# Math Portfolio: trigonometry investigation (circle trig)

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Introduction

Math Honors 1: Trigonometry Investigation Task

This investigation will present an analysis on initial problem by setting patterns and establishing mathematical relationships between the parameters in the problem. In this specific investigation, I will find to see the relationship between radius R and point X and Y in a coordinate plane. The center of the circle will be (0, 0) or the origin and the radius R will be unknown. Point P with the coordinate (x,y) will always be on the circumference of the circle, and will always be perpendicular from the X axis to the point.

Part A: Circle Trigonometry

The diagram above, the radius r and the point (x, y) will form a right triangle. Therefore we can state that the equation to find the relationship can be described as Pythagorean Theorem. Wherever the coordinates (x, y) lies on the circle it will always form right triangle.

Therefore we can use the equation distance between the two points.

After we can state that the endpoints of the radius are set in the origin meaning that one end point will always set in the origin (0, 0). So we can simplify our equation further more.

Then after squaring this will result in Pythagorean Theorem

The diagram above with point R and Q is just to back up or example the idea of relationship use of right triangle to use the equation of Pythagorean Theorem.

As a result, due to Pythagorean Theorem, r squared will always be positive. And by using the relation equation the x and y value can be positive and negative thus 0. Negative and positive values meaning that they can be in any exact quadrants but one endpoint being at origin creating a right triangle.

Middle

-40

-0.6427876

0.766044

-0.8390996

-0.83909963

-20

-0.3420201

0.939693

-0.3639702

-0.36397023

0

0

1

0

0

20

0.34202014

0.939693

0.36397023

0.363970234

40

0.64278761

0.766044

0.83909963

0.839099631

60

0.8660254

0.5

1.73205081

1.732050808

80

0.98480775

0.173648

5.67128182

5.67128182

90

1

6.13E-17

undefined

1.63246E+16

100

0.98480775

-0.17365

-5.6712818

-5.67128182

120

0.8660254

-0.5

-1.7320508

-1.73205081

140

0.64278761

-0.76604

-0.8390996

-0.83909963

160

0.34202014

-0.93969

-0.3639702

-0.36397023

180

1.2251E-16

-1

-1.225E-16

-1.2251E-16

200

-0.3420201

-0.93969

0.36397023

0.363970234

220

-0.6427876

-0.76604

0.83909963

0.839099631

240

-0.8660254

-0.5

1.73205081

1.732050808

260

-0.9848078

-0.17365

5.67128182

5.67128182

270

-1

-1.8E-16

undefined

5.44152E+15

280

-0.9848078

0.173648

-5.6712818

-5.67128182

300

-0.8660254

0.5

-1.7320508

-1.73205081

320

-0.6427876

0.766044

-0.8390996

-0.83909963

340

-0.3420201

0.939693

-0.3639702

-0.36397023

360

-2.45E-16

1

-2.45E-16

-2.4503E-16

As can be clearly seen in the table above, the values for tanθ and are identical.

Further to prove this, I used TI 83 calculator for better explanation such as random numbers.

First for quadrant 1, I will choose a value or a degree for tangent theta between 0-90 degrees and -360 to -270. Which are 76 and -298 degrees.

For other revolutions

For quadrant two the range between 90 to 180 and -270 to -180

With the example of 127 and -222

For quadrant 3, the range of tan theta will be 180-270 and -180 to -90.

With the example of 199 and -101

Finall for quadrant four with the range of 270 to 360 and -90 to 0

With the example of 354 and -4 degrees.

Therefore, through this further examples we again can see that the conjecture between tanθ and are identical.

a) Cos θ= x/r

sin θ= y/r

tan θ= y/x

sin θ/ Cos θ=(y/r)/(x/r) so y/r divided by x/r equals y/r times r/x. Which r gets simplified equaling y/x. So if we see tan it will become y/x which means sin θ/ Cos θ equals to tan θ.

Equation:

b) Expressing Cosθ,tan θ and tan θ in terms of x,y and r

then for x value it will be x= Cosθ( r ), y value y= sin θ( r ) and tan equaling Tan θ= y/x equaling sin θ( r ) divided by Cosθ( r ) then as what I said in the above, the r gets simplified just leaving.

y= sin θ( r )

Tan θ= y/x equaling sin θ( r ) divided by Cosθ( r ) then as what I said in the above, the r gets simplified just leaving

sin²θ+cos²θ, the relationship between sin²θ and cos²θ is whenever I add them sin²θ+cos²θ this becomes one and even though whatever angle we put for θ the sum sin²θ+cos²θ will always be 1.

