In section 8.1, we began by introducing the sine function using a circle in the coordinate plane:

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1 Chapter 8.: Degrees and Radians, Reference Angles In section 8.1, we began by introducing the sine function using a circle in the coordinate plane: y (3,3) θ x 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 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: 1 S ection 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 0 0 and All of the angles below are examples of acute angles. Obtuse angles measure between 90 0 and All of the angles below are examples of obtuse angles. S ection 8.

3 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: bbbbbbbb θθ: tttttttttt 3 S ection 8.

4 8.. Measuring Angles in The Unit Circle: Radians and Degrees y QII (0,1) QI (-1,0) (0,0) (1,0) x QIII (0,-1) QIV The unit circle has four quadrants, usually named going counterclockwise around the circle using Roman numerals. QI is quadrant 1, QII is quadrant, 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 180 0, S ection 8. 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 = ππππ. The unit circle has a radius of 1, so the circumference of the unit circle is C = ππ(1) = ππ. 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 ππ.

5 Knowing that one full circle is ππ, we can see that half of a circle is ππ, which is in the same place as 180 0, and that a quarter of a circle is then ππ, which is the same measure as To get to 70, we need to add another quarter of a circle, so we have ππ + ππ = ππ + ππ = 3ππ ππ QII QI ππ = ππ 180 0, 360 0, ππ QIII QIV ππ Since ππ = 360, divide both sides by 360 to get /360 /360 ππ = 1 which simplifies to ππ = Thus, to switch back and forth between radians and degrees we use the following formulas: To change from degrees to radians: multiply by ππ 180 To change from radians to degrees, do the opposite: divide by ππ in other 180 words, multiply by 180 ππ 8.. Example 1. Convert 3ππ into degrees. We multiply 3ππ 180 ππ = 540ππ 540ππ. The pi s cancel out, and simplifying we get = ππ ππ 540 = 70 Does this make sense? Look at the above circle! 70 is the same as 3ππ! 8.. Example. Convert 7ππ 4 go to the next page. into degrees. See if you can do this yourself before you 5 S ection 8.

6 Multiply 7ππ ππ = 160ππ 4ππ = Example 3. Convert 45 into radians. Multiply 45 ππ 180 = 45ππ 180 = ππ Example 4. Convert 40 into radians: Multiply 40 ππ = 40ππ = 7ππ Reference Angles and Positive and Negative Trigonometric Values Now that we can measure degrees or radians that are 90 or ππ and higher, we can examine what happens to trigonometric functions in each of the four quadrants Example 1. Find the six trigonometric functions of the angle (in radians) ππ. Begin by locating ππ on the unit circle. Recall that one full circle is ππ rraaaaaaaaaaaa, then half of a circle is ππ, which is in the same place as 180 0, and that a quarter of a circle is then ππ, which is the same measure as So, ππ 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: sin ππ = sin 90 = yy rr = 1 1 = 1 cos ππ = cos 90 = xx rr = 0 1 = 0 tan ππ = tan 90 = yy xx = 1 0 = uuuuuuuuuuuuuuuuuu csc ππ = csc 90 = rr yy = 1 1 = 1 (the cosecant is the reciprocal of the sine) sec ππ = sec 90 = rr xx = 1 0 = uuuuuuuuuuuuuuuuuu (the secant is the reciprocal of the cosine) cot ππ = cot 90 = xx yy = 0 1 = 0 (the cotangent is the reciprocal of the tangent) 8..3 Example. 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. 6 S ection 8.

7 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 = yy rr = 0 1 = 0 cos ππ = cos 180 = xx rr = 1 1 = 1 tan ππ = tan 180 = yy xx = 0 1 = 0 csc ππ = csc 180 = rr yy = 1 0 = uuuuuuuuuuuuuuuuuu sec ππ = sec 180 = rr xx = 1 1 = 1 cot ππ = cot 180 = xx yy = 1 0 = uuuuuuuuuuuuuuuuuu Thus, we can see that as we go around the unit circle, some trigonometric values will be negative 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 = 7ππ ππ radians and 110 = 11ππ On your calculator, find sin 70 and sin 110 with your calculator in degree mode. Make sure you can get the same results with sin 7ππ 11ππ and sin with your calculator in radians. Next, try the same experiment with cos 70 and cos 110. Then look at cosine and sine of 50, and cosine and sine of 90. Write your results in a table and look for a pattern. Angle degrees Angle -- Radians Sine ??? Cosine 0.340??? Complete the table before you go to the next page S ection 8.

8 Your completed table should look like this: Angle degrees Angle -- Radians ππ 18 11ππ 18 5ππ 18 9ππ 18 Sine Cosine Wait, why are some of these values the same? But some are positive and some are negative? 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. θ = 110 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. 8 S ection 8.

