MHB Approximating Trig Values using the Unit Circle

  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    Trig
AI Thread Summary
The discussion focuses on approximating trigonometric values using a unit circle, specifically for cos 160° and sin 160°. The user expresses confusion over the explanation provided in a textbook and seeks clarity on how to derive these values from a diagram of the unit circle. They describe measuring distances in pixels to estimate the coordinates of the point at 160° on the circle, leading to calculated values of sin(160°) and cos(160°). The conversation emphasizes the method of using pixel measurements to derive the necessary trigonometric ratios, highlighting the importance of understanding the relationship between the radius and the distances from the axes. Clear steps for using pixel measurements to approximate trigonometric values are requested for better comprehension.
mathdad
Messages
1,280
Reaction score
0
The following question, in my opinion, is not well-explained in Section 6.2 by David Cohen. I went over the question several times but it is not clear at all.

Use the picture to approximate the following trig values to within successive tenths. Then use a calculator to check your answers. Round the calculator answers to two decimal places.

1. cos 160° and sin 160°

2. cos 3 and sin 3

Note: If your explanation is clear and simple enough to follow, I may try question 2 on my own (or at least try) showing my work as always.

Note: The MHB does not allow me to upload the picture because it is too big. I will try describing the picture.

The picture is that of the unit circle: x^2 + y^2 = 1.

I will now describe each quadrant.

In Quadrant 1:

Along the edge of the circle several degrees are listed from 10° to 80°. Between 50° and 60°, there is a number 1.

In Quadrant 2:

Along the edge of the circle several degrees are listed from 100° to 170°. Between 110° and 120°, there is a number 2. Between 170° and the x-axis there is a number 3.

In Quadrant 3:

Along the edge of the circle several degrees are listed from 190° to 260°. Between 220° and 230°, there is a number 4.

In Quadrant 4:

Along the edge of the circle several degrees are listed from 280° to 350°. Between 280° and 290°, there is a number 5.
Between 340° and 350° there is a number 6.

Note: The degrees on the unit circle are in increments of 10 degrees.
 
Mathematics news on Phys.org
Here is an image of a unit circle with degrees marked off on the inside and radians on the outside:

View attachment 7925

Look at where $160^{\circ}$ is marked on the circle, what would you estimate the $x$ and $y$ coordinates of that point to be?
 

Attachments

  • Degree-Radian_Conversion_tau.png
    Degree-Radian_Conversion_tau.png
    87.8 KB · Views: 132
I found 160° on the circle but cannot estimate what the x and y coordinates of that point are, honestly. Can you answer this question? I can then use your steps to try the rest.
 
RTCNTC said:
I found 160° on the circle but cannot estimate what the x and y coordinates of that point are, honestly. Can you answer this question? I can then use your steps to try the rest.

It would be easier to use your printed diagram and a ruler, but I can't locate that problem in section 6.2 of my copy of the book. I would begin by measuring the radius of the circle, from the origin, to the circle along one of the axes, we'll call this $r$. Next, measure the horizontal distance from the $y$-axis to the point on the circle at $160^{\circ}$, which we'll call $x$. And finally, measure the vertical distance from the point to the $x$-axis and call this $y$.

Then:

$$\sin\left(160^{\circ}\right)=\frac{y}{r}\approx\,?$$

$$\cos\left(160^{\circ}\right)=-\frac{x}{r}\approx\,?$$
 
I will try tonight after work.
 
RTCNTC said:
I will try tonight after work.

I realized I could use the above image and measure the pixels, and I found:

r = 428
x = 404
y = 148

Using these measurements, I find:

$$\sin\left(160^{\circ}\right)\approx0.3458$$

$$\cos\left(160^{\circ}\right)\approx-0.9439$$

Using a calculator, we find:

$$\sin\left(160^{\circ}\right)\approx0.3420$$

$$\cos\left(160^{\circ}\right)\approx-0.9397$$
 
I sure will try later tonight.
 
MarkFL said:
I realized I could use the above image and measure the pixels, and I found:

r = 428
x = 404
y = 148

Using these measurements, I find:

$$\sin\left(160^{\circ}\right)\approx0.3458$$

$$\cos\left(160^{\circ}\right)\approx-0.9439$$

Using a calculator, we find:

$$\sin\left(160^{\circ}\right)\approx0.3420$$

$$\cos\left(160^{\circ}\right)\approx-0.9397$$

I am not getting this at all.

How did you conclude sin (160°) is approximately 0.3458 reading the unit circle provided? Write down the steps?
 
RTCNTC said:
I am not getting this at all.

How did you conclude sin (160°) is approximately 0.3458 reading the unit circle provided? Write down the steps?

Using the image I posted, I found the radius of the circle to be 428 pixels, and I found the distance of the point on the circle at 160° to the $x$-axis to be 148 pixels. And so:

$$\sin\left(160^{\circ}\right)\approx\frac{148}{428}\approx0.3458$$
 
  • #10
MarkFL said:
Using the image I posted, I found the radius of the circle to be 428 pixels, and I found the distance of the point on the circle at 160° to the $x$-axis to be 148 pixels. And so:

$$\sin\left(160^{\circ}\right)\approx\frac{148}{428}\approx0.3458$$

Explain this pixel stuff.

I understand this:

sin (a given degree) = d/r, where d = distance of the point on the circle at the given degree and r = radius of circle.

I am confused about the pixel reading. Good night. I will read your next reply tomorrow.
 
  • #11
RTCNTC said:
Explain this pixel stuff.

I understand this:

sin (a given degree) = d/r, where d = distance of the point on the circle at the given degree and r = radius of circle.

I am confused about the pixel reading. Good night. I will read your next reply tomorrow.

I loaded the image into MS-Paint, and then used the line drawing tool to measure the distances. It was simply a means for me to measure the distances without having a printed image that I could measure with a ruler.
 
  • #12
Ok.
 
Back
Top