Trig functions

Domain and Range

Based on Professor Leonard, Trig 8(Precalculus 81)

$f(\theta )$	domain	range
$\sin \theta$	$\mathrm{\Re }$	$[1,1]$
$\cos \theta$	$\mathrm{\Re }$	$[1,1]$
$\tan \theta$	$\theta \ne \frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},...$	$(-\mathrm{\infty },\mathrm{\infty })$
$\csc \theta$	$\theta \ne 0,\pi ,2\pi ,3\pi ...$	$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$
$\sec \theta$	$\theta \ne \frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},...$	$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$
$\cot \theta$	$\theta \ne 0,\pi ,2\pi ,3\pi ...$	$(-\mathrm{\infty },\mathrm{\infty })$

Periods and Related Identities

$f(\theta )$	period	related identity
$\sin (\theta )$	$2\pi$	$\sin (\theta \pm 2\pi )=\sin (\theta )$
$\cos (\theta )$	$2\pi$	$\cos (\theta \pm 2\pi )=\cos (\theta )$
$\tan (\theta )$	$\pi$	$\tan (\theta \pm \pi )=\tan (\theta )$
$\csc (\theta )$	$2\pi$	$\csc (\theta \pm 2\pi )=\csc (\theta )$
$\sec (\theta )$	$2\pi$	$\sec (\theta \pm 2\pi )=\sec (\theta )$
$\cot (\theta )$	$\pi$	$\cot (\theta \pm \pi )=\cot (\theta$)

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:

$$x^2+y^2=r^2$$

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) $$$\begin{gathered} {13}^2=169 \\ {12}^2=144 \\ 169-144=25=x^2 \\ \therefore x=\pm 5 \end{gathered}$$$

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, $\tan \theta =\frac{y}{x}=\frac{\sin \theta }{\cos \theta }$