1 | S e c t i o n 8 . 2

Chapter 8.2: Degrees and Radians, Reference Angles In section 8.1, we began by introducing the sine function using a circle in the coordinate plane: We now return to the coordinate plane, but we will use the unit circle, a circle with radius 1, with its center at (0,0). We examine what happens to trigonometric functions as we go past 90 degrees, to larger angles in the unit circle. 8.2.1 Angles

Before we talk about angles in the unit circle, here are some definitions involving angles.

An angle is formed when two rays have a common endpoint. The common endpoint is called the vertex of the angle, and the rays are called the sides of the angle.

The side at which the angle starts is called the initial side, and the side at which the angle ends is called the terminal side. We use an arrow to determine which is the initial side and which is the terminal side as below:

y x

θ

(3,3)

2 | S e c t i o n 8 . 2

When an angle opens up in a counterclockwise manner, the angle is considered to be a positive angle. When it opens up in a clockwise manner, the angle is negative. Acute Angles measure between 00 and 900. All of the angles below are examples of acute angles.

Obtuse angles measure between 900 and 1800. All of the angles below are examples of obtuse angles.

3 | S e c t i o n 8 . 2

When we are studying angles on the unit circle, we often name the angles using Greek letters. This is to avoid confusing the angles with the sides, in which we typically use the English letters. Some Greek letters that we will use to represent angles are:

𝜶𝜶: 𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 𝜷𝜷:𝒃𝒃𝒃𝒃𝒃𝒃𝒂𝒂 𝜽𝜽: 𝒃𝒃𝒂𝒂𝒃𝒃𝒃𝒃𝒂𝒂

4 | S e c t i o n 8 . 2

8.2.2 Measuring Angles in The Unit Circle: Radians and Degrees QII QI QIII QIV The unit circle has four quadrants, usually named going counterclockwise around the circle using Roman numerals. QI is quadrant 1, QII is quadrant 2, QIII is quadrant 3 and QIV is quadrant 4. They are named in the same direction as the increasing positive angles in the unit circle.

We measure angles on the unit circle starting at the ordered pair, (1,0) or zero degrees, and moving around the circle counterclockwise. QII QI QIII QIV A quarter of the way around the circle, we are at 90 degrees, and halfway is 180 degrees. One complete circle is 360°.

In addition to measuring angles in degrees, we can measure them in radians. Recall that the formula for the circumference (a fancy word for the perimeter) of a circle is C = 2𝜋𝜋𝜋𝜋. The unit circle has a radius of 1, so the circumference of the unit circle is C = 2𝜋𝜋(1) = 2𝜋𝜋. This means the distance from the beginning of the unit circle at point (1,0) all the way around to the end of the unit circle is 2𝜋𝜋.

x

y

(0,0)

90°

45° 135°

180°

270°

225° 315°

(1,0)

(0,1)

(-1,0)

(0,-1)

0°, 360°

5 | S e c t i o n 8 . 2

Knowing that one full circle is 2𝜋𝜋, we can see that half of a circle is 𝜋𝜋, which is in the same place as 1800, and that a quarter of a circle is then 𝜋𝜋

2, which is the same

measure as 900. To get to 270°, we need to add another quarter of a circle, so we have 𝜋𝜋 + 𝜋𝜋

2= 2𝜋𝜋

2+ 𝜋𝜋

2= 3𝜋𝜋

2

QII QI QIII QIV Since 2𝜋𝜋 = 360°, divide both sides by 360 to get /360 /360

2𝜋𝜋360

= 1° which simplifies to 𝜋𝜋180

= 1° Thus, to switch back and forth between radians and degrees we use the following formulas:

• To change from degrees to radians: multiply by 𝜋𝜋180

• To change from radians to degrees, do the opposite: divide by 𝜋𝜋180

− in other

words, multiply by 180𝜋𝜋

8.2.2 Example 1. Convert 3𝜋𝜋

2 into degrees.

