}\cdot \cos 75^{\circ}=\dfrac{1}{2}\{\cos(45^\circ-75^\circ)+\cos(45^\circ+75^\circ)\}=\dfrac{1}{2}\Big(\dfrac{\sqrt3}{2}-\dfrac{1}{2}\Big)=\dfrac{\sqrt{3}-1}{4}$
5. $\sin 75^{\circ}+\sin 15^{\circ}=2\sin\dfrac{75^\circ+15^\circ}{2}\cos\dfrac{75^\circ-15^\circ}{2}=2\cdot\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt{6}}{2}$
6. $\sin 75^{\circ}-\sin 15^{\circ}=2\sin\dfrac{75^\circ-15^\circ}{2}\cos\dfrac{75^\circ+15^\circ}{2}=2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt{2}}{2}$
7. $\cos 105^{\circ}+\cos 15^{\circ}=2\cos\dfrac{105^\circ+15^\circ}{2}\cos\dfrac{105^\circ-15^\circ}{2}=2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt{2}}{2}$
8. $\cos 105^{\circ}-\cos 15^{\circ}=-2\sin\dfrac{105^\circ+15^\circ}{2}\sin\dfrac{105^\circ-15^\circ}{2}=-2\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt2}{2}=-\dfrac{\sqrt{6}}{2}$