sin cos 4 quadrants

trigonometric identities Link to heading

Thera identities in 1st - 4th quadrants, map sine and cosine function to 1st quadrant.

1st quadrant Link to heading

$$\tag{4.Q1.1} \cos (2\pi + \beta) = \cos \beta $$ $$\tag{4.Q1.2} \sin (2\pi + \beta) = \sin \beta $$ $$\tag{4.Q1.3} \cos (\tfrac12 \pi - \beta) = \sin \beta $$ $$\tag{4.Q1.4} \sin (\tfrac12 \pi - \beta) = \cos \beta $$

2nd quadrant Link to heading

$$\tag{4.Q2.1} \cos (\pi - \beta) = -\cos \beta $$ $$\tag{4.Q2.2} \sin (\pi - \beta) = \sin \beta $$ $$\tag{4.Q2.3} \cos (\tfrac12 \pi + \beta) = -\sin \beta $$ $$\tag{4.Q2.4} \sin (\tfrac12 \pi + \beta) = \cos \beta $$

3rd quadrant Link to heading

$$\tag{4.Q3.1} \cos (\pi + \beta) = -\cos \beta $$ $$\tag{4.Q3.1} \sin (\pi + \beta) = -\sin \beta $$ $$\tag{4.Q3.3} \cos (\tfrac32 \pi - \beta) = -\sin \beta $$ $$\tag{4.Q3.4} \sin (\tfrac32 \pi - \beta) = -\cos \beta $$

4th quadrant Link to heading

$$\tag{4.Q4.1} \cos (2\pi - \beta) = \cos \beta $$ $$\tag{4.Q4.1} \sin (2\pi - \beta) = -\sin \beta $$ $$\tag{4.Q4.3} \cos (\tfrac32 \pi + \beta) = \sin \beta $$ $$\tag{4.Q4.4} \sin (\tfrac32 \pi + \beta) = -\cos \beta $$

The two Eqns (4.Q4.1) and (4.Q4.2) also mean that

$\cos (-\beta) = \cos \beta$,
$\sin (-\beta) = -\sin \beta$.

examples Link to heading

$\sin 60 \degree$ Link to heading

sin 60 = sin (90 - 30) = cos 30

flowchart LR A --> B --> C --> D --> A A(["sin 60"]) B(["sin (90 - 30)"]) C(["cos 30"]) D(["cos (90 - 60)"])

$\cos 120 \degree$ Link to heading

cos 120 = cos (180 - 60) = - cos 60
cos 120 = cos (90 + 30) =  - sin 30 = - cos 60

flowchart LR A --> B --> C A --> D --> E --> C A(["cos 120"]) B(["cos (180 — 60)"]) C(["— cos 60"]) D(["cos (90 + 30) "]) E(["— sin 30"])

$\cos 210 \degree$ Link to heading

cos 210 = cos (180 + 30) = - cos 30
cos 210 = cos (270 - 60) = - sin 60 = - cos 30

flowchart LR A --> B --> C A --> D --> E --> C A(["cos 210"]) B(["cos (180 + 30)"]) C(["— cos 30"]) D(["cos (270 — 60) "]) E(["— sin 60"])

$\sin (-330 \degree)$ Link to heading

sin(-330) = sin(-360 + 30) = sin(30)
sin(-330) = -sin(330) = -sin(360-30) = sin(30)
sin(-330) = -sin(330) = -sin(270+60) = cos(60)

flowchart LR A --> B --> C A --> D --> E --> C D --> F --> G --> C A(["sin (—330)"]) B(["sin (—360 + 30)"]) C(["sin 30"]) D(["—sin 330 "]) E(["—sin (360 — 30)"]) F(["—sin (270 + 60)"]) G(["cos 60"])

$\sin (-150 \degree)$ Link to heading

sin (-150) = -sin 150 = -sin (180 - 30) = -sin 30
sin (-150) = -sin 150 = -sin (90 + 60) = -cos 60 = -sin 30
sin (-150) = sin (360-210) = sin (-210) = -sin 210 = -sin (180 + 30) = -sin 30
sin (-150) = sin (360-210) = sin (-210) = -sin 210 = -sin (270-60) = -cos 60 = -sin 30

flowchart LR A --> B --> C --> D B --> E --> F --> D A --> G --> H --> I --> D H --> J --> K --> D A(["sin (—150)"]) B(["— sin 150"]) C(["— sin (180 — 30)"]) D(["— sin 30"]) E(["— sin (90 + 60)"]) F(["— cos 60"]) G(["sin (—360 + 210)"]) H(["sin 210"]) I(["sin (180 + 30)"]) J(["sin (270 - 60)"]) K(["— cos 60"])

Test results

sin(-150) = -0.5000
-sin(150) = -0.5000
-sin(180-30) = -0.5000
-sin(30) = -0.5000
-sin(90+60) = -0.5000
-cos(60) = -0.5000
sin(-360+210) = -0.5000
sin(210) = -0.5000
sin(180+30) = -0.5000
sin(270-60) = -0.5000

Python code https://onecompiler.com/python/3zrb5qd5e

import math

def sin(x):
  return math.sin(x*math.pi/180);

def cos(x):
  return math.cos(x*math.pi/180);

s = [
  "sin(-150)", 
  "-sin(150)",
  "-sin(180-30)",
  "-sin(30)",
  "-sin(90+60)",
  "-cos(60)",
  "sin(-360+210)",
  "sin(210)",
  "sin(180+30)",
  "sin(270-60)"
]

for e in s:
  print(e, "=", f'{eval(e):.4f}')

challenges Link to heading

Try your own expression, e.g. represent $\cos (\theta + \tfrac14 \pi)$ in 2nd, 3rd, and 4th quadrants.