We multiply 3𝜋𝜋2

× 180𝜋𝜋

= 540𝜋𝜋2𝜋𝜋

. The pi’s cancel out, and simplifying we get 540𝜋𝜋2𝜋𝜋

=

5402

= 270°

Does this make sense? Look at the above circle! 270° is the same as 3𝜋𝜋2

! 8.2.2 Example 2. Convert 7𝜋𝜋

4 into degrees. See if you can do this yourself before you

go to the next page.

𝜋𝜋2

90°

2𝜋𝜋2

= 𝜋𝜋 180° 0°, 360° 0, 2𝜋𝜋

45° 135°

270° 3𝜋𝜋2

225° 315°

6 | S e c t i o n 8 . 2

Multiply 7𝜋𝜋4

× 180𝜋𝜋

= 1260𝜋𝜋4𝜋𝜋

= 315° 8.2.2 Example 3. Convert 45° into radians.

Multiply 45° × 𝜋𝜋180

= 45𝜋𝜋180

= 𝜋𝜋4

8.2.2 Example 4. Convert 420° into radians:

Multiply 420° × 𝜋𝜋180

= 420𝜋𝜋180

= 7𝜋𝜋3

8.2.3 Reference Angles and Positive and Negative Trigonometric Values Now that we can measure degrees or radians that are 90° or 𝜋𝜋

2 and higher, we can

examine what happens to trigonometric functions in each of the four quadrants. 8.2.3 Example 1. Find the six trigonometric functions of the angle (in radians) 𝜋𝜋

2 .

Begin by locating 𝜋𝜋2

on the unit circle. Recall that one full circle is 2 𝜋𝜋 𝜋𝜋𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, then half of a circle is 𝜋𝜋, which is in the same place as 1800, and that a quarter of a circle is then 𝜋𝜋

2, which is the same

measure as 900. So, 𝜋𝜋2

corresponds to the ordered pair (0,1). We have our x value as 0, and our y value as 1. What is our r? Well, recall that we are on the unit circle, therefore our radius is 1! So, for our 6 trigonometric functions, we have:

sin𝜋𝜋2 = sin 90° =

𝑦𝑦𝜋𝜋 =

11 = 1

cos𝜋𝜋2 = cos 90° =

𝑥𝑥𝜋𝜋 =

01 = 0

tan𝜋𝜋2 = tan 90° =

𝑦𝑦𝑥𝑥 =

10 = 𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑎𝑎

csc𝜋𝜋2 = csc 90° =

𝜋𝜋𝑦𝑦 =

11 = 1

(the cosecant is the reciprocal of the sine)

sec𝜋𝜋2 = sec 90° =

𝜋𝜋𝑥𝑥 =

10 = 𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑎𝑎

(the secant is the reciprocal of the cosine)

cot𝜋𝜋2 = cot 90° =

𝑥𝑥𝑦𝑦 =

01 = 0

(the cotangent is the reciprocal of the tangent)

8.2.3 Example 2. Find the six trigonometric functions of the angle 𝜋𝜋 (in radians). See if you can find the values yourself before going to the next page.

7 | S e c t i o n 8 . 2

We begin by locating 𝜋𝜋 on the unit circle, at 180°, the straight line going across the circle. Notice that now our point is (-1,0), which has a negative value for x. The 6 trigonometric functions are:

sin 𝜋𝜋 = sin 180° =𝑦𝑦𝜋𝜋 =

01 = 0

cos𝜋𝜋 = cos 180° =𝑥𝑥𝜋𝜋 =

−11 = −1

tan 𝜋𝜋 = tan 180° =𝑦𝑦𝑥𝑥 =

01 = 0

csc 𝜋𝜋 = csc 180° =𝜋𝜋𝑦𝑦 =

10 = 𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑎𝑎

sec 𝜋𝜋 = sec 180° =𝜋𝜋𝑥𝑥 =

1−1 = − 1

cot𝜋𝜋 = cot 180° =𝑥𝑥𝑦𝑦 =

−10 = 𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑢𝑢𝑎𝑎𝑎𝑎𝑢𝑢𝑎𝑎

Thus, we can see that as we go around the unit circle, some trigonometric values will be negative. 8.2.3 Example 3. When do trigonometric values come out positive and when do they come out negative? Look at the angle 70° and compare the sine and cosine of this angle with the sine and cosine of 110°. Note that these angles are the same as 70 × 𝜋𝜋

