Rather than serving as a text or treatise, the book focuses on the essentials of trigonometry. All fourteen sections are organized in a manner that allows readers to advance sequentially or to skip around. The approach encourages memorization of ratios and formulas, and the practice problems offer ample opportunities to become comfortable with applying the trig ratios to a variety of settings.
Rather than serving as a text or treatise, the book focuses on the essentials of trigonometry. All fourteen sections are organized in a manner that allows readers to advance sequentially or to skip around. The approach encourages memorization of ratios and formulas, and the practice problems offer ample opportunities to become comfortable with applying the trig ratios to a variety of settings.
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Overview
Rather than serving as a text or treatise, the book focuses on the essentials of trigonometry. All fourteen sections are organized in a manner that allows readers to advance sequentially or to skip around. The approach encourages memorization of ratios and formulas, and the practice problems offer ample opportunities to become comfortable with applying the trig ratios to a variety of settings.
Product Details
| ISBN-13: | 9780486804798 |
|---|---|
| Publisher: | Dover Publications |
| Publication date: | 03/30/2015 |
| Series: | Dover Books on Mathematics |
| Sold by: | Barnes & Noble |
| Format: | eBook |
| Pages: | 176 |
| File size: | 12 MB |
| Note: | This product may take a few minutes to download. |
About the Author
Read an Excerpt
Attacking Trigonometry Problems
By David S. Kahn
Dover Publications, Inc.
Copyright © 2015 David S. KahnAll rights reserved.
ISBN: 978-0-486-80479-8
CHAPTER 1
UNIT ONE
The Basic Trig Ratios
Trigonometry consists of learning how to use six different functions, or ratios, which show up in a surprisingly large number of places. Where do they come from? A good place to start is with some basic geometry. Remember similar triangles? If two triangles are similar, then they have equal angles and the ratio of their sides is the same. For example,
If the two triangles above are similar, then A = [??]D, [??]B = [??]E, and [??]C = [??]F, and a/b = b/e = c/f,
Let's look at two similar right triangles:
Notice that ΔABC is similar to ΔADE because each contains a right angle (C & E) and the same angle (A), and so the third angle must also be the same (because the measures of the angles in a triangle add to 180°). This means that AD/AB=AE/AC=DE/BC. If we had the following set of right triangles, the corresponding ratios would all be equal.
In fact, for any right triangle that has an angle with a measure equal to the measure of angle A, the ratios are the same as those of any other right triangle that has an angle with measure equal to that of A. This is the essential fact of Trigonometry and can be used in many powerful ways.
For the triangle below, these are the three basic trigonometric ratios to learn:
The sine of angle A is the ratio a/c.
The cosine of angle A is the ratio b/c.
The tangent of angle A is the ratio a/b.
We usually abbreviate sine as sin, cosine as cos, and tangent as tan, and write these ratios using the following notation:
sin A = a/c, cos A = b/c, tan A = a/b.
You will want to get comfortable with these ratios. There is an easy way to memorize them.
There are three sides from the perspective of angle A: a is the side opposite angle A, b is the side adjacent to angle A, and c is the hypotenuse. Therefore, we can think of:
sin A as opposite/hypotenuse
sin A as adjacent/hypotenuse
tan A as opposite/adjacent
This gives the traditional mnemonic:
which stands for: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent
In the triangle above, sin A is the side opposite angle A (3) divided by the hypotenuse (5), so sin A = 3/5. Similarly, cos A is the side adjacent to angle A (4) divided by the hypotenuse (5), so cos A = 4/5. What is tan A? tan . Got the idea? Of course, we could also find the trig ratios of the other acute angle, B. Now, sin B is the side opposite angle B (4) divided by the hypotenuse (5), so sin . Similarly, cos B is the side adjacent to angle B (3) divided by the hypotenuse (5), so cos and tan .
Example 2: Figure 7
What are the three trig ratios of angle A?
sin , cos , and tan What if we asked for the three trig ratios of angle B?
They are sin , cos , and tan .
Notice how in these examples, sin A = cos B and cos A = sin B? This is not a coincidence! Remember that the sum of the two acute angles in a right triangle is 90°.
This means that angle A = (90° - B) and angle B = (90° - A). Therefore, in any right triangle, sin A = cos(90° - A) and cos A = sin(90° - A).
Example 3: Figure 8
For this triangle, sin , and sin .
