# Trig functions

# Domain and Range

Based on Professor Leonard, Trig 8(Precalculus 81)

\(f(\theta)\) |
domain |
range |
---|---|---|

\(\sin{\theta}\) |
\(\Re\) |
\([1,1]\) |

\(\cos{\theta}\) |
\(\Re\) |
\([1,1]\) |

\(\tan{\theta}\) |
\(\theta\ne\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, ... \) |
\((-\infty,\infty)\) |

\(\csc{\theta}\) |
\(\theta\ne0,\pi,2\pi,3\pi ... \) |
\((-\infty,-1]\space\cup\space[1, \infty)\) |

\(\sec{\theta}\) |
\(\theta\ne\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, ... \) |
\((-\infty,-1]\space\cup\space[1, \infty)\) |

\(\cot{\theta}\) |
\(\theta\ne0,\pi,2\pi,3\pi ... \) |
\((-\infty,\infty)\) |

# Reciprocal Identities

(from Trig 9 precalc 82)

Because (for unit circle) \(\sin\theta = y\) and \(\csc\theta = \frac{1}{y}\), therefore \(\csc\theta = \frac{1}{\sin\theta}\)

Etc. for other recip functions. Easy. Skip

But then here’s equation for circle:

Looks like Pythagorean theorem, because it is! Can always inscribe a right triangle onto a circle this way.

So given: \(\sin\theta = \frac{12}{13}\), find y, r, and x.

So y is 12, r = 13. OR can make it \(y = \frac{12}{13}\) with a radius of one! Because with similar triangles, can multiply all our sides by same number to get a similar triangle in same proportion. Can therefore also multiply by 13 to get rid of fraction. For Pythagorean theorem, it’s better to stay with y = 12, r = 13.

So let’s find x (I did) $\( 13^2 = 169 \\ 12^2 = 144 \\ 169 - 144 = 25 = x^2 \\ \therefore x = \pm 5 \)$

Important that \(\pm\), because we’re taking a square root here. To know sign, we need two trig functions or quadrant info, because triangle can flip on y axis, putting it into quadrant II, so there – x is negative.

Once again, review, \(\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta}\)