180= 7𝜋𝜋

18 radians and 110 × 𝜋𝜋

180= 11𝜋𝜋

18

On your calculator, find sin 70 and sin 110 with your calculator in degree mode. Make sure you can get the same results with sin7𝜋𝜋

18 and sin 11𝜋𝜋

18 with your calculator

in radians. Next, try the same experiment with cos 70 and cos 110. Then look at cosine and sine of 250, and cosine and sine of 290. Write your results in a table and look for a pattern.

Angle – degrees 70 110 250 290

Angle -- Radians 𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟕𝟕𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏

𝟐𝟐𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏

Sine 0.9396… ? ? ?

Cosine 0.3420… ? ? ?

Complete the table before you go to the next page.

8 | S e c t i o n 8 . 2

Your completed table should look like this:

Angle – degrees

70 110 250 290

Angle -- Radians

7𝜋𝜋18

11𝜋𝜋18

25𝜋𝜋18

29𝜋𝜋18

Sine

0.9396… 0.9396… -0.9396… -0.9396…

Cosine

0.3420… -0.3420… -0.3420… 0.3420…

To answer this question, let’s look at the triangle created by each of these angles on the unit circle. The triangle will include the reference angle. The reference angle is the shortest related angle from the terminal side of θ to the x-axis. Find the reference angle for 110°, then use a right triangle drawn to the x-axis to determine why the sine of 110° is positive, while the cosine is negative.

The above sketch shows the location of a 110° angle. Before going to the next page, see if you can draw a right triangle from the terminal side of the angle to the x-axis. Then use the fact that the measure halfway around the circle is 180° to find the angle inside the triangle.

θ = 110°

Wait, why are some of these values the same? But some are positive and some are negative?

9 | S e c t i o n 8 . 2

The triangle drawn to the x-axis gives us a reference angle of 70°. We get this because we know the whole horizontal line is 180°, so we subtract from 180: 180 − 110 = 70.

The sin of 70° is 0.9397… The cos of 70° is 0.342 The sin of 110° is also 0.9397… BUT the cos of 110° is -0.342

WHY? Because in the two different quadrants the x values change from positive to negative, but the y values stay positive. Recall that sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜= 𝑂𝑂

𝐻𝐻= 𝑦𝑦

𝑟𝑟

and cos θ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 = 𝐴𝐴

𝐻𝐻= 𝑥𝑥

𝑟𝑟

On the unit circle, since the radius, r, is 1, we end up with sin θ = 𝑦𝑦

𝑟𝑟= 𝑦𝑦

1= 𝑦𝑦 and cos θ = 𝑥𝑥

𝑟𝑟 = 𝑥𝑥

1= 𝑥𝑥.

In QII, the side adjacent to angle θ is the x value, and it is negative, so cosθ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 will always be a negative number. Thus, cos 110° is the same

(absolute) value as cos 70°, but is negative.

Now let’s look at 250°. Draw a picture of roughly where on the unit circle you can find 250°, and then draw the triangle from the point on the circle to the x-axis.

θ

QI QII In QII, the side opposite angle θ is the y value, and it is positive, so sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜

will always be a positive number. Thus, sin 110° is the same as sin 70, and is positive.

+

-

θ = 110° 70°

10 | S e c t i o n 8 . 2

Recall that the reference angle is the shortest related angle from the terminal side of θ to the x-axis. With θ = 250°, we again have a reference angle of 70°, because 250 – 180 = 70. Now draw the angle for 290°. See if you can draw the triangle from the point on the circle to the x-axis and determine which sides are positive or negative before you turn the page.

x

y

x

θ

QI QII This is roughly 250°, with a triangle drawn to the x-axis.

QIII

250°

QIII

250°

θ=70°

QI QII In QIII, the side opposite angle θ is the y value, and it is negative, so sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜

will always be a negative number. Thus, sin 250° is the same (absolute) value as sin 70°, but is negative.