By the way, the tangent of one of the acute angles is the reciprocal of the tangent of the other acute angle. In other words, (and vice versa). Notice that in Figure 8, , and .
Notice that we have not been finding the trig ratios for the right angle. Right now, we only know how to find the trig ratios for an angle between 0° and 90°. Later, we will learn how to find the trig ratios for an angle of any measure.
Time to practice!
Practice Problems Practice problem 1: Figure 9
Practice problem 2: Figure 10
Practice problem 3: Figure 11
Practice problem 4: If sin(2x - 5) = cos(5x + 25), find x.
Practice problem 5: If cos(x2 - 20) = sin(12 + x2), find x, where the angle is in degrees.
Solutions to the Practice Problems Solution to practice problem 1: Remember our mnemonic SOH CAH TOA! Here, the side opposite angle A is 9, the side adjacent to angle A is 12, and the hypotenuse is 15. Therefore, we get:
Similarly, the side opposite angle B is 12, the side adjacent to angle B is 9, and the hypotenuse is 15. We get:
Solution to practice problem 2: Here, the side opposite angle x is 24, the side adjacent to angle x is 7, and the hypotenuse is 25. Therefore, we get:
Now, the side opposite angle y is 7, the side adjacent to angle y is 24, and the hypotenuse is 25. We get:
Solution to practice problem 3: Here, the side opposite angle P is 15, the side adjacent to angle P is 8, and the hypotenuse is 17. Therefore, we get:
Now, the side opposite angle Q is 8, the side adjacent to angle Q is 15, and the hypotenuse is 17. We get:
Solution to practice problem 4: Remember our rule that sin A = cos(90° - A). This means that we can rewrite the equation:
If the two sines are equal, then the angles must be equal. This gives us:
Now we can use some simple algebra!
Solution to practice problem 5: Again, we use our rule that sin A = cos (90° - A). We can rewrite the equation:
If the two cosines are equal, then the angles must be equal. This gives us:
Algebra time!
UNIT TWO Special Triangles Now that we have learned the three basic trig ratios, let's learn how to find the sine, cosine, and tangent of angles in special triangles. You should remember from Geometry that an equilateral triangle has some special properties. First, the measures of all three angles are 60°. Second, the lengths of all the sides are equal. If we drop a perpendicular line from a vertex of an equilateral triangle to the opposite side, the triangle is cut into two congruent triangles. In the figure below, we can see that the measures of the angles of triangle ABD and triangle CBD are then 30° 60° - 90°: Figure 1
Next, if the sides of the triangle each have length 2x, then side AD has length x, and we can use the Pythagorean Theorem to find that side BD has length : Figure 2
Now we can find the trig functions for angles of measures 30° and 60°:
You should try to commit these to memory because they show up often in many types of math problems. In a little bit, we will learn an easy way to memorize these. Now, let's look at another special triangle. A square has some special properties as well. First, all of the sides are congruent. Second, all of the angles are right angles. If you take a square of side x and draw a diagonal, you get two isosceles right triangles.
Figure 3
We can use the Pythagorean Theorem to find the length of the diagonal, which is . This triangle is often referred to as a 45° -45° -90° triangle. Now we can find the trig ratios of a 45°-angle:
Note that sin 45° = cos 45°. Why should this make sense? By the way, many people rationalize the denominator of by multiplying the numerator and denominator by . This makes memorizing the sines and cosines of the special angles a little easier.
Let's make a table of the trig functions that we just learned:
The reason that these are called the special angles is that we can find the exact values of the trig functions for these angles. For almost all other angles, however, we approximate the trig ratios with decimal values.
UNIT THREE Trig Ratios for Other Angles Now let's learn how to find the trig values for other angles. Suppose that you draw a circle of radius 1 (the "unit circle"), centered at the origin. Pick a point in Quadrant I on the circle and draw the radius from the origin to that point. Figure 1
We can construct a right triangle using that radius. The lengths of the legs are equal to the coordinates x and y, respectively, and the hypotenuse of the right triangle is 1. If we call the angle between the positive x-axis and the radius ?, then we can see that , , . In other words, the coordinates of the point (x, y) are (cos ?, sin ?).
Figure 2
Why is this useful? We can now use this information to find the sine, cosine, and tangent of any angle.
Example 1: Let's find the sine, cosine, and tangent of 0°.
These ratios are simply the coordinates when the angle between the radius and the positive x-axis is ? = 0°. The coordinates on the unit circle are (1, 0): Figure 3
Thus, sin 0° = 0, cos 0° = 1, and .
