# Trigonometry Identity Question

1. Feb 23, 2016

For the given triangle as below:

I can obtain trigonometry identities as below:

$\sin α = \frac{y}{r} = \cos (90° - α)$
$\cos α = \frac{x}{r} = \sin (90° - α)$
$\tan α = \frac{y}{x} = \cot (90° - α)$
$\cot α = \frac{x}{y} = \tan (90° - α)$
$\sec α = \frac{r}{x} = \csc (90° - α)$
$\csc α = \frac{r}{y} = \sec (90° - α)$

$\sin (90° + α) = \sin (90° - (-α)) = \cos (-α) = \cos α$
$\cos (90° + α) = \cos (90° - (-α)) = \sin (-α) = - \sin α$
$\tan (90° + α) = \tan (90° - (-α)) = \cot (-α) = - \cot α$
$\cot (90° + α) = \cot (90° - (-α)) = \tan (-α) = - \tan α$
$\sec (90° + α) = \sec (90° - (-α)) = \csc (-α) = - \csc α$
$\csc (90° + α) = \csc (90° - (-α)) = \sec (-α) = \sec α$

$\sin (180° - α) = \sin (90° + (90° - α)) = \cos (90° - α) = \sin α$
$\cos (180° - α) = \cos (90° + (90° - α)) = - \sin (90° - α) = - \cos α$
$\tan (180° - α) = \tan (90° + (90° - α)) = - \cot (90° - α) = - \tan α$
$\cot (180° - α) = \cot (90° + (90° - α)) = - \tan (90° - α) = - \cot α$
$\sec (180° - α) = \sec (90° + (90° - α)) = - \csc (90° - α) = - \sec α$
$\csc (180° - α) = \csc (90° + (90° - α)) = \sec (90° - α) = \csc α$

$\sin (180° + α) = \sin (90° + (90° + α)) = \cos (90° + α) = - \sin α$
$\cos (180° + α) = \cos (90° + (90° + α)) = - \sin (90° + α) = - \cos α$
$\tan (180° + α) = \tan (90° + (90° + α)) = - \cot (90° + α) = - (-\tan α) = \tan α$
$\cot (180° + α) = \cot (90° + (90° + α)) = - \tan (90° + α) = - (- \cot α) = \cot α$
$\sec (180° + α) = \sec (90° + (90° + α)) = - \csc (90° + α) = - \sec α$
$\csc (180° + α) = \csc (90° + (90° + α)) = \sec (90° + α) = - \csc α$

$\sin (270° - α) = \sin (180° + (90° - α)) = - \sin (90° - α) = - \cos α$
$\cos (270° - α) = \cos (180° + (90° - α)) = - \cos (90° - α) = - \sin α$
$\tan (270° - α) = \tan (180° + (90° - α)) = \tan (90° - α) = \cot α$
$\cot (270° - α) = \cot (180° + (90° - α)) = \cot (90° - α) = \tan α$
$\sec (270° - α) = \sec (180° + (90° - α)) = - \sec (90° - α) = - \csc α$
$\csc (270° - α) = \csc (180° + (90° - α)) = - \csc (90° - α) = - \sec α$

$\sin (270° + α) = \sin (180° + (90° + α)) = - \sin (90° + α) = - \cos α$
$\cos (270° + α) = \cos (180° + (90° + α)) = - \cos (90° + α) = - (- \sin α) = \sin α$
$\tan (270° + α) = \tan (180° + (90° + α)) = \tan (90° + α) = - \cot α$
$\cot (270° + α) = \cot (180° + (90° + α)) = \cot (90° + α) = - \tan α$
$\sec (270° + α) = \sec (180° + (90° + α)) = - \sec (90° + α) = - ( - \csc α) = \csc α$
$\csc (270° + α) = \csc (180° + (90° + α)) = - \csc (90° + α) = - \sec α$

$\sin (360° - α) = \sin (270° + (90° - α)) = - \cos (90° - α) = - \sin α$
$\cos (360° - α) = \cos (270° + (90° - α)) = \sin (90° - α) = \cos α$
$\tan (360° - α) = \tan (270° + (90° - α)) = - \cot (90° - α) = - \tan α$
$\cot (360° - α) = \cot (270° + (90° - α)) = - \tan (90° - α) = - \cot α$
$\sec (360° - α) = \sec (270° + (90° - α)) = \csc (90° - α) = \sec α$
$\csc (360° - α) = \csc (270° + (90° - α)) = - \sec (90° - α) = - \csc α$

$\sin (360° + α) = \sin (270° + (90° + α)) = - \cos (90° + α) = - (- \sin α) = \sin α$
$\cos (360° + α) = \cos (270° + (90° + α)) = \sin (90° + α) = \cos α$
$\tan (360° + α) = \tan (270° + (90° + α)) = - \cot (90° + α) = - (- \tan α) = \tan α$
$\cot (360° + α) = \cot (270° + (90° + α)) = - \tan (90° + α) = - (- \cot α) = \cot α$
$\sec (360° + α) = \sec (270° + (90° + α)) = \csc (90° + α) = \sec α$
$\csc (360° + α) = \csc (270° + (90° + α)) = - \sec (90° + α) = - (- \csc α) = \csc α$

This is very tedious work. Is there any simply method to memorize these identities?

2. Feb 23, 2016

### blue_leaf77

Sum of angle rules:
$$\sin(\alpha\pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha \\ \cos(\alpha\pm \beta) = \cos\alpha\cos\beta \mp \sin\beta\sin\alpha \\$$

Last edited: Feb 23, 2016
3. Feb 23, 2016

### Staff: Mentor

You should also be familiar with specific angles like 30, 60, 90, 120, 150, 180...

and 0, 45, 90, 135, 180, ...

and how to get their sin, cos and tan values.

Commonly found on the unit circle:

https://en.wikipedia.org/wiki/Unit_circle