QIII

250°

In QIII, the side adjacent to angle θ is the x value, and it is also negative, so cosθ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 will

always be a negative number. Thus, cos 250° is the same (absolute) value as cos 70°, but is negative.

QI QII Careful not to draw a triangle to the y-axis!

11 | S e c t i o n 8 . 2

Once more, we have a reference angle of 70°, but this time we don’t use 180 to get the reference angle. Instead, we see that 290° is very close to 360° (all the way around the circle), so we subtract 360 − 290 to see what is left for the angle inside the triangle. These rules for reference angles are summarized on the next page. Summary of Signs for Reference Angles Quadrant I: both x and y values are positive, thus both cosine and sine are positive. Quadrant II: x values are negative, and y values are positive, thus cosine is negative and sine is positive. Quadrant III, both x and y values are negative, so both cosine and sine are negative. Quadrant IV, x values are positive, while y values are negative, so cosine is negative and sine is positive.

Now recall that tan θ = 𝑦𝑦

𝑥𝑥 and see if you can tell what sign the tangent will have in

each quadrant! Try this on your own before you go to the next page!

x

x

QI QII In QIV, the side opposite angle θ is the y value, and it is negative, so sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜

will always be a negative number. Thus, sin 290° is the same (absolute) value as sin 70°, but is negative.

QIII

290°

In QIV, the side adjacent to angle θ is the x value, and it is positive, so cosθ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 will

always be a positive number. Thus, cos 290° is the same value as cos 70° (and positive).

QIV

θ=70°

QI QII

QIII

(−, +) (+, +) cos − sin + cos + sin +

QIV

(−, −) (+, −) cos − sin − cos + sin −

12 | S e c t i o n 8 . 2

The values for sine, cosine and tangent:

This leads us to the mnemonic: All Students Take Calculus, telling us which of the three trigonometric values is positive in each quadrant! (All positive in Q1, Sine positive in QII, etc.) But if you understand the relationship between the trigonometric values and x and y, you should also be able to figure out whether each value is positive or negative, even if you have not memorized this mnemonic! 8.2.3 Example 4 If you know that sin 20°≈ 0.342 and sin 40°≈ 0.643, find the values for sin 200°, and sin 320°, using reference angles and without using a calculator.

To find the reference angle for 200°, first draw the angle, then draw a right triangle to the x-axis. We know that 200 is just a bit more than 180, the straight line, so we draw our angle just past 180. Next, we draw a line straight up to the x-axis: The angle inside the triangle is 200 − 180 = 20, so we can use the fact that sin 20°≈ 0.342 to find sin 200°. We just have to decide if sin 200°≈ 0.342 or sin 200°≈ −0.342.

We can use the mnemonic, above, to see that in QIII, only the tangent is positive, so sin 200°≈ −0.342. Or we can use the fact that sin 𝜃𝜃 = 𝑂𝑂

𝐻𝐻. The side opposite θ is the y-

value, which is negative. The hypotenuse is always positive, so we have sin𝜃𝜃 = 𝑂𝑂𝐻𝐻

=𝑦𝑦𝑜𝑜𝑛𝑛𝑎𝑎𝑜𝑜𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛𝑜𝑜

= 𝑎𝑎𝑢𝑢𝑛𝑛𝑎𝑎𝑛𝑛𝑎𝑎𝑛𝑛𝑢𝑢. Check on your calculator to see that you are right!

x

QIII

Sine positive All positive (−, +) (+, +)

cos − sin + tan − cos + sin + tan +

(−, −) (+, −) cos − sin − tan + cos + sin − tan −

Tangent positive Cosine positive

θ

QI QII

QIV

13 | S e c t i o n 8 . 2

Sometimes people think that to find the value of θ here, you need to subtract from 270°.

But that would give you the angle here:

This is an angle we will never need to get, since it is the angle to the y-axis, not the x-axis.

Now, let’s find the reference angle for sin 320°

We know that 320° is past 270° and close to 360°, so we draw our angle almost all the way around the circle. Next, we draw a line straight up to the x-axis: The angle inside the triangle is 360 − 320 = 40, so we can use the fact that sin 40°≈ 0.643 to find sin 320°. We just have to decide if sin 320°≈ 0.643 or sin 320°≈ −0.643.

