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Теория: Значения синуса, косинуса и формулы приведения

Задание

Заполните таблицу:

  \(\displaystyle \frac{2\pi}{3}\) \(\displaystyle \frac{3\pi}{4}\) \(\displaystyle \frac{5\pi}{6}\) \(\displaystyle \frac{7\pi}{6}\) \(\displaystyle \frac{5\pi}{4}\) \(\displaystyle \frac{4\pi}{3}\)
\(\displaystyle \sin\) \(\displaystyle \frac{\sqrt{3}}{2}\) \(\displaystyle \frac{\sqrt{2}}{2}\) \(\displaystyle \frac{1}{2}\) \(\displaystyle -\frac{1}{2}\) \(\displaystyle -\frac{\sqrt{2}}{2}\) \(\displaystyle -\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\) \(\displaystyle -\frac{1}{2}\) \(\displaystyle -\frac{\sqrt{2}}{2}\) \(\displaystyle -\frac{\sqrt{3}}{2}\) \(\displaystyle -\frac{\sqrt{3}}{2}\) \(\displaystyle -\frac{\sqrt{2}}{2}\) \(\displaystyle -\frac{1}{2}\)

 

  \(\displaystyle \frac{5\pi}{3}\) \(\displaystyle \frac{7\pi}{3}\) \(\displaystyle \frac{11\pi}{3}\)
\(\displaystyle \sin\) \(\displaystyle -\frac{\sqrt{3}}{2}\) \(\displaystyle \frac{\sqrt{3}}{2}\) \(\displaystyle -\frac{\sqrt{3}}{2}\)
\(\displaystyle \cos\) \(\displaystyle \frac{1}{2}\) \(\displaystyle \frac{1}{2}\) \(\displaystyle \frac{1}{2}\)

 

Решение

\(\displaystyle \sin\frac{5\pi}{6}=\frac{1}{2}\)

\(\displaystyle \sin\frac{7\pi}{6}=-\frac{1}{2}\)

\(\displaystyle \cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}\)

\(\displaystyle \cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}\)

\(\displaystyle \sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}.\)

\(\displaystyle \sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\)

\(\displaystyle \cos\frac{4\pi}{3}=-\frac{1}{2}\)

\(\displaystyle \sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}\)

\(\displaystyle \cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}\)

\(\displaystyle \sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\)

\(\displaystyle \cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\)

\(\displaystyle \sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}\)

\(\displaystyle \cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}\)

\(\displaystyle \sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}\)

\(\displaystyle \cos\frac{5\pi}{3}=\frac{1}{2}\)

\(\displaystyle \sin\frac{11\pi}{6}=-\frac{1}{2}\)

 \(\displaystyle \cos\frac{11\pi}{6}=\frac{\sqrt{3}}{2}\)