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Теория: Построение графика функции \(\displaystyle y=\sin(x)\)

Задание

На отрезке \(\displaystyle [0;\, \pi]\) заданы зеленые точки, лежащие на графике функции \(\displaystyle y=\sin(x){\small .}\)

Используя табличные значения \(\displaystyle \sin(x){ \small ,}\) постройте \(\displaystyle 8\) точек на промежутке \(\displaystyle (\pi;\, 2\pi]\) так, чтобы получился график функции \(\displaystyle y=\sin(x)\) на отрезке \(\displaystyle [0;\, 2\pi]{\small .}\)

 

Решение

\(\displaystyle \sin \left( \frac{7\pi}{6} \right)=\sin \left(\pi+ \frac{\pi}{6} \right)=-\sin \left( \frac{\pi}{6} \right)=-\frac{1}{2}{\small ,}\)

 

\(\displaystyle \sin \left( \frac{5\pi}{4} \right)=\sin \left(\pi+ \frac{\pi}{4} \right)=-\sin \left( \frac{\pi}{4} \right)=-\frac{\sqrt{2}}{2}{\small ,}\)

 

\(\displaystyle \sin \left( \frac{4\pi}{3} \right)=\sin \left(\pi+ \frac{\pi}{3} \right)=-\sin \left( \frac{\pi}{3} \right)=-\frac{\sqrt{3}}{2}{\small ,}\)

 

\(\displaystyle \sin \left( \frac{3\pi}{2} \right)=\sin \left(\pi+ \frac{\pi}{2} \right)=-\sin \left( \frac{\pi}{2} \right)=-1{\small ,}\)

 

\(\displaystyle \sin \left( \frac{5\pi}{3} \right)=\sin \left(\pi+ \frac{2\pi}{3} \right)=-\sin \left( \frac{2\pi}{3} \right)=-\frac{\sqrt{3}}{2}{\small ,}\)

 

\(\displaystyle \sin \left( \frac{7\pi}{4} \right)=\sin \left(\pi+ \frac{3\pi}{4} \right)=-\sin \left( \frac{3\pi}{4} \right)=-\frac{\sqrt{2}}{2}{\small ,}\)

 

\(\displaystyle \sin \left( \frac{11\pi}{6} \right)=\sin \left(\pi+ \frac{5\pi}{6} \right)=-\sin \left( \frac{5\pi}{6} \right)=-\frac{1}{2}{\small ,}\)

 

\(\displaystyle \sin \left( 2\pi \right)=0{\small .}\)