You Can Go to the Solution for Each Problem by Clicking on the Problem Letter

Pre-Class Problems15 for Wednesday, October 24, and Monday, October 29

These are the type of problems that you will be working on in class. These problems are fromLesson 9.

Solution to Problems on the Pre-Exam.

You can go to the solution for each problem by clicking on the problem letter.

Objective of the following problems: To find the value of the inverse sine of a given number.

1.Find the exact value of the following.

a. b. c.

d. e. f.

g. h. i.

Objective of the following problems: To find the value of the inverse tangent of a given number.

2.Find the exact value of the following.

a. b. c.

d. e. f.

Additional problems available in the textbook: Page 551 …7 – 16, 18, 20, 23, 25, 26, 29. Examples 1 and 2bcstarting on page 541.

Solutions:

Definition The inverse sine function, denoted by , is defined by if and only if , where and .

Notation: Sometimes, is denoted by . That is, = .

The restriction that is put on the sine function in order to make it be one-to-one means that your angle answer must be in the closed interval . NOTE: The angles in the open interval are in the first quadrant. The angles in the open interval are in the fourth quadrant. The angle 0 is on the positive x-axis, the angle is on the positive y-axis, and the angle is on the negative y-axis.

1a.Answer:

NOTE: Since is positive and not maximum positive of 1, then the angle answer is in the first quadrant. That is, the angle answer is in the open interval . We know that either by Unit Circle Trigonometry or Right Triangle Trigonometry. Note that the point is on the graph of the sine function.

Back to Problem 1.

1b.Answer:

Back to Problem 1.

1c.Answer:

NOTE: Since is negative and not minimum negative of , then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We will show that in Problem 1g.

We know that using reference angles. Note that the point is on the graph of the sine function.

Back to Problem 1.

1d.Answer:

NOTE: Since is negative and not minimum negative of , then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We showed that in Problem 1b.

We know that using reference angles. Note that the point is on the graph of the sine function.

Back to Problem 1.

1e.Answer:

NOTE: Since 1 is the maximum positive of 1, then the angle answer is not in the first quadrant. So, the angle answer is on one of the coordinate axes, namely the positive y-axis. The angle which is in the closed interval and lies on the positive y-axis is the angle . We know that by Unit Circle Trigonometry. Note that the point is on the graph of the sine function.

Back to Problem 1.

1f.Answer: 0

NOTE: Since 0 is not positive, then the angle answer is not in the first quadrant. Since 0 is not negative, then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the positive x-axis. The angle which is in the closed interval and lies on the positive x-axis is the angle 0. We know that by Unit Circle Trigonometry. Note that the point is on the graph of the sine function.

Back to Problem 1.

1g.Answer:

Back to Problem 1.

1h.Answer:

NOTE: Since is the minimum negative of , then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the negative y-axis. The angle which is in the closed interval and lies on the negative y-axis is the angle . We know that by Unit Circle Trigonometry. Note that the point is on the graph of the sine function.

Back to Problem 1.

1i.Answer:

NOTE: Since is negative and not minimum negative of , then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We showed that in Problem 1a.

We know that using reference angles. Note that the point is on the graph of the sine function.

Back to Problem 1.

Definition The inverse tangent function, denoted by , is defined by if and only if , where an y is any real number.

Notation: Sometimes, is denoted by . That is, = .

The restriction that is put on the tangent function in order to make it be one-to-one means that your angle answer must be in the open interval . NOTE: The angles in the open interval are in the first quadrant. The angles in the open interval are in the fourth quadrant. The angle 0 is on the positive x-axis.

From Lesson 2, you had the following diagram to help you find the value of the tangent of the three Special Angles in the I quadrant.

Tangent

Now, reverse the arrows to have a diagram to help you find the value of the inverse tangent of these three numbers of 1, , and :

Inverse Tangent

2a.Answer:

NOTE: Since is positive, then the angle answer is in the first quadrant. That is, the angle answer is in the open interval . We know that either by memorization or Right Triangle Trigonometry. Note that the point is on the graph of the tangent function.

Back to Problem 2.

2b.Answer:

Back to Problem 2.

2c.Answer:

NOTE: Since is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We will show that in Problem 2f.

We know that using reference angles. Note that the point is on the graph of the tangent function.

Back to Problem 2.

2d.Answer:

NOTE: Since is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We showed that in Problem 2b.

We know that using reference angles. Note that the point is on the graph of the tangent function.

Back to Problem 2.

2e.Answer: 0

NOTE: Since 0 is not positive, then the angle answer is not in the first quadrant. Since 0 is not negative, then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the positive x-axis. The angle which is in the open interval and lies on the positive x-axis is the angle 0. We know that by Unit Circle Trigonometry. Note that the point is on the graph of the tangent function.

Back to Problem 2.

2f.Answer:

NOTE: Since 1 is positive, then the angle answer is in the first quadrant. That is, the angle answer is in the open interval . We know that either by memorization or Right Triangle Trigonometry. Note that the point is on the graph of the tangent function.

Back to Problem 2.

2g.Answer:

NOTE: Since is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

NOTE: We showed that in Problem 2a.

We know that using reference angles. Note that the point is on the graph of the tangent function.

Back to Problem 2.

Solution to Problems on the Pre-Exam:Back to Page 1.

16.Find the exact value of each of the following. (3 pts. each)

a.Answer:

c.Answer:

NOTE: Since is negative and not minimum negative of , then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

We know that using reference angles. Note that the point is on the graph of the sine function.

d.Answer:

NOTE: Since is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval .

If we let , then . Thus,

. That is, .

We know that using reference angles. Note that the point is on the graph of the tangent function.