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Figure 426 : Petals of R = sin(p*A)

    Petals of R = sin(p*A)


  • Q01 | - Diagram : Petals of R = sin(p*A)
  • Q02 | - Prove that the graph of R = sin(p*A) has p petals if p is odd
  • Q03 | - Prove that the graph of R = sin(p*A) has p petals if p is odd
  • Q04 | - Graphic solution is easy and clear
  • Q05 | - Prove that the graph of R = cos(p*A) has 2*p petals if p is even
  • Q06 | - Prove that R = sin(A) is a circle
  • Q07 | - References

    Q01. Diagram : graph of R = sin(p*A)


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    Q02. Prove that the graph of R = sin(p*A) has p petals if p is odd

    Graphic solution
    • As shown in Figure 426, the proof is completed
    Numerical proof Numerical proof : Find direction of petals of R = sin(3*A)
    • Number 3*A ..... Angle A .. Value R .. Plot angle .. (Direction of Petal)
    • -------------------------------------------------------------------------
    • 1 .... 090 ..... 030 ...... +1 ....... 030 ......... 30
    • 2 .... 270 ..... 090 ...... -1 ....... 270 ......... (090 + 180) = 270
    • 3 .... 450 ..... 150 ...... +1 ....... 150 ......... 150
    • 4 .... 630 ..... 210 ...... -1 ....... 030 ......... 210 + 180 - 360 = 30
    • 5 .... 810 ..... 270 ...... +1 ....... 270 ......... 270
    • Since petal number 1 and petal number 4
      • The petal direction = 30 degrees with petal length = 1
      • Hence it has 3 petals
      • The cycle domain is (210 - 30) = pi
      • Angle A = 0 to 360 degrees
    • Similary we can prove
      • sin(5*A) has 05 petals with A = 0 to pi
      • sin(7*A) has 07 petals with A = 0 to pi
      • etc.
    Petal of R = sin(A) is a circle
    • Number ..A .... Angle A ... Value R ... Plot angle .. (Direction of Petal)
    • --------------------------------------------------------------------------
    • 1 .... 090 .... 090 ....... +1 ........ 090 ......... 90
    • 2 .... 270 .... 270 ....... -1 ........ 090 ......... 270 + 180 - 360 = 90
    • Since petal 1 and petal 2 are same direction
    • Hence it has only one petal with cycle domain pi
    • The petal of R = sin(A) is a circle
      • R = Sqr(x^2 + y^2) and sin(A) = y/R
      • Hence R = y/R
      • R^2 = y
      • x^2 + y^2 - y = 0
      • x^2 + (y^2 - y + 1/4 - 1/4) = 0
      • x^2 + (y - 1/2)^2 = (1/2)^2
      • This is a circle with radius = 1/2 and center at (0, 1/2)

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    Q03. Prove that the graph of R = sin(p*A) has 2*p petals if p is even

    Graphic solution
    • As shown in Figure 426, the proof is completed
    Numerical proof : Find direction of petals of R = sin(2*A)
    • Number 2*A .... Angle A .. Value R .. Plot angle .. (Direction of Petal)
    • -------------------------------------------------------------------------
    • 1 .... 090 .... 045 ...... +1 ....... 045 ......... 45
    • 2 .... 270 .... 135 ...... -1 ....... 315 ......... (135 + 180) = 315
    • 3 .... 450 .... 225 ...... +1 ....... 225 ......... 225
    • 4 .... 630 .... 315 ...... -1 ....... 090 ......... 315 + 180 - 360 = 135
    • 5 .... 810 .... 405 ...... +1 ....... 045 ......... 405 - 360 = 45
    • Since petal number 1 and petal number 5
      • The petal direction = 45 degrees with petal length = 1
      • Hence it has 4 petals
      • The cycle domain is (405 - 45) = 2*pi
      • Angle A = 0 to 360 degrees
    • Similary we can prove
      • sin(4*A) has 08 petals with A = 0 to 2*pi
      • sin(6*A) has 12 petals with A = 0 to 2*pi
      • etc.

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    Q04. Graphic solution is easy and clear

    Graphic solution
    • The graphic solution is clear

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    Q05. Prove that the graph of R = cos(p*A) has 2*p petals if p is even

    Numerical proof : Find direction of petals of R = cos(2*A)
    • Number 2*A ...... Angle A ... Value R ... Plot angle ... (Direction of Petal)
    • -----------------------------------------------------------------------------
    • 1 .... 000 ...... 000 ....... +1 ........ 000 .......... 0
    • 2 .... 180 ...... 090 ....... -1 ........ 270 .......... (090 + 180) = 270
    • 3 .... 360 ...... 180 ....... +1 ........ 180 .......... 180
    • 4 .... 540 ...... 270 ....... -1 ........ 090 .......... 270 + 180 - 360 = 90
    • 5 .... 720 ...... 360 ....... +1 ........ 000 .......... 0
    • Since petal number 1 and petal number 5
      • The value R = 1 and petal direction = 0 degrees
      • Hence it has 4 petals
      • The cycle domain is (360 - 0) = 2*pi
      • Angle A = 0 to 360 degrees
    • Similary we can prove
      • sin(4*A) has 08 petals with A = 0 to 2*pi
      • sin(6*A) has 12 petals with A = 0 to 2*pi
      • etc.
    Compare the direction of petals of R = cos(2*A) and R = sin(2*A)
    • R = cos(2*A) 000 090 180 270 (TR 10 06)
    • R = sin(2*A) 045 135 225 315 (TR 10 03)
    • Hence the petals have 45 degrees differenct

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    Q06. Prove that the graph of R = sin(A) is a circle

    Graphic solution
    • As shown in Figure 426, the proof is completed
    Petal of R = sin(A) is a circle
    • Number ..A ...... Angle A ... Value R ... Plot angle ... (Direction of Petal)
    • -----------------------------------------------------------------------------
    • 1 .... 090 ...... 090 ....... +1 ........ 090 .......... 90
    • 2 .... 270 ...... 270 ....... -1 ........ 090 .......... 270 + 180 - 360 = 90
    • Since petal 1 and petal 2 are same direction
    • Hence it has only one petal with cycle domain pi
    • The petal of R = sin(A) is a circle
      • R = Sqr(x^2 + y^2) and sin(A) = y/R
      • Hence R = y/R
      • R^2 = y
      • x^2 + y^2 - y = 0
      • x^2 + (y^2 - y + 1/4 - 1/4) = 0
      • x^2 + (y - 1/2)^2 = (1/2)^2
      • This is a circle with radius = 1/2 and center at (0, 1/2)

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    Q07. References

    Book
    • Chapter 2 : Graphs of R = sin(n*A)
    • Pictorial Mathematics : Patterns by Dr. Shih
    • P13 - P32
    Computer porgram
    • Mathgraph on personal computer by Dr. Shih
    On internet     Compare y = sin(x) and R = sin(A)
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    • Show Room of MD2002 Contact Dr. Shih Math Examples Room
    Copyright © Dr. K. G. Shih. Nova Scotia, Canada.
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