9 The triangle drawn to the x-axis gives us a reference angle of θ = 110 We get this because we know the whole horizontal line is 180, so we subtract from 180: = 70. The sin of 70 is The cos of 70 is 0.34 The sin of 110 is also BUT the cos of 110 is 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 = OO = yy hyyyyyyyyyyyyyyyyyy HH rr and cos θ = aaaaaaaaaaaaaaaa hyyyyyyyyyyyyyyyyyy = AA HH = xx rr On the unit circle, since the radius, r, is 1, we end up with sin θ = yy = yy = yy and cos θ = xx = xx = xx. rr 1 rr 1 In QII, the side opposite angle θ is the y value, and it is positive, so sin θ = oooooooooooooooo hyyyyyyyyyyyyyyyyyy will always be a positive number. Thus, sin 110 is the same as sin 70, and is positive. QII + - θ QI In QII, the side adjacent to angle θ is the x value, and it is negative, so cos θ = aaaaaaaaaaaaaaaa will always be a negative number. Thus, cos 110 is the same hyyyyyyyyyyyyyyyyyy (absolute) value as cos 70, but is negative. Now let s look at 50. Draw a picture of roughly where on the unit circle you can find 50, and then draw the triangle from the point on the circle to the x-axis. 9 S ection 8.

10 This is roughly 50, with a triangle drawn to the x- axis. x QII 50 θ QI QIII Careful not to draw a triangle to the y-axis! QII 50 QI QIII Recall that the reference angle is the shortest related angle from the terminal side of θ to the x-axis. With θ = 50, we again have a reference angle of 70, because = 70. In QIII, the side opposite angle θ is the y value, and it is negative, so sin θ = oooooooooooooooo hyyyyyyyyyyyyyyyyyy will always be a negative number. Thus, sin 50 is the same (absolute) value as sin 70, but is negative. x QIII QII 50 θ=70 Now draw the angle for 90. 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. y QI In QIII, the side adjacent to angle θ is the x value, and it is also negative, so cos θ = aaaaaaaaaaaaaaaa hyyyyyyyyyyyyyyyyyy will always be a negative number. Thus, cos 50 is the same (absolute) value as cos 70, but is negative. 10 S ection 8.

11 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 90 is very close to 360 (all the way around the circle), so we subtract to see what is left for the angle inside the triangle. These rules for reference angles are summarized on the next page. In QIV, the side opposite angle θ is the y value, and it is negative, so sin θ = oooooooooooooooo hyyyyyyyyyyyyyyyyyy will always be a negative number. Thus, sin 90 is the same (absolute) value as sin 70, but is negative. x QIII QII 90 QI θ=70 QIV In QIV, the side adjacent to angle θ is the x value, and it is positive, so cos θ = aaaaaaaaaaaaaaaa hyyyyyyyyyyyyyyyyyy will always be a positive number. Thus, cos 90 is the same value as cos 70 (and positive). 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. QII QI x (, +) (+, +) cos sin + cos + sin + (, ) (+, ) cos sin cos + sin QIII QIV Now recall that tan θ = yy and see if you can tell what sign the tangent will have in xx each quadrant! Try this on your own before you go to the next page! 11 S ection 8.

12 The values for sine, cosine and tangent: Sine positive All positive QII (, +) (+, +) QI cos sin + tan cos + sin + tan + x (, ) (+, ) cos sin tan + cos + sin tan Tangent positive Cosine positive QIII QIV 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..3 Example 4 If you know that sin and sin , find the values for sin 00, and sin 30, using reference angles and without using a calculator. To find the reference angle for 00, first draw the angle, then draw a right triangle to the x-axis. We know that 00 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: θ The angle inside the triangle is = 0, so we can use the fact that sin to find sin 00. We just have to decide if sin or sin We can use the mnemonic, above, to see that in QIII, only the tangent is positive, so sin Or we can use the fact that sin θθ = OO. The side opposite θ is the y- HH value, which is negative. The hypotenuse is always positive, so we have sin θθ = OO = HH nnnnnnnnnnnnnnnn pppppppppppppppp = nnnnnnnnnnnnnnnn. Check on your calculator to see that you are right! 1 S ection 8.

13 CAUTION! Sometimes people think that to find the value of θ here, you need to subtract from 70. θ But that would give you the angle here: 70 θ This is an angle we will never need to get, since it is the angle to the y-axis, not the x-axis. 70 Now, let s find the reference angle for sin 30 We know that 30 is past 70 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: θ The angle inside the triangle is = 40, so we can use the fact that sin to find sin 30. We just have to decide if sin or sin We can use the mnemonic, above, to see that in QIV, only the cosine is positive, so sin Or we can use the fact that sin θθ = OO. The side opposite θ is the y- HH value, which is negative. The hypotenuse is always positive, so we have sin θθ = OO = HH nnnnnnnnnnnnnnnn pppppppppppppppp = nnnnnnnnnnnnnnnn. Check on your calculator to see that you are right! 13 S ection 8.

14 8..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.. 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 70). 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 Example 1 Find the reference angle for Sketch Remember is in between 90 and 180, which puts us in the second quadrant:. Draw a line to the x-axis and label θ. Notice that θ is inside the original unit circle. θ 3. Since we are closer to 180 than 360, subtract from 180: = 45. Thus, the reference angle is S ection 8.

