Crash Course on Trigonometry by Dr. Don Spickler, Lecture notes of Engineering Mathematics

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Crash Course in Trigonometry

Dr. Don Spickler

September 5, 2003

Contents

• 1 Trigonometric Functions
  • 1.1 Introduction
  • 1.2 Right Triangle Trigonometry
    • 1.2.1 Definitions
    • 1.2.2 Trigonometric Functions of Standard Angles
    • 1.2.3 Trigonometric Identities
  • 1.3 Unit Circle Trigonometry
    • 1.3.1 Setting up the Relationship
    • 1.3.2 More Trigonometric Identities
  • 1.4 Functions
  • 1.5 Quick Reference

side of the triangle. For this reason it is sometimes labeled opp. The side labeled b is the side adjacent to the angle θ since it is on the same side of the triangle. For this reason it is sometimes labeled adj. The side labeled c is the hypotenuse and is sometimes labeled hyp. There are six trigonometric functions that are defined as ratios of the lengths of this triangle. The sine of θ is denoted as sin(θ) and it is the length of the opposite side a divided by the hypotenuse c, that is,

sin(θ) =

opp hyp

a c

The cosine of θ is denoted as cos(θ) and it is the length of the adjacent side b divided by the hypotenuse c, that is,

cos(θ) =

adj hyp

b c

The remaining four trigonometric functions can be represented by the sine and the cosine as well as a, b and c. The tangent of θ is denoted as tan(θ) and it is the length of the opposite side a divided by the length of the adjacent side b, that is,

tan(θ) =

opp adj

a b

We can also represent it as the sin(θ) divided by the cos(θ). Note that with a little algebra we do obtain ab as well.

tan(θ) =

sin(θ) cos(θ)

a c b c

a b

The cotangent is the reciprocal of the tangent. So we can represent like that or by the cosine divided by the sine or by the adjacent side divided by the opposite side.

cot(θ) =

tan(θ)

sin(θ) cos(θ)

cos(θ) sin(θ)

b c a c

b a

The secant of θ is the reciprocal of the cosine. So we can represent like that or by the hypotenuse divided by the adjacent side.

sec(θ) =

cos(θ)

b c

c b

Finally, the cosecant of θ is the reciprocal of the sine. So we can represent like that or by the hypotenuse divided by the opposite side.

csc(θ) =

sin(θ)

a c

c a

Example 1.2.1 : Say we have the following triangle

A quick calculation using the Pythagorean theorem verifies that this is indeed a right triangle.

√ 32 + 4^2 =

The trigonometric functions on θ are as follows.

sin(θ) =

a c

cos(θ) =

b c

tan(θ) =

a b

cot(θ) =

b a

sec(θ) =

c b

csc(θ) =

c a

1.2.2 Trigonometric Functions of Standard Angles

Most of the time when we want the sine or cosine of an angle we will get out our calculator and just type it in. Unless we have a TI-89 or 92, the calculator will give us an approximation to the value. In most cases this approximation is perfectly valid and in general finding the exact values of trigonometric functions can be rather difficult. There are some angles that, with a little work, we can find the exact values to the trigonometric functions. These will come in handy when we look at the other two formulations of trigonometry. The angles in question are 30◦, 45◦^ and 60◦. We will call these the standard angles. First we will take care

side of the 30-60-90 triangle is

√ 3

2. All in all we have the following diagram.

Now if we consider just one of the 30-60-90 triangles we have the following trigonometric functions for the 30◦^ and 60◦^ angles.

sin(30◦) =

a c

1 2 1

cos(30◦) =

b c

√ 3 2 1

tan(30◦) =

a b

1 √^2 3 2

cot(30◦) =

b a

√ 3 2 1 2

sec(30◦) =

c b

√ 3 2

csc(30◦) =

c a

1 2

and

sin(60◦) =

a c

√ 3 2 1

cos(60◦) =

b c

1 2 1

tan(60◦) =

a b

√ 3 2 1 2

cot(60◦) =

b a

1 √^2 3 2

sec(60◦) =

c b

1 2

csc(60◦) =

c a

√ 3 2

When we do the unit circle trigonometry we will expand upon this list. For the other angles that are not as nice as these we can simply use a calculator to find approximations.

