№ 2.25 Геометрія = № 2.25 Математика
Знайдіть значення виразу:
1) $\sin^{2}150^{\circ}+\cos^{2}120^{\circ}-\text{tg}^{2}150^{\circ}$
2) $\frac{\sin 150^{\circ}}{\text{tg}150^{\circ}}-cos150^{\circ}$
Розв’язок:
1) $\sin^{2}150^{\circ}+\cos^{2}120^{\circ}-\text{tg}^{2}150^{\circ}=$
$=(sin30^{\circ})^{2}+(-cos60^{\circ})^{2}-(-\text{tg}30^{\circ})^{2}=$
$=\left( \frac{1}{2} \right)^{2}+\left(-\frac{1}{2} \right)^{2}-\left(-\frac{1}{\sqrt{3}} \right)^{2}=$
$=\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$
2) $\frac{\sin 150^{\circ}}{\text{tg}150^{\circ}}-cos150^{\circ}=$
$=\frac{\sin 150^{\circ}}{\frac{\sin 150^{\circ}}{\cos 150^{\circ}}}-cos150^{\circ}=$
$=sin150^{\circ}\cdot\frac{\cos 150^{\circ}}{\sin 150^{\circ}}-cos150^{\circ}=$
$=cos150^{\circ}-cos150^{\circ}=0$