Example 2: Let's do it again for ? = 90°. Here, the coordinates for the angle between the radius and the positive x-axis are (0, 1): Figure 4
Thus, sin 90° = 1, cos 90° = 0, and . Notice that the tangent is undefined at 90°. You will find that the trig functions other than sine and cosine are undefined at certain angles.
These will occur with trig functions that are formed with the sine and cosine functions in their denominators. They will be undefined at those angles where the sine or cosine is zero because you can't have zero in the denominator of a fraction. You will learn more about that in the next unit.
Example 3: Let's keep going to 180°: Figure 5
Here, the coordinates of the intersection are (-1, 0). So, sin 180° = 0, cos 180° = -1, and .
Example 4: Finally, let's find the ratios for 270°: Figure 6
Here, the coordinates of the intersection are (0, -1). So, sin 270° = -1, cos 270° = 0, and .
You should memorize the sine, cosine, and tangent for 0°, 90°, 180°, and 270°. We can make a table to help us memorize these values.
Example 5: Let's find the trig values for ? = 150°. If we draw the unit circle, we are looking for the coordinates of the point where the radius intersects the circle when the angle between the radius and the positive x-axis is 150°: Figure 7
But however, notice that the angle between the radius and the negative x-axis is 30°. This is called the reference angle and we can use it, rather than 150°, to find the trig ratios. We can use the reference angle to make a 30°-60°-90° triangle with the negative x-axis. The coordinates of the point of intersection are . This means that , , and . Notice that these are the same values as sin 30°, cos 30°, and tan 30°, except that cosine and tangent are negative.
It is important to understand what the reference angle is. When we are looking for the trig ratio of any angle other than an acute one, we find it by using the unit circle. We then draw the radius that is the measure of the angle we are evaluating. The radius will always make an acute angle with either the positive or negative x-axis. That angle is the reference angle, and we use the trig values of that acute angle. Remember that the reference angle is always an acute angle, and it is always formed between the radius and the x-axis, never the y-axis.
Example 6: Now, let's find sine, cosine, and tangent of 225°.
Figure 8
Here, we use the angle between the radius and the negative x-axis to find the reference angle of 225° - 180° = 45°. We can use the reference angle to make a 45°-45°-90° triangle with the x-axis. This time, the coordinates are . (Why?) Therefore, , , and tan 225° = 1. Again, these are the same values as sin 45°, cos 45°, and tan 45°, except that the sine and cosine are negative.
Why do the signs change? In Quadrant I, the x- and y-coordinates of a point will be both positive. However, in Quadrant II the x-coordinate will be negative while the y-coordinate stays positive. Because cosine is equal to the value of the x-coordinate, cosine is negative, while sine stays positive. Furthermore, because tangent is , and one coordinate is negative, tangent is negative. In Quadrant III, both coordinates are negative, so sine and cosine are both negative. The tangent is positive because it is the ratio of two negative numbers. Finally, in Quadrant IV, the x-coordinate is positive and the ycoordinate is negative. So, sine is negative, cosine is positive, and tangent is negative.
There is an easy way to remember the signs of the trig ratios in the various quadrants.
Look at the figure below: Figure 9
The mnemonic is.
All Students Try Candy which stands for: All trig ratios are positive in Quadrant I.
Sine is positive in Quadrant II (and the others are negative).
Tangent is positive in Quadrant III (and the others are negative).
Cosine is positive in Quadrant IV (and the others are negative).
Let's do another, just to make sure that you get the idea.
Example 7: Let's find sine, cosine, and tangent of 300°. Notice that we stopped drawing the unit circle. The circle isn't really necessary because the acute angle that the radius makes with the x-axis is what we are interested in.
Figure 10
Here, we use the angle between the radius and the positive x-axis. This is the reference angle and has a measure of 360° - 300° = 60°. We can use the reference angle to make a 30°-60°-90°-triangle with the x-axis. This time, the coordinates are . Therefore, , , and .
These are the same values as sin 60°, cos 60°, and tan 60°, except that the sine and tangent are negative, which we expected from our mnemonic.
What if we used an angle greater than 360°? Well, we would just go around the axes in full circles as many times as needed and then around in part of a circle until we get to our destination.
Example 8: Suppose we wanted to find sine, cosine, and tangent of 420°. We would go once around the x-axis, and another 60°.
