Задание
Найдите значение выражения:
\(\displaystyle 18 \sqrt{2} \, \sin {30^\circ} \cos {45^\circ}=\)
Решение
Подставим табличные значения и упростим полученное числовое выражение:
Подсказка - табличные значения
| \(\displaystyle \alpha\) | \(\displaystyle \frac{\pi}{6}=30^{\circ}\) | \(\displaystyle \frac{\pi}{4}=45^{\circ}\) | \(\displaystyle \frac{\pi}{3}=60^{\circ}\) |
| \(\displaystyle \sin \alpha\) | \(\displaystyle \frac{1}{2}\) | \(\displaystyle \frac{\sqrt{2}}{2}\) | \(\displaystyle \frac{\sqrt{3}}{2}\) |
| \(\displaystyle \cos \alpha\) | \(\displaystyle \frac{\sqrt{3}}{2}\) | \(\displaystyle \frac{\sqrt{2}}{2}\) | \(\displaystyle \frac{1}{2}\) |
| \(\displaystyle \tg \alpha\) | \(\displaystyle \frac{\sqrt{3}}{3}\) | \(\displaystyle 1\) | \(\displaystyle \sqrt{3}\) |
| \(\displaystyle \ctg \alpha\) | \(\displaystyle \sqrt{3}\) | \(\displaystyle 1\) | \(\displaystyle \frac{\sqrt{3}}{3}\) |
\(\displaystyle 18 \sqrt{2} \, \color{blue}{\sin {30^\circ}} \color{orange}{ \cos {45^\circ}}=18 \sqrt{2}\cdot \color{blue}{\frac {1}{2}}\cdot \color{orange}{\frac {\sqrt{2}}{2}}= \frac {18 \sqrt{2}\cdot\sqrt{2}}{2\cdot 2}=9.\)
Ответ: \(\displaystyle 9 {\small.} \)