trigonometric identities Link to heading

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

1st quadrant Link to heading

cos⁡(2π+β)=cos⁡β(4.Q1.1)\tag{4.Q1.1} \cos (2\pi + \beta) = \cos \beta sin⁡(2π+β)=sin⁡β(4.Q1.2)\tag{4.Q1.2} \sin (2\pi + \beta) = \sin \beta cos⁡(12π−β)=sin⁡β(4.Q1.3)\tag{4.Q1.3} \cos (\tfrac12 \pi - \beta) = \sin \beta sin⁡(12π−β)=cos⁡β(4.Q1.4)\tag{4.Q1.4} \sin (\tfrac12 \pi - \beta) = \cos \beta

2nd quadrant Link to heading

cos⁡(π−β)=−cos⁡β(4.Q2.1)\tag{4.Q2.1} \cos (\pi - \beta) = -\cos \beta sin⁡(π−β)=sin⁡β(4.Q2.2)\tag{4.Q2.2} \sin (\pi - \beta) = \sin \beta cos⁡(12π+β)=−sin⁡β(4.Q2.3)\tag{4.Q2.3} \cos (\tfrac12 \pi + \beta) = -\sin \beta sin⁡(12π+β)=cos⁡β(4.Q2.4)\tag{4.Q2.4} \sin (\tfrac12 \pi + \beta) = \cos \beta

3rd quadrant Link to heading

cos⁡(π+β)=−cos⁡β(4.Q3.1)\tag{4.Q3.1} \cos (\pi + \beta) = -\cos \beta sin⁡(π+β)=−sin⁡β(4.Q3.1)\tag{4.Q3.1} \sin (\pi + \beta) = -\sin \beta cos⁡(32π−β)=−sin⁡β(4.Q3.3)\tag{4.Q3.3} \cos (\tfrac32 \pi - \beta) = -\sin \beta sin⁡(32π−β)=−cos⁡β(4.Q3.4)\tag{4.Q3.4} \sin (\tfrac32 \pi - \beta) = -\cos \beta

4th quadrant Link to heading

cos⁡(2π−β)=cos⁡β(4.Q4.1)\tag{4.Q4.1} \cos (2\pi - \beta) = \cos \beta sin⁡(2π−β)=−sin⁡β(4.Q4.1)\tag{4.Q4.1} \sin (2\pi - \beta) = -\sin \beta cos⁡(32π+β)=sin⁡β(4.Q4.3)\tag{4.Q4.3} \cos (\tfrac32 \pi + \beta) = \sin \beta sin⁡(32π+β)=−cos⁡β(4.Q4.4)\tag{4.Q4.4} \sin (\tfrac32 \pi + \beta) = -\cos \beta

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

  • cos⁡(−β)=cos⁡β\cos (-\beta) = \cos \beta,
  • sin⁡(−β)=−sin⁡β\sin (-\beta) = -\sin \beta.

examples Link to heading

sin⁡60°\sin 60 \degree Link to heading

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

cos⁡120°\cos 120 \degree Link to heading

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

cos⁡210°\cos 210 \degree Link to heading

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

sin⁡(−330°)\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)
sin (—330)
sin (—360 + 30)
sin 30
—sin 330
—sin (360 — 30)
—sin (270 + 60)
cos 60

sin⁡(−150°)\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
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)
— 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⁡(θ+14π)\cos (\theta + \tfrac14 \pi) in 2nd, 3rd, and 4th quadrants.