Figure 11
Thus, the sine, cosine, and tangent of 420° are the same as for 60°, namely, , , and . They are all positive because we are back in Quadrant I.
Finally, what about if we measure the angle going clockwise instead of counterclockwise? We call this a negative angle.
Example 9: Let's find the sine, cosine, and tangent of -210°. Now we find the angle by going around the axes in the opposite direction: Figure 12
Thus, going to -210° is the same as going to 150° because -210° + 360° = 150°. Thus, the trig ratios are , , and .
Let's do some practice problems.
Practice Problems Practice problem 1: Find the sine, cosine, and tangent of 120°.
Practice problem 2: Find the sine, cosine, and tangent of 135°.
Practice problem 3: Find the sine, cosine, and tangent of 210°.
Practice problem 4: Find the sine, cosine, and tangent of 240°.
Practice problem 5: Find the sine, cosine, and tangent of 315°.
Practice problem 6: Find the sine, cosine, and tangent of 330°.
Practice problem 7: Find the sine, cosine, and tangent of 480°.
Practice problem 8: Find the sine, cosine, and tangent of 870°.
Practice problem 9: Find the sine, cosine, and tangent of -45°.
Practice problem 10: Find the sine, cosine, and tangent of -300°.
Solutions to the Practice Problems Solution to practice problem 1: Find the sine, cosine, and tangent of 120°.
First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 60° (the reference angle), so we need to find the sine, cosine, and tangent of 60°, and then remember that the sine will be positive and the cosine and tangent will be negative because we are in Quadrant II. We get:
Solution to practice problem 2: Find the sine, cosine, and tangent of 135°. First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 45° (the reference angle), so we need to find the sine, cosine, and tangent of 45°. Remember that the sine will be positive and the cosine and tangent will be negative because we are in Quadrant II. We get:
Solution to practice problem 3: Find the sine, cosine, and tangent of 210°. First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 30° (the reference angle), so we need to find the sine, cosine, and tangent of 30°. Remember that the tangent will be positive and the sine and cosine will be negative because we are in Quadrant III. We get:
Solution to practice problem 4: Find the sine, cosine, and tangent of 240°. First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 60° (the reference angle), so we need to find the sine, cosine, and tangent of 60°. Remember that the tangent will be positive and the sine and cosine will be negative because we are in Quadrant III. We get:
Solution to practice problem 5: Find the sine, cosine, and tangent of 315°. First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 45° (the reference angle), so we need to find the sine, cosine, and tangent of 45°. Remember that the cosine will be positive and the sine and tangent will be negative because we are in Quadrant IV. We get:
Solution to practice problem 6: Find the sine, cosine, and tangent of 330°. First, let's draw a picture so that we can find the reference angle:
We can see that the acute angle with the x-axis is 30° (the reference angle), so we need to find the sine, cosine, and tangent of 30°. Remember that the cosine will be positive and the sine and tangent will be negative because we are in Quadrant IV. We get:
Solution to practice problem 7: Find the sine, cosine, and tangent of 480°. First, let's draw a picture so that we can find the reference angle:
Here, we are going around the axes more than once (360°), and then continue another 120°, so we end up with the acute angle made with the x-axis as 60° (the reference angle). So we need to find the sine, cosine, and tangent of 60°. Remember that the sine will be positive and the cosine and tangent will be negative because we are in Quadrant II. We get:
Solution to practice problem 8: Find the sine, cosine, and tangent of 870°. First, let's draw a picture so that we can find the reference angle:
Here, we are going around the axes twice (2 : 360° = 720°) and then continue another
(Continues...)
(Continues...)
Excerpted from Attacking Trigonometry Problems by David S. Kahn. Copyright © 2015 David S. Kahn. Excerpted by permission of Dover Publications, Inc..
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
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Table of Contents
Contents
Unit One The Basic Trig Ratios,Unit Three Trig Ratios for Other Angles,
Unit Four Degrees and Radians,
Unit Five The Reciprocal Functions,
Unit Six Some Basic Trigonometry Problems,
Unit Seven Sine and Cosine Graphs,
Unit Eight Graphing Tangent, Cotangent, Secant, and Cosecant,
Unit Nine Inverse Trigonometric Functions,
Unit Ten Basic Trigonometric Identities and Equations,
Unit Eleven More Trigonometric Identities,
Unit Twelve Trigonometric Angle Formulas,
Unit Thirteen The Law of Sines,
Unit Fourteen The Law of Cosines and Area Formulas,