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Math Portfolio: trigonometry investigation (circle trig)

Extracts from this document...

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 Trigonometryimage00.pngimage01.png

  1. image14.png

image07.pngimage08.png

image09.png

image10.png

image11.png

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.

image15.png

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.

image78.png

Then after squaring this will result in Pythagorean Theorem

image125.png

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.

...read more.

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 image136.png 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.

image138.pngimage137.png

For other revolutions

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

With the example of 127 and -222

image141.pngimage140.png

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

With the example of 199 and -101image142.png

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

With the example of 354 and -4 degrees.image143.png

image144.png

Therefore, through this further examples we again can see that the conjecture between tanθ and image136.png 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:

image145.png

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.

image47.png

image33.png

image147.png

image148.png

image149.png

=(image150.png)+ (image151.png)

=image152.png

=image153.png

=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

...read more.

Conclusion

image102.png, which is how long the cycle takes to repeat while the value of b in image94.png is image103.png so it is image104.pngand image105.png when calculated, which is how long the cycle takes to repeat.

-image19.png≤bθ≤image19.png

If b is positive, the above inequality is 0≤θ≤image106.png. When 0<b<1, the period of y=sinbx is greater than image19.png and represents a horizontal stretching of the graph of y=asinx. Similarly, if b>1, the period of y=sinbx is less than image19.png and represents a horizontal compression of the graph of y=asinx. When b is negative, the above inequality becomes image100.png≤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 ofimage107.png. Then, the frequency, the reciprocal of the period, isimage108.png.

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

image109.png

image110.png

The maxima of the image111.png graph is 1 and the minima of the image91.png graph is -1. The amplitude in the image111.png graph is calculated by  resulting to 1. The period in theimage111.pnggraph is image112.pngwhich approximately 3.6276 when expressed to radian are. The frequency is image113.png which is approximately 0.2757 when expressed to radian.

The maxima of the image114.png graph is 1 and the minima of the image114.png graph is -1. The amplitude in the image114.png graph is calculated by image115.png resulting to 1. The period in theimage114.pnggraph is image116.pngwhich approximately 10.8828 when expressed to radian are. The frequency is image117.pngwhich is approximately 0.9069 when expressed to radian.

image118.pngimage119.png

The maxima of the image120.png graph is 1 and the minima of the image120.png graph is -1. The amplitude in the image120.png graph is calculated by image115.png resulting to 1. The period in theimage120.pnggraph is image121.pngwhich approximately 4.4429 when expressed to radian are.

The frequency is image122.pngwhich is approximately 0.5642 when expressed to radian.

The maxima of the image123.png graph is 1 and the minima of the image123.png graph is -1.

The amplitude in the image123.png graph is calculated by image115.png resulting to 1.

The period in theimage123.pnggraph is image124.pngwhich is approximately image126.pngwhen expressed to radian. The frequency is image127.pngwhich is image128.png when expressed to radian.

Conclusion:

...read more.

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