# 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:
$$\sin(\alpha\pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha \\ \cos(\alpha\pm \beta) = \cos\alpha\cos\beta \mp \sin\beta\sin\alpha \\$$

Last edited:
jim mcnamara
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