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Mathematics Dictionary
Dr. K. G. Shih
Functions in various quadrants
Subjects


  • TR 04 01 | - Signs of functions in various quadrants
  • TR 04 02 | - Angles in 1st quadrant
  • TR 04 03 | - Angles in 2nd quadrant
  • TR 04 04 | - Angles in 3rd quadrant
  • TR 04 05 | - Angles in 4th quadrant
  • TR 04 06 | - Examples : Find values of sin(A) if A > 90 degrees
  • TR 04 07 | - Examples : Find values of cos(A) if A > 90 degrees
  • TR 04 08 | - Examples : Find values of tan(A) if A > 90 degrees
  • TR 04 09 | - Show that sin(1) + sin(2) + sin(3) + ... + sin(359) + sin(360) = 0
  • TR 04 10 | - Show that cos(1) + cos(2) + cos(3) + ... + cos(179) + cos(180) = -1
    • Answers


    TR 04 01. Trigonometric ratio in rectangular coordinate

    Defintion : Coordinate P(x,y) and center C(0,0). PC makes angle A with x-axis.
    • Sin(A) = y/r and csc(A) = r/y.
    • cos(A) = x/r and sec(A) = r/x.
    • tan(x) = y/x and cot(A) = x/y.
    The sign of the functions in different quadrants
      sin(A) = (+) | sin(A) = (+)
      cos(A) = (-) | cos(A) = (+)
      tan(A) = (-) | tan(A) = (+)
      -------------|--------------
      sin(A) = (-) | sin(A) = (-)
      cos(A) = (-) | cos(A) = (+)
      tan(A) = (+) | tan(A) = (-)

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    TR 04 02. Functions and co-functions

    In 1st quadarnt : Angle = 90-A or angle = 360+A
    • sin(90-A) = cos(A).
    • cos(90-A) = sin(A).
    • tan(90-A) = cot(A).
    • csc(90-A) = sec(A).
    • sec(90-A) = csc(A).
    • cot(90-A) = tan(A).
    Angle 360+A
    • sin(360+A) = +sin(A).
    • cos(360+A) = +cos(A).
    • tan(360+A) = +tan(A).
    • csc(360+A) = +csc(A).
    • sec(360+A) = +sec(A).
    • cot(360+A) = +cot(A).

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    TR 04 03. Angles in 2nd quadrant

    Angle 90+A and A is acute angle
    • sin(90+A) = +cos(A).
    • cos(90+A) = -sin(A).
    • tan(90+A) = -cot(A).
    • csc(90+A) = +sec(A).
    • sec(90+A) = -csc(A).
    • cot(90+A) = -tan(A).
    Angle 180-A and A is acute angle
    • sin(180-A) = +sin(A).
    • cos(180-A) = -cos(A).
    • tan(180-A) = -tan(A).
    • csc(180-A) = +csc(A).
    • sec(180-A) = -sec(A).
    • cot(180-A) = -cot(A).
    Note
    • The formulae are also true if A is not acute
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    Q04 Angles in 3rd quadrant

    Angle 270-A and A is acute angle
    • sin(270-A) = -cos(A).
    • cos(270-A) = -sin(A).
    • tan(270-A) = +cot(A).
    • csc(270-A) = -sec(A).
    • sec(270-A) = -csc(A).
    • cot(270-A) = +tan(A).
    Angle 180+A and A is acute angle
    • sin(180+A) = -sin(A).
    • cos(180+A) = -cos(A).
    • tan(180+A) = +tan(A).
    • csc(180+A) = -csc(A).
    • sec(180+A) = -sec(A).
    • cot(180+A) = +cot(A).
    Note
    • The formulae are also true if A is not acute

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    TR 04 05. Angles in 4th quadrant

    Angle 270+A and A is acute angle
    • sin(270+A) = -cos(A).
    • cos(270+A) = +sin(A).
    • tan(270+A) = -cot(A).
    • csc(270+A) = -sec(A).
    • sec(270+A) = +csc(A).
    • cot(270+A) = -tan(A).
    Angle 360-A and A is acute angle
    • sin(360-A) = -sin(A).
    • cos(360-A) = +cos(A).
    • tan(360-A) = -tan(A).
    • csc(360-A) = -csc(A).
    • sec(360-A) = +sec(A).
    • cot(360-A) = -cot(A).
    Note
    • The formulae are also true if A is not acute

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    TR 04 06. Examples : Find values of sin(A) if A > 90 degrees

    Angle in 2nd quadrant
    • 1. sin(120) = sin(180-60) = sin(60) = Sqr(3)/2.
    • 2. sin(135) = sin(180-45) = sin(45) = Sqr(2)/2.
    • 3. sin(150) = sin(180-30) = sin(30) = 1/2.
    Angle in 3rd quadrant
    • 1. sin(210) = sin(180+30) = -sin(30) = -1/2.
    • 2. sin(225) = sin(180+45) = -sin(45) = -Sqr(2)/2.
    • 3. sin(240) = sin(180+60) = -sin(60) = -Sqr(3)/2.
    Angle in 4th quadrant
    • 1. sin(300) = sin(360-60) = -sin(60) = -Sqr(3)/2.
    • 2. sin(315) = sin(360-45) = -sin(45) = -Sqr(2)/2.
    • 3. sin(330) = sin(360-30) = -sin(30) = -1/2.
    Angle greater than 360
    • 1. sin(0405) = sin(0360+45) = sin(45) = Sqr(2)/2
    • 2. sin(0750) = sin(0720+30) = sin(30) = 1/2
    • 3. sin(1110) = sin(1080+30) = sin(30) = 1/2

