Монгол Бодлогын Сан

Заавар: $288^\circ-18^\circ=270^\circ$, $32^\circ+148^\circ=180^\circ$, $302^\circ-122^\circ=180^\circ$ байгааг анхаар.

$$\cos\alpha\cos\beta=\dfrac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$$ $$\sin\alpha\sin\beta=\dfrac12[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$$

ба $\tg\alpha=\dfrac{\sin\alpha}{\cos\alpha}$ болохыг ашигла.

Бодолт: $$\tg18^{\circ}\tg288^{\circ}=\dfrac{\sin18^\circ}{\cos18^\circ}\cdot\dfrac{\sin 288^\circ}{\cos288^\circ}=$$ $$=\dfrac{\frac12[\cos(288^\circ-18^\circ)+\cos(288^\circ+18^\circ)]}{\frac12[\cos(288^\circ-18^\circ)-\cos(288^\circ+18^\circ)]}=$$ $$=\dfrac{\cos270^\circ+\cos306^\circ}{\cos270^\circ-\cos306^\circ}=-1$$ Түүнчлэн $$\sin32^\circ\sin 148^\circ=\dfrac12[\cos(148^\circ-32^\circ)+\cos(148^\circ+32^\circ)]=\dfrac{\cos116^\circ+1}{2}$$ $$\sin302^\circ\sin 122^\circ=\dfrac12[\cos(302^\circ-122^\circ)+\cos(302^\circ+122^\circ)]=$$ $$=\dfrac{1+\cos424^\circ}{2}=\dfrac{1+\cos64^\circ}{2}$$ Иймд $$\tg18^{\circ}\tg288^{\circ}+\sin32^{\circ}\sin148^{\circ}-\sin302^{\circ}\sin122^{\circ}=$$ $$=-1+\dfrac{1+\cos116^\circ}{2}+\dfrac{1+\cos64^\circ}{2}=$$ $$=\dfrac12(\cos116^\circ+\cos64^\circ)=0$$