15 8..4 Example 3 Find the reference angle for Sketch the angle is in between and 70 0, so it is in Quadrant III.. 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 And the distance from the starting point to where the purple route ends is Therefore, our reference angle is = Example 3 Find the reference angle for Sketch the angle 00. Recall if our angle is negative, then we still start in the same place, but we travel clockwise! Travelling would take us to (- 1,0), and we still have to go 0 more, so we are now in Quadrant II:. Next, draw the triangle and the angle in it: θ 3. Finally, we find the angle in the triangle. Unfortunately, subtracting 180 doesn t work! = 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 0, since in step 1, we noted, Travelling would take us to (- 1,0), and we still have to go 0 more. 15 S ection 8.

16 8..5 Exact Values and Reference Angles There are two special right triangles, the triangle, and the triangle, that allow us to find certain trigonometric values without a calculator. In the 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 =, then 3, to see the pattern. Write the hypotenuse in simplified radical form, not as a decimal. 45 Side, t Side, t Hypotenuse t t For t=1, we use the Pythagorean theorem to find the hypotenuse: aa + bb = cc Gives us = cc = cc = cc Thus cc = Now try t = 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. 16 S ection 8.

17 For t=, we get + = cc = cc 8 = cc Thus cc = 8 = 4 = For t=3, we get = cc = cc 18 = cc Thus cc = 18 = 9 = 3 We can predict any value now, using the pattern: Side, t Side, t Hypotenuse t t tt We can also work it out algebraically: For t in general, we get tt + tt = cc tt = cc Thus cc = tt = tt Since sin θ = oooooooooooooooo hyyyyyyyyyyyyyyyyyy = OO HH, the sine of 45 will always be OO HH = tt = 1 tt Similarly, cos θ = aaaaaaaaaaaaaaaa hyyyyyyyyyyyyyyyyyy = AA HH, the cosine of 45 will always be AA HH = We rationalize the denominator to get sin 45 = cos 45 = 1 =. tt = 1 tt 17 S ection 8.

18 To figure out the sides of a triangle 30 Place two of the triangles back to back. Call the original base of the triangle, t. 60 You now have a triangle where each angle is t t t 60 Since all the angles are equal, all the sides are equal. Thus, the sides are each t. t t Going back to our original triangle, we now have? 30 t t 60 t This means that if the side opposite 30 in a triangle is 1, the hypotenuse would be twice that, or. The hypotenuse is always twice the smaller leg. Fill in the table, below. Use the Pythagorean theroem to find the missing leg. t, side opposite 30 side opposite 60 Hypotenuse, t 1?? 4 3? 6 t? t Careful! Now the leg is missing, not the hypotenuse! For t=1, we get 18 S ection bb = Finish this before going to the next page!

19 1 + bb = 1 + bb = bb = 3 Thus bb = 3 For t=, we get + bb = bb = bb = 1 Thus bb = 1 = 4 3 = 3 You can probably now predict the table! t, side opposite 30 side opposite 60 Hypotenuse, t t tt 3 t We can also work it out algebraically: For t in general, we get tt + bb = (tt) tt + bb = 4tt Thus bb = 4tt tt = 4tt 1tt = 3tt Since bb = 3tt, if we take the square root of both sides, we get bb = 3tt = tt 3 oooooooooooooooo Since sin θ = = OO, the sine of 30 will always be OO = tt = 1 hyyyyyyyyyyyyyyyyyy HH HH tt Similarly, cos θ = aaaaaaaaaaaaaaaa = AA, the cosine of 60 will always be AA = tt 3 = 3 hyyyyyyyyyyyyyyyyyy HH HH tt 19 S ection 8.

20 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.. 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 Example 1 Find the exact value for sin Sketch the angle and find the reference angle. 300 is in between 70 and 360, hence in Q IV. The red arc represents the quickest distance back to the x axis. We are at 300 0, and we have to get to 360 0, so the reference angle is = The sine of 60 in the table is 3. 0 S ection 8.

21 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 = Finding one trig value given another Example 1 find the remaining trigonometric functions of αα if cos αα = 3, and 5 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, so the fact that cos αα = 3 is telling us rr 5 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. Note that because we are in Quadrant II, our x values 5 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: y + (-3) = 5 y + 9 = 5 y = 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: 1 S ection 8.

22 cos αα = 3 5 secαα = 5 3 sin αα = 4 5 csc αα = 5 4 tan αα = 4 3 cotαα = Example Find the remaining trig functions if sin θθ = 1, and θθ terminates in quadrant III. The sine function represents yy. This is telling us that our y value is -1 and our r rr value is. We need to find our x value. -1 Again, we can use the Pythagorean theorem to find the x value. x + (-1) = x + 1 = 4 x = 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 = So, cos θθ = 3 sin θθ = 1 tan θθ = 1 = secθθ = = csc θθ = 1 = cotθθ = 3 1 = 3 S ection 8.

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