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    TR 04 07. Examples : Find values of cos(A) if A > 90 degrees

    Angle in 2nd quadrant
    • 1. cos(120) = cos(180-60) = -cos(60) = -1/2.
    • 2. cos(135) = cos(180-45) = -cos(45) = -Sqr(2)/2.
    • 3. cos(150) = cos(180-30) = -cos(30) = -Sqr(3)/2.
    Angle in 3rd quadrant
    • 1. cos(210) = cos(180+30) = -cos(30) = -Sqr(3)/2.
    • 2. cos(225) = cos(180+45) = -cos(45) = -Sqr(2)/2.
    • 3. cos(240) = cos(180+60) = -cos(60) = -1/2.
    Angle in 4th quadrant
    • 1. cos(300) = cos(360-60) = +cos(60) = +1/2.
    • 2. cos(315) = cos(360-45) = +cos(45) = +Sqr(2)/2.
    • 3. cos(330) = cos(360-30) = +cos(30) = +Sqr(3)/2.
    Angle greater than 360
    • 1. cos(0405) = cos(0360+45) = cos(45) = Sqr(2)/2
    • 2. cos(0750) = cos(0720+30) = cos(30) = Sqr(3)/2
    • 3. cos(1110) = cos(1080+30) = cos(30) = Sqr(3)/2

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    TR 04 08. Examples : Find values of tan(A) if A > 90 degrees

    Angle in 2nd quadrant
    • 1. tan(120) = tan(180-60) = -tan(60) = -Sqr(3).
    • 2. tan(135) = tan(180-45) = -tan(45) = -1.
    • 3. tan(150) = tan(180-30) = -tan(30) = -Sqr(3)/3.
    Angle in 3rd quadrant
    • 1. tan(210) = tan(180+30) = +tan(30) = +Sqr(3)/3.
    • 2. tan(225) = tan(180+45) = +tan(45) = +1.
    • 3. tan(240) = tan(180+60) = +tan(60) = +Sqr(3).
    Angle in 4th quadrant
    • 1. tan(300) = tan(360-60) = -tan(60) = -Sqr(3).
    • 2. tan(315) = tan(360-45) = -tan(45) = -1.
    • 3. tan(330) = tan(360-30) = -tan(30) = -Sqr(3)/3.
    Angle greater than 360
    • 1. tan(0405) = tan(0360+45) = tan(45) = 1.
    • 2. tan(0750) = tan(0720+30) = tan(30) = Sqr(3)/3.
    • 3. tan(1110) = tan(1080+30) = tan(30) = Sqr(3)/3.

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    TR 04 09. Show that sin(1) + sin(2) + sin(3) + ... + sin(359) + sin(360) = 0

    Hint
    • 1. Formula required
      • sin(180-A) = +sin(A).
      • sin(180+A) = -sin(A).
      • sin(360-A) = -sin(A).
    • 2. Special angles : sin(90)=1, sin(180)=0,sin(270)=-1 and sin(360)=0
    • 3. Special method
      • sin(1) + sin(359) = 0.
      • sin(2) + sin(358) = 0.
    Proof
    • Prove that sin(1)+sin(359)+sin(2)+sin(358)+...+sin(89)+sin(271) = 0
      • Since sin(359) = sin(360-1) = -sin(1) and hence sin(1) + sin(359) = 0.
      • Since sin(358) = sin(360-2) = -sin(2) and hence sin(2) + sin(358) = 0.
      • Hence sin(1) + sin(2) + ... + sin(89) + sin(271) + ... + sin(359) = 0.
    • And sin(0) + sin(90) + sin(180) + sin(270) = 0
    • Prove that sin(91)+sin(269)+sin(92)+sin(268)+...+sin(179)+sin(181) = 0.
      • Since sin(269) = sin(180+89) = -sin(89) and sin(91) = sin(180-89) = +sin(89).
      • Hence sin(91)+sin(269) = 0.
      • Since sin(268) = sin(180+88) = -sin(88) and sin(92) = sin(180-88) = +sin(88).
      • Hence sin(92)+sin(268) = 0.
      • Hence the proof is completed.

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    TR 04 10. Show that cos(1) + cos(2) + cos(3) + ... + cos(179) + cos(180) = -1

    Hint
    • 1. Formula
      • cos(180 - A) = -cos(A).
    • 2. Facts
      • cos(90) = 0 and cos(180) = -1.
    • 2. Method
      • Prove that cos(1) + cos(180) = 0.
    • Proof
      • cos(1) + cos(179) = cos(1) + cos(180-1) = cos(1) - cos(1) = 0.
      • cos(2) + cos(178) = cos(2) + cos(180-2) = cos(2) - cos(2) = 0.
      • ....
      • cos(89) + cos(91) = cos(89) + cos(180-89) = cos(89) - cos(89) = 0.
      • cos(90) = 0 and cos(180) = -1.
      • Hence the proof is completed.

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