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Теория: 11 Табличные значения тригонометрических функций

Задание

Найдите тангенсы углов.

\(\displaystyle \tg\left(\frac{\pi}{6}\right)=\)
\frac{\sqrt{3}}{3}

 
\(\displaystyle \tg\left(\frac{\pi}{4}\right)=\)
1

 
\(\displaystyle \tg\left(\frac{\pi}{3}\right)=\)
\sqrt{3}
Решение

По определению тангенса:

\(\displaystyle \tg\left(\alpha\right)=\frac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}{\small.}\)


Воспользуемся таблицей значений синуса и косинуса в точках \(\displaystyle \frac{\pi}{6}{\small,}\) \(\displaystyle \frac{\pi}{4}\) и \(\displaystyle \frac{\pi}{3}{\small:}\)

 

\(\displaystyle 30^{\circ}\)\(\displaystyle \left( \frac{\pi}{6}\right)\)

\(\displaystyle 45^{\circ}\)\(\displaystyle \left(\frac{\pi}{4}\right)\)\(\displaystyle 60^{\circ}\)\(\displaystyle \left( \frac{\pi}{3}\right)\)
\(\displaystyle {\sin}\)\(\displaystyle \frac{1}{2}\)\(\displaystyle {\frac{\sqrt{2}}{2}}\)\(\displaystyle {\frac{\sqrt{3}}{2}}\)
\(\displaystyle {\cos}\)\(\displaystyle \frac{\sqrt{3}}{2}\)\(\displaystyle {\frac{\sqrt{2}}{2}}\)\(\displaystyle {\frac{1}{2}}\)


Тогда

  • \(\displaystyle \tg\left(\frac{\pi}{6}\right)=\frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}=\frac{\,\,\,\frac{1}{2}\,\,\,}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} {\small,}\)
     
  • \(\displaystyle \tg\left(\frac{\pi}{4}\right)=\frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}=\frac{\,\,\,\frac{\sqrt{2}}{2}\,\,\,}{\frac{\sqrt{2}}{2}}=1{\small,}\)
     
  • \(\displaystyle \tg\left(\frac{\pi}{3}\right)=\frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)}=\frac{\,\,\,\frac{\sqrt{3}}{2}\,\,\,}{\frac{1}{2}}=\sqrt{3}{\small.}\)