We can use the mnemonic, above, to see that in QIV, only the cosine is positive, so sin 320°≈ −0.643. Or we can use the fact that sin 𝜃𝜃 = 𝑂𝑂

𝐻𝐻. The side opposite θ is the y-

value, which is negative. The hypotenuse is always positive, so we have sin𝜃𝜃 = 𝑂𝑂𝐻𝐻

=𝑦𝑦𝑜𝑜𝑛𝑛𝑎𝑎𝑜𝑜𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛𝑜𝑜

= 𝑎𝑎𝑢𝑢𝑛𝑛𝑎𝑎𝑛𝑛𝑎𝑎𝑛𝑛𝑢𝑢. Check on your calculator to see that you are right!

CAUTION!

θ

270°

θ

270°

θ

14 | S e c t i o n 8 . 2

8.2.4 More Practice Finding Reference Angles To find the reference angle our steps are:

1. Sketch the angle. While it doesn’t have to be exact, it does have to be in the correct quadrant to yield correct results.

2. Draw a line straight to the x-axis to create a right triangle, and label the angle inside the triangle (and inside the middle of the unit circle).

3. Subtract from 180 or subtract 180 from your angle if you are closest to 180; subtract from 360 if you are closest to 360 (we never use 270).

Note that the reference angle is defined to be the shortest distance back to the x axis, once you sketch your angle. In the examples below, the angle is black, and the reference angle is in red.

8.2.4 Example 1 Find the reference angle for 1350.

1. Sketch 1350. Remember 1350 is in between 90° and 180°, which puts us in the second quadrant:

2. Draw a line to the x-axis and label θ. Notice that θ is inside the original unit circle.

3. Since we are closer to 180 than 360, subtract from 180: 180 −135 = 45.

Thus, the reference angle is 45.

θ

15 | S e c t i o n 8 . 2

8.2.4 Example 3 Find the reference angle for 210°.

1. Sketch the angle 210°. 2100 is in between 1800 and 2700, so it is in Quadrant III.

2. There are two routes to get back to the x axis, the purple and the orange. I hope it is clear that the purple route is the quicker route to get back to the x-axis! So create the triangle straight up to the x-axis.

3. Now, we have to find the distance of the purple route. The angle that we made was 2100. And the distance from the starting point to where the purple route ends is 1800. Therefore, our reference angle is 210 – 180 = 300.

8.2.4 Example 3 Find the reference angle for −200°.

1. Sketch the angle −200°. Recall if our angle is negative, then we still start in the same place, but we travel clockwise! Travelling -1800 would take us to (-1,0), and we still have to go 20° more, so we are now in Quadrant II:

2. Next, draw the triangle and the angle in it:

3. Finally, we find the angle in the triangle. Unfortunately, subtracting 180 doesn’t work! −200−180 = −380, which is not a reference angle, since reference angles are always between 0 and 90. Also, because we think of reference angles in terms of distances, we don’t have negative reference angles! But we actually already know that the reference angle is 20°, since in step 1, we noted, “Travelling -1800 would take us to (-1,0), and we still have to go 20° more.”

θ

16 | S e c t i o n 8 . 2

8.2.5 Exact Values and Reference Angles There are two special right triangles, the 45-45-90 triangle, and the 30-60-90 triangle, that allow us to find certain trigonometric values without a calculator. In the 45-45-90 triangle, because two of the angles are equal, the corresponding sides are equal. t t This means that if one side is t, the other side is also t. What about the hypotenuse? To see what happens to the hypotenuse, fill in the table, below, first for t = 1, then t = 2, then 3, to see the pattern. Write the hypotenuse in simplified radical form, not as a decimal.

Side, t Side, t Hypotenuse 1 1

2 2

3 3

t t

For t=1, we use the Pythagorean theorem to find the hypotenuse:

𝑎𝑎2 + 𝑏𝑏2 = 𝑐𝑐2 Gives us

12 + 12 = 𝑐𝑐2

1 + 1 = 𝑐𝑐2 2 = 𝑐𝑐2

Thus 𝑐𝑐 = √2

Now try t =2 and t = 3 before you turn the page. See if the patterns helps you to predict how the hypotenuse will always be related to the two sides.