θ | sin²θ | cos²θ | sin²θ+cos²θ | |||

-360 | 6.00395E-32 | 1 | 1 | |||

-340 | 0.116977778 | 0.883022 | 1 | |||

-320 | 0.413175911 | 0.586824 | 1 | |||

-300 | 0.75 | 0.25 | 1 | |||

-280 | 0.96984631 | 0.030154 | 1 | |||

-270 | 1 | 3.38E-32 | 1 | |||

-260 | 0.96984631 | 0.030154 | 1 | |||

-240 | 0.75 | 0.25 | 1 | |||

-220 | 0.413175911 | 0.586824 | 1 | |||

-360 | 6.00395E-32 | 1 | 1 | |||

-200 | 0.116977778 | 0.883022 | 1 | |||

-180 | 1.50099E-32 | 1 | 1 | |||

-160 | 0.116977778 | 0.883022 | 1 | |||

-140 | 0.413175911 | 0.586824 | 1 | |||

-120 | 0.75 | 0.25 | 1 | |||

-100 | 0.96984631 | 0.030154 | 1 | |||

-90 | 1 | 3.75E-33 | 1 | |||

-80 | 0.96984631 | 0.030154 | 1 | |||

-60 | 0.75 | 0.25 | 1 | |||

-40 | 0.413175911 | 0.586824 | 1 | |||

-20 | 0.116977778 | 0.883022 | 1 | |||

0 | 0 | 1 | 1 | |||

20 | 0.116977778 | 0.883022 | 1 | |||

40 | 0.413175911 | 0.586824 | 1 | |||

60 | 0.75 | 0.25 | 1 | |||

80 | 0.96984631 | 0.030154 | 1 | |||

90 | 1 | 3.75E-33 | 1 | |||

100 | 0.96984631 | 0.030154 | 1 | |||

120 | 0.75 | 0.25 | 1 | |||

140 | 0.413175911 | 0.586824 | 1 | |||

160 | 0.116977778 | 0.883022 | 1 | |||

180 | 1.50099E-32 | 1 | 1 | |||

200 | 0.116977778 | 0.883022 | 1 | |||

220 | 0.413175911 | 0.586824 | 1 | |||

240 | 0.75 | 0.25 | 1 | |||

260 | 0.96984631 | 0.030154 | 1 | |||

270 | 1 | 3.38E-32 | 1 | |||

280 | 0.96984631 | 0.030154 | 1 | |||

300 | 0.75 | 0.25 | 1 | |||

320 | 0.413175911 | 0.586824 | 1 | |||

340 | 0.116977778 | 0.883022 | 1 | |||

360 | 6.00395E-32 | 1 | 1 |

In Part A when expressing sin θ, cos θ and tan θ in terms of x, y and r. Which sin θ=y/r, cos θ= x/r and cos² θ= x²/r² so when squaring cosine theta and squaring sine theta sin² θ will equal= y²/r² after this when adding sin² θ + cos² θ this will equal x²/r² + y²/r² which the sum is x²+ y²/r² then since x ² + y² is r² we will substitute r² in the equation of x²+ y²/r² giving r²/ r² simplifying to 1. As a result x²+ y²=r² so in number 1 Part A

sin² θ + cos² θ= 1.

The value of Sinθ and Cos θ in the first quadrant 0<= θ <= 90. The value of sine theta equals the value of y divided by the value of r and the value of cosine theta equals the value of x divided by the value of r. Therefore, the square of the value of sine theta equals the square of the value of y divided by the square of the value of r and the square of the value of cosine theta equals the square of the value of x divided by the square of the value of r. When the square of the value of cosine theta is added to the square of the value of sine theta, the result of the square of the value of x divided by the square of the value of r is added to the result of the square of the value of y divided by the square of the value of r.

=()+ ()

=

=

=1

The values of sinθ and cosθ within the range of 0°≤θ≤90° are analyzed in order to conjecture another relationship between sinθ and cosθ for any angle θ.

A portion of Table 1 within the range of 0°≤θ≤90° showing the relationship between sinθ and cosθ for any angle θ

θ | sinθ | cosθ |

0 | 0 | 1 |

10 | 0.17364818 | 0.984808 |

20 | 0.34202014 | 0.939693 |

30 | 0.5 | 0.866025 |

40 | 0.64278761 | 0.766044 |

50 | 0.76604444 | 0.642788 |

60 | 0.8660254 | 0.5 |

70 | 0.93969262 | 0.34202 |

80 | 0.98480775 | 0.173648 |

90 | 1 | 0 |

Conclusion

-≤bθ≤

If b is positive, the above inequality is 0≤θ≤. When 0<b<1, the period of y=sinbx is greater than and represents a horizontal stretching of the graph of y=asinx. Similarly, if b>1, the period of y=sinbx is less than and represents a horizontal compression of the graph of y=asinx. When b is negative, the above inequality becomes ≤θ≤0.

For either a positive or negative value of b, one cycle (period) of the graph of y=sinbθ and y=cosbθ are obtained respectively on an interval of. Then, the frequency, the reciprocal of the period, is.

The conjecture is now verified by considering further examples of b.

The maxima of the graph is 1 and the minima of the graph is -1. The amplitude in the graph is calculated by resulting to 1. The period in thegraph is which approximately 3.6276 when expressed to radian are. The frequency is which is approximately 0.2757 when expressed to radian.

The maxima of the graph is 1 and the minima of the graph is -1. The amplitude in the graph is calculated by resulting to 1. The period in thegraph is which approximately 10.8828 when expressed to radian are. The frequency is which is approximately 0.9069 when expressed to radian.

The maxima of the graph is 1 and the minima of the graph is -1. The amplitude in the graph is calculated by resulting to 1. The period in thegraph is which approximately 4.4429 when expressed to radian are.

The frequency is which is approximately 0.5642 when expressed to radian.

The maxima of the graph is 1 and the minima of the graph is -1.

The amplitude in the graph is calculated by resulting to 1.

The period in thegraph is which is approximately when expressed to radian. The frequency is which is when expressed to radian.

Conclusion:

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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