# Trigonometry Identity Question

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?

#### blue_leaf77

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

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#### jedishrfu

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