45°

45°

45°

17 | S e c t i o n 8 . 2

For t=2, we get 22 + 22 = 𝑐𝑐2 4 + 4 = 𝑐𝑐2

8 = 𝑐𝑐2 Thus 𝑐𝑐 = √8 = √4 × 2 = 2√2

For t=3, we get 32 + 32 = 𝑐𝑐2

9 + 8 = 𝑐𝑐2 18 = 𝑐𝑐2

Thus 𝑐𝑐 = √18 = √9 × 2 = 3√2 We can predict any value now, using the pattern:

Side, t Side, t Hypotenuse 1 1 √2

2 2 2√2

3 3 3√2

t t 𝑛𝑛√2

We can also work it out algebraically: For t in general, we get 𝑛𝑛2 + 𝑛𝑛2 = 𝑐𝑐2

2𝑛𝑛2 = 𝑐𝑐2 Thus 𝑐𝑐 = √2𝑛𝑛2 = 𝑛𝑛√2

Since sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜= 𝑂𝑂

𝐻𝐻, the sine of 45° will always be 𝑂𝑂

𝐻𝐻= 𝑜𝑜

𝑜𝑜√2= 1

√2

Similarly, cos θ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 = 𝐴𝐴

𝐻𝐻, the cosine of 45° will always be 𝐴𝐴

𝐻𝐻= 𝑜𝑜

𝑜𝑜√2= 1

√2

We rationalize the denominator to get sin 45 = cos 45 = 1 ∙ √2

√2 ∙ √2= √2

2.

18 | S e c t i o n 8 . 2

To figure out the sides of a 30-60-90 triangle… Since all the angles are equal, all the sides are equal. Thus, the sides are each 2t. Going back to our original triangle, we now have This means that if the side opposite 30 in a 30-60-90 triangle is 1, the hypotenuse would be twice that, or 2. The hypotenuse is always twice the smaller leg. Fill in the table, below. Use the Pythagorean theroem to find the missing leg.

Careful! Now the leg is missing, not the hypotenuse! For t=1, we get 12 + 𝑏𝑏2 = 22 Finish this before going to the next page!

t, side opposite 30° side opposite 60° Hypotenuse, 2t 1 ? 2

2 ? 4

3 ? 6

t ? 2t

60°

30°

?

Place two of the 30-60-90 triangles back to back. Call the original base of the triangle, t. You now have a triangle where each angle is 60°.

t

30° 30°

60° 60°

60°

t t

2t

2t 2t

60°

30°

t

2t

19 | S e c t i o n 8 . 2

12 + 𝑏𝑏2 = 22 1 + 𝑏𝑏2 = 4

-1 -1 𝑏𝑏2 = 3

Thus 𝑏𝑏 = √3 For t=2, we get 22 + 𝑏𝑏2 = 42

4 + 𝑏𝑏2 = 16 -4 -4

𝑏𝑏2 = 12 Thus 𝑏𝑏 = √12 = √4 × 3 = 2√3

You can probably now predict the table!

We can also work it out algebraically: For t in general, we get 𝑛𝑛2 + 𝑏𝑏2 = (2𝑛𝑛)2

𝑛𝑛2 + 𝑏𝑏2 = 4𝑛𝑛2 Thus 𝑏𝑏2 = 4𝑛𝑛2 − 𝑛𝑛2 = 4𝑛𝑛2 − 1𝑛𝑛2 = 3𝑛𝑛2

Since 𝑏𝑏2 = 3𝑛𝑛2, if we take the square root of both sides, we get 𝑏𝑏 = �3𝑛𝑛2 = 𝑛𝑛√3 Since sin θ = 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜= 𝑂𝑂

