Trigonometry: Trigonometric functions
Computing function values
If you know the values of the cosine and sine of special angles between #0# and \(\tfrac{1}{2}\!\pi\), then you can calculate the values of special angles between \(\tfrac{1}{2}\!\pi\) and \(2\pi\) by using mirror symmetry.
Given \(\cos(\tfrac{1}{6}\!\pi)=\tfrac{1}{2}\!\sqrt{3}\), what is \(\cos(\tfrac{11}{6}\!\pi)\)?
Given \(\cos(\tfrac{1}{6}\!\pi)=\tfrac{1}{2}\!\sqrt{3}\), what is \(\cos(\tfrac{11}{6}\!\pi)\)?
| \(\cos(\tfrac{11}{6}\!\pi)={}\) |
Unlock full access
