Trigonometry Identity Question

  • Thread starter askor
  • Start date
  • #1
110
2
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?
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
784
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:
  • Like
Likes jim mcnamara
  • #3
12,244
5,953
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
 

Related Threads on Trigonometry Identity Question

  • Last Post
Replies
9
Views
675
Replies
15
Views
3K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
14K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
832
  • Last Post
Replies
11
Views
2K
Top