Тапсырма
Бұрыштардың тангенсін табыңыз.
\(\displaystyle \tg\left(\frac{\pi}{6}\right)=\)
\(\displaystyle \tg\left(\frac{\pi}{4}\right)=\)
\(\displaystyle \tg\left(\frac{\pi}{3}\right)=\)
Шешім
Тангенстің анықтамасы бойынша:
\(\displaystyle \tg\left(\alpha\right)=\frac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}{\small.}\)
\(\displaystyle \frac{\pi}{6}{\small,}\) \(\displaystyle \frac{\pi}{4}\) және \(\displaystyle \frac{\pi}{3}{\small}\) нүктелерінде синус пен косинус мәндерінің кестесін қолданайық:
\(\displaystyle 30^{\circ}\)\(\displaystyle \left( \frac{\pi}{6}\right)\) | \(\displaystyle 45^{\circ}\)\(\displaystyle \left(\frac{\pi}{4}\right)\) | \(\displaystyle 60^{\circ}\)\(\displaystyle \left( \frac{\pi}{3}\right)\) | |
| \(\displaystyle {\sin}\) | \(\displaystyle \frac{1}{2}\) | \(\displaystyle {\frac{\sqrt{2}}{2}}\) | \(\displaystyle {\frac{\sqrt{3}}{2}}\) |
| \(\displaystyle {\cos}\) | \(\displaystyle \frac{\sqrt{3}}{2}\) | \(\displaystyle {\frac{\sqrt{2}}{2}}\) | \(\displaystyle {\frac{1}{2}}\) |
Сонда
- \(\displaystyle \tg\left(\frac{\pi}{6}\right)=\frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}=\frac{\,\,\,\frac{1}{2}\,\,\,}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} {\small,}\)
- \(\displaystyle \tg\left(\frac{\pi}{4}\right)=\frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}=\frac{\,\,\,\frac{\sqrt{2}}{2}\,\,\,}{\frac{\sqrt{2}}{2}}=1{\small,}\)
- \(\displaystyle \tg\left(\frac{\pi}{3}\right)=\frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)}=\frac{\,\,\,\frac{\sqrt{3}}{2}\,\,\,}{\frac{1}{2}}=\sqrt{3}{\small.}\)