1.2.3 Trigonometric Identities

Trigonometric identities are very useful whenever you are simplifying or solving trigonometric expressions. The nifty thing is that most of the identities come directly from the Pythagorean Theorem, and a little algebra. First, sin^2 (θ) + cos^2 (θ) = 1 for any angle θ whatsoever. To see this just go back to the definitions of sine and cosine.

sin^2 (θ) + cos^2 (θ) =

(a

c

b c

a^2 c^2

b^2 c^2 =

a^2 + b^2 c^2 =

c^2 c^2 = 1

This also gives us

sin^2 (θ) = 1 − cos^2 (θ) and cos^2 (θ) = 1 − sin^2 (θ)

We can use these to get identities for the other four trigonometric functions as well. For

usually use radian measure rather than degree measurement. Radians are simply another unit for measuring the size of an angle. Just like feet and meters are two different ways to measure length. It turns out that there is an easy way to convert from degrees to radians and back. The formulas are,

x◦^ = x

( (^) π

180

radians and x radians = x

π

degrees

We will see how these were derived shortly. The radian measures the angle by using the length of the portion of the unit circle that is cut by the angle.

In other words, in radians, the measure of the angle ∠CAB is the length of the arc BC. So if the length of BC was one then the angle ∠CAB would be one radian. Since the circumference of a circle is 2πr and r in this case is 1, the length of the entire circle is 2π. So 2π is equivalent to 360◦, which means

2 π radians = 360 ◦ 1 radian =

2 π 1 radian =

π

and

2 π radians = 360 ◦ 2 π 360

radians = 1 ◦ π 180

radians = 1 ◦

Now let’s convert the standard angles to radians. Take 30◦, this is 121 of the entire circle. So in radians this would be 212 π = π 6. Now take 45◦, this is 18 of the entire circle. So in radians

this would be 28 π = π 4. Finally, take 60◦, this is 16 of the entire circle. So in radians this would be 26 π = π 3.

Degree Radians 30 ◦^ π 6 45 ◦^ π 4 60 ◦^ π 3

We will now extend these angles around the circle.

Degree Radians 0 ◦^0 30 ◦^ π 6 45 ◦^ π 4 60 ◦^ π 3 90 ◦^ π 2 120 ◦^23 π 135 ◦^34 π 150 ◦^56 π

Degree Radians 180 ◦^ π 210 ◦^76 π 225 ◦^54 π 240 ◦^43 π 270 ◦^32 π 300 ◦^53 π 315 ◦^74 π 330 ◦^116 π

Of course, all of the other angles between 0◦^ and 360◦^ have corresponding radian measures, they just are not as nice as the ones above. For example, the angle 29◦^ corresponds to 29 · 180 π = 29180 π radians. Since this is not a particularly nice angel we tend to convert it to decimal form and simply say that it is 0.506145483078 radians. We will find that the decimal representation comes in handy when we shift over to the function representation of the trigonometric functions. Before we move to the function representation let’s continue with the unit circle. We have two major things to do. First we need to get a correspondence between the (x, y) points on the unit circle and the trigonometry of the triangle. Second we need to use symmetry to find the trigonometric functions for each of the nice angles above. The correspondence between the (x, y) points on the unit circle and the trigonometry of the triangle is probably the most important relationship in trigonometry. To get this relationship we will place a right triangle inside the unit circle so that its hypotenuse is a radius. We will call the angle between the positive x axis and the hypotenuse θ and our trigonometric functions will be of the angle θ.

So every point on the unit circle has the coordinates (cos(θ), sin(θ)). Now if we put this observation together with some symmetry we can obtain the trigonometric functions at the nice angles. First, let’s place all of the nice angles on the unit circle. We will start with the three nice angles from the triangle representation (30◦, 45◦^ and 60◦) along with 0◦^ and 90◦. These angles correspond to 0, π 6 , π 4 , π 3 and π 2 radians respectively.

This takes care of the first quadrant. In the second quadrant, we take the nice angles (30◦, 45◦, 60◦^ and 90◦) and add 90◦^ to each. Similarly, in radian measure we take the nice angles (π 6 , π 4 , π 3 and π 2 ) and add π 2 to each.

This takes care of the second quadrant. In the third quadrant, we can think of the nice angles in a couple ways. First, we could take the nice angles from the first quadrant (30◦, 45 ◦, 60◦^ and 90◦) and add 180◦^ to each. Similarly, in radian measure we would take (π 6 , π 4 , π 3 and π 2 ) and add π to each. Another way to think about the nice angles in the third quadrant is to take the nice angles in the second quadrant (120◦, 135◦, 150◦^ and 180◦) and add 90◦ to each. Similarly, in radian measure we would take (^23 π , 34 π , 56 π and π) and add π 2 to each. In either case we get the angles, 210◦, 225◦, 240◦^ and 270◦, which corresponds to 76 π , 54 π , 43 π and 32 π respectively.

