Trigonometry Identity Question

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For the given triangle as below:

Triangle.png


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

Science Advisor
Homework Helper
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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:
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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
 

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