Inverse Trig With Notes

Analyzing Circles on the Coordinate Plane

Comment [SS1]: This is to be used BEFORE trigonometry on the unit circle is introduced. This gets students thinking about the unit circle and coordinates without layering on sine/cosine/tangent… Just playing around and thinking conceptually and concretely.

Directions: On the previous page is a unit circle. You’re going to practice using a protractor and estimating coordinates on the circle. Let the point (1,0) represent 0 degrees, (0,1) represent 90 degrees, (-1,0) represent 180 degrees, and (0,-1) represent 270 degrees.

The forward problem Since you are estimating, and the diagram is labeled to the tenths, you are going to estimate to the hundredths. Estimate the following coordinates/slopes: We will compare the true answers to your estimations in a couple days – and the persons with the most accurate answers will get a prize! angle

xcoordinate

ycoordinate

slope of angle

angle

23o

x-coordinate

ycoordinate

slope of angle

154o

48o

185o

75o

227o

105o

300o

The backward problem Below we have table of values. You are given one piece of information. It turns out that with that one piece of information, you can get all the other values. In fact, with that one piece of information, you can get two different sets of answers! If you don’t believe me, do the first problem below – and be sure to carefully use the unit circle diagram! angle

xcoordinate

(a)

ycoordinate

slope of angle

angle

x-coordinate

(a)

(a)

(a)

(a)

(b)

(b)

(b)

(b)

ycoordinate (a)

0.30 (b) (a)

-0.62

(a)

(a)

(a)

(b)

(a)

(b)

(b)

(b) (a)

1 (b) (a)

(a)

(b)

(b)

(a)

(a)

(b)

(b)

-0.89 (b)

(a)

(a)

-1.23 (b)

(a)

(b)

(a)

0.77 (b)

(a)

(a)

(a)

(a)

(b)

(b)

0.30 (b)

(b)

slope of angle

-0.2 (b)

(b)

Now I’d like for you to make some observations and answer some questions. (1) Did you use vertical and horizontal lines to help you? If so, how? If not, that’s totes fine too!

(2) What did you notice while filling out the table?

(3) What did you wonder when you were filling out the table?

(4) Is there a y-coordinate that I could have filled in that would have given only one answer to the y-coordinate? If so, what is it? Are there any others? If not, what is your reasoning?

(5) Is there a slope that I could have filled in that would have given only one answer to the x-coordinate? Are there any others? If so, what is it? Are there any others? If not, what is your reasoning?

(6) If I gave you the x-coordinate and no unit circle diagram, can you think of a way to get the y-coordinate? If you need a hint, look at the footnote.1

Comment [SS2]: The idea behind 6, 7 is to see if any kids are thinking algebraically… the x^2+y^2=1 relationship for the unit circle…

(7) If I gave you the slope of the angle and no unit circle diagram, can you think of a way to get the x-coordinates and the y-coordinates?

(8) If I gave you the x-coordinate and no unit circle diagram, can you think of a way to get both angles?

1

The sum of the squares of…

Comment [SS3]: The idea behind this question is to how that getting the angle algebraically is hard/impossible with what they know now… but this open question will be resolved later with their calculators and knowledge of inverse trig.

Inverse Trigonometry via the Unit Circle

Comment [SS4]: This part of the worksheet is to be done AFTER unit circle trig is introduced… the point is to get students to think about inverse trigonometry more concretely.

Directions: On the previous page is a unit circle. Comment [SS5]: Obviously in class discussion it should be brought up that the answers are the same as in the PREVIOUS packet on the unit circle… as these are the same angles… same for the backward problem.

The forward problem You now now about sine, cosine, and tangent. Use your knowledge to fill in this table.

angle

sine

cosine

23o

tangent

angle

sine

cosine

tangent

154o

48o

185o

75o

227o

105o

300o

The backward problem Find all angles (between 0 o and 360o ) using the unit circle diagram and a protractor that satisfy these trig equations, okay?: Equation

cos   0.30

Solution(s)

Equation

sin   0.62

sin   0.77

cos   1.23

tan   1

cos   0.89

sin   0.30

tan   0.2

Solution(s)

Above are graphs of sine and cosine. Use the graphs to come up with the solutions to the following equations (between Comment [SS6]: A relationship between the unit circle and the graphs should be emphasized…

0 o and 360o ) Equation

cos   0.30

Solution(s)

Equation

sin   0.62

sin   0.77

cos   1.23

sin   0.30

cos   0.89

Solution(s)

(1) Let’s extend things! What if we looked at all possible angles, not just the angles between 0 o and 360o … can you find all solutions to sin   0.77 ?

(2) Similarly… can you find all solutions to cos   0.30 ?

Calculators! Your calculators have the ability to give you the solutions that you’re approximating by hand (using the protractor or graphs). If you have x 2  4 , we know to get the variable alone, we “undo” the squaring by the square root. So we get

x  4  2 , because the opposite of squaring is square-rooting. They are inverse operations. Similarly for trig equations, we might have cos   0.30 , and we can “undo” the cosine operation by using the inverse cosine operation. That means   cos1 (0.30) which when we use our calculators is about 72.54o . Equation

cos   0.30

Calculator Solution

Equation

sin   0.62

sin   0.77

cos   1.23

sin   0.30

cos   0.89

Calculator Solution

However we saw above that there are multiple solutions to cos   0.30 (between 0 o and 360o ), not just one! Let’s reconsider x 2  4 … There is a second solution to this, which our calculator did not give us. It is x  2 . Similarly, our calculator only gave us one solution to cos   0.30 , but there is another! One solution ( 72.54o ) is in the first quadrant, but the second solution is in the ________ quadrant! I know this because:

And this solution is:

Now go back to the table above and determine what quadrant the second solution will be in, and then determine the second solution to all the problems!

Extensions and Generalizations (1) Write a trig equation which has only one solution for  (between 0 o and 360o ). Explain how you came up with your equation and how you know it only has one solution.

(2) Write a trig equation which has no solutions for  (between 0 o and 360o ). Explain how you came up with your equation and how you know it hasn’t any solutions.

(3) Write a trig equation which has exactly three solutions for  (between 0 o and 360o ). Explain how you came up with your equation and how you know it only has exactly three solutions.

(4) Write a trig equation which has only one solution for  (for all angles). Explain how you came up with your equation and how you know it only has one solution.

(5) Write a trig equation which has no solutions for  (for all angles). Explain how you came up with your equation and how you know it hasn’t any solutions.

(6) Write a trig equation which has exactly three solutions for  (for all angles). Explain how you came up with your equation and how you know it only has exactly three solutions.

(7) Write a trig equation which has exactly two solutions for  (between 0 o and 360o ) in quadrants II and IV. Explain how you came up with your equation and how you know it only has the solutions in the correct quadrants.

(8) Write a trig equation which has exactly two solutions for  (between 0 o and 360o ) in quadrants III and IV. Explain how you came up with your equation and how you know it only has the solutions in the correct quadrants.

(9) Write a trig equation which has exactly two solutions for  (between 0 o and 360o ) in quadrants II and III. Explain how you came up with your equation and how you know it only has the solutions in the correct quadrants.

(10)Write a trig equation which has exactly two solutions for  (between 0 o and 360o ) in quadrants I and IV. Explain how you came up with your equation and how you know it only has the solutions in the correct quadrants.

(11)Write a trig equation which has exactly two solutions for  (between 0 o and 360o ) in quadrants I and II. Explain how you came up with your equation and how you know it only has the solutions in the correct quadrants.