Finally, in the fourth quadrant, we can construct the nice angles in three different ways from the angles we already have. First, we could take the nice angles from the first quadrant and add 270◦^ (^32 π radians). Second, we could take the nice angles from the second quadrant

the y-axis. When you reflect the point (x, y) about the y-axis you get the point (−x, y). So to fill in the chart for the second quadrant we simply change the signs of the x coordinates.

To get the third quadrant values we could either reflect the second quadrant points about the x-axis (effectively changing all of the signs of the y coordinates) or by reflecting the first quadrants points about the origin (changing the signs of both coordinates).

Finally, to finish it all up we simply need to fill in the fourth quadrant. This can be done in any one of three ways. We could reflect the first quadrant over the x-axis or reflect the third quadrant over the y-axis or reflect the second quadrant over the origin.

That completes the unit circle chart of the trigonometric functions of the nice angles. Memorizing the entire chart is not necessary if one simply remembers the way we constructed the chart from just the three points in the first quadrant. In fact, this is overkill as well. To construct the chart you need only remember three values 12 ,

√ 2 2 and^

√ 3 2 and the fact that 1 2 <^

√ 2 2 <^

√ 3 2.^ Then the chart can be constructed by imagining where the angle lies and then placing the corresponding values in the x and y coordinates, finally put the signs on the coordinates that correspond with the quadrant the point is in.

1.3.2 More Trigonometric Identities

With unit circle trigonometry we can derive many more useful trigonometric identities. We will look at only a few of the more important ones here. The identities that spawn most of the others are the formulas for the sine or cosine of angle sums and differences. We will derive those now. Consider the following diagram where A and B are two points on the unit circle at angles u and v respectively.

Note that since we rotated clockwise by an angle of v the angle to the point A is now u − v. If we take the distance between A and B, noting that B is at (1, 0), we get.

AB =

(cos(u − v) − 1)^2 + (sin(u − v) − 0)^2

Doing a little algebra we have,

AB =

(cos(u − v) − 1)^2 + (sin(u − v) − 0)^2

=

cos^2 (u − v) − 2 cos(u − v) + 1 + sin^2 (u − v)

=

2 − 2 cos(u − v)

Setting these two equations equal to each other gives,

2 − 2 cos(u − v) =

2 − 2(cos(u) cos(v) + sin(u) sin(v)) 2 − 2 cos(u − v) = 2 − 2(cos(u) cos(v) + sin(u) sin(v)) −2 cos(u − v) = −2(cos(u) cos(v) + sin(u) sin(v)) cos(u − v) = cos(u) cos(v) + sin(u) sin(v)

Which is our first difference angle formula. Using the fact that sin(−θ) = − sin(θ) and cos(−θ) = cos(θ), both of which can be seen from the unit circle constructions above.

cos(u + v) = cos(u − (−v)) = cos(u) cos(−v) + sin(u) sin(−v) = cos(u) cos(v) − sin(u) sin(v)

Also, we can get the first of our double angle formulas as

cos(2u) = cos(u + u) = cos(u) cos(u) − sin(u) sin(u) = cos^2 (u) − sin^2 (u)

Let’s move onto the sine by converting the cosine formulas into sine formulas. Note that

cos

(π 2

− θ

= cos

(π 2

cos(θ) + sin

(π 2

sin(θ) = sin(θ)

and thus

sin

(π 2

− θ

= cos

(π 2

(π 2

− θ

= cos(θ)

So,

sin(u + v) = cos

(π

2

− (u + v)

= cos

((π

2

− u

− v

= cos

(π

2

− u

cos(v) + sin

(π

2

− u

sin(v) = sin(u) cos(v) + cos(u) sin(v)

and

sin(u − v) = sin(u) cos(−v) + cos(u) sin(−v) = sin(u) cos(v) − cos(u) sin(v)

Which also gives a double angle formula of,

sin(2u) = sin(u + u) = sin(u) cos(u) + cos(u) sin(u) = 2 sin(u) cos(u)

We can use these to construct formulas for the other trig functions as well. For example,