### Trigonometry: Angles with sine, cosine and tangent

### Special values of trigonometric functions

We have seen that the unit circle is *symmetrical* and that we therefore only need to know the first eighth. We will now look at special values for the first quarter. We then see the symmetry in the sine and cosine as we saw earlier. It is important to memorize these values.

#{\bf \alpha}# |
#{\bf 0}# |
#{\bf \dfrac{\pi}{6}}# |
#{\bf \dfrac{\pi}{4}}# |
#{\bf \dfrac{\pi}{3}}# |
#{\bf \dfrac{\pi}{2}}# |

#{\bf \sin(\alpha)}# |
#0# |
#\dfrac{1}{2}# |
#\dfrac{1}{\sqrt{2}}# |
#\dfrac{\sqrt{3}}{2}# |
#1# |

#{\bf\cos(\alpha)}# |
#1# |
#\dfrac{\sqrt{3}}{2}# |
#\dfrac{1}{\sqrt{2}}# |
#\dfrac{1}{2}# |
#0# |

#{\bf \tan(\alpha)}# |
#0# |
#\dfrac{\sqrt{3}}{3}# |
#1# |
#\sqrt{3}# |
- |

Determine #\sin(\frac{5 \pi}{3})# without using your calculator.

#\sin(\frac{5 \pi}{3})=# #-\frac{1}{2}\sqrt{3}#

By use of reflections across the #x#-axis, we find #\sin(\frac{5 \pi}{3})=-\sin(\frac{\pi}{3})=-\dfrac{\sqrt{3}}{2}#.

By use of reflections across the #x#-axis, we find #\sin(\frac{5 \pi}{3})=-\sin(\frac{\pi}{3})=-\dfrac{\sqrt{3}}{2}#.

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