Approximating Trig Values using the Unit Circle

[mathdad]

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.

 

[MarkFL]

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?

 

[mathdad]

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.

 

[MarkFL]

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:

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

\(\displaystyle \cos\left(160^{\circ}\right)=-\frac{x}{r}\approx\,?\)

 

[mathdad]

I will try tonight after work.

 

[MarkFL]

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:

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

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

Using a calculator, we find:

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

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

 

[mathdad]

I sure will try later tonight.

 

[mathdad]

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:

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

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

Using a calculator, we find:

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

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

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?

 

[MarkFL]

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:

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

 

[mathdad]

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:

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

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.

 

[MarkFL]

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.

 

[mathdad]

Ok.