𝐻𝐻, the sine of 30° will always be 𝑂𝑂

𝐻𝐻= 𝑜𝑜

2𝑜𝑜= 1

2

Similarly, cos θ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑦𝑦𝑜𝑜

ℎ𝑦𝑦𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑦𝑦𝑦𝑦𝑜𝑜𝑜𝑜 = 𝐴𝐴

𝐻𝐻, the cosine of 60° will always be 𝐴𝐴

𝐻𝐻= 𝑜𝑜√3

2𝑜𝑜= √3

2

t, side opposite 30° side opposite 60° Hypotenuse, 2t 1 √3 2

2 2√3 4

3 3√3 6

t 𝑛𝑛√3 2t

20 | S e c t i o n 8 . 2

In summary, we know the following values without using a calculator:

We can use this table and the reference angles to find exact values for certain angles. To find the exact value, use the following steps:

1. Sketch the angle and find the reference angle. 2. Find the trig value for that angle in the exact values table. 3. Figure out whether the trig function is positive or negative, based on the

quadrant. 8.2.5 Example 1 Find the exact value for sin 300°.

1. Sketch the angle and find the reference angle. 300 is in between 270 and 360, hence in Q IV.

The red arc represents the quickest distance back to the x axis. We are at 3000, and we have to get to 3600, so the reference angle is 360 – 300 = 600

2. The sine of 60 in the table is √32

.

21 | S e c t i o n 8 . 2

3. Because we are in quadrant IV, our x values are positive, and our y values are negative. The sine value represents the y value, therefore our answer has to be negative.

Putting it all together, sin 3000 = −√3

2

8.2.6 Finding one trig value given another. 8.2.6 Example 1 find the remaining trigonometric functions of 𝛼𝛼 if cos 𝛼𝛼 = − 3

5, and

𝛼𝛼 terminates in Quadrant II Recall that if we need to find the trigonometric functions of a certain value, we have to find x, y, and r. Remember that cos 𝛼𝛼 = 𝑥𝑥

𝑟𝑟, so the fact that cos 𝛼𝛼 = − 3

5 is telling us

that the x value is 3 and the r value is 5. However, one of those values HAS to be negative, because cos 𝛼𝛼 = − 3

5. Note that because we are in Quadrant II, our x values

are negative, and our y values are positive! And remember r is ALWAYS positive. Therefore, the x value is -3, and the r value is 5. Now, we have to find the y value. Consider a picture below based on this information:

5

-3 Based on the picture above, is it clear that we can use the Pythagorean theorem to find our y value? Let’s try it: y2 + (-3)2 = 52 y2 + 9 = 25 y2 = 16 y = ±√16 = ± 4 The question is, which value is y equal to, positive 4, or negative 4? As stated above, because we are in the second quadrant, all of the y values in this quadrant are positive! Therefore, our y value = 4. So, we know x = -3, y = 4, and r = 5, and for our trigonometric functions, we have:

22 | S e c t i o n 8 . 2

cos 𝛼𝛼 =− 35 sec𝛼𝛼 = −5

3

sin 𝛼𝛼 = 4

5 csc 𝛼𝛼 = 5

4

tan 𝛼𝛼 = − 4

3 cot𝛼𝛼 = − 3

4

8.2.6 Example 2 Find the remaining trig functions if sin 𝜃𝜃 = − 1

2, and 𝜃𝜃 terminates in

quadrant III. The sine function represents 𝑦𝑦

𝑟𝑟. This is telling us that our y value is -1 and our r

value is 2. We need to find our x value. -1 2

Again, we can use the Pythagorean theorem to find the x value. x2 + (-1)2 = 22 x2 + 1 = 4 x2 = 3 x = ±√3

Is the answer positive or negative √3? Well, we are in quadrant III, so this means that our answer must be negative, because our x values are negative in this quadrant.

Therefore, x = −√3, y = -1 and r = 2 So, cos 𝜃𝜃 =−√3

2 sec𝜃𝜃 = − 2

√3= − 2√3

3

sin 𝜃𝜃 = − 1

2 csc 𝜃𝜃 = − 2

1= −2

tan 𝜃𝜃 = −1

−√3 = √3

3 cot𝜃𝜃 = −√3

−1= √3