Chapter 14 - Trigonometry

This chapter discusses trigonometry, which is the study of triangles. It contains the following sections:

section 14.1 - In this section we introduce the trigonometry of right triangles, then oblique triangles, then vectors, and then rotating objects.
section 14.2 - In this section we derive and list a set of trigonometric identities and show how they can be used to simplify trigonometric expressions.
section 14.3 - In this section we explain how to solve trigonometric equations. Often trigonometric identities must be used to solve them.

14.1 - Introduction to Trigonometry

The Cartesian Plane

A cartesian plane is a plane on which an x axis is drawn horizontally and a y axis is drawn vertically. Any point in the plane is referred to by using ordered pair notation like this:

This is the point at x = 1.5 and y = 2. It is indicated in the picture to the right.

This method of using an x value and a y value to locate a point is called the rectangular coordinate system (notice the dotted rectangle shown in the picture). Another coordinate system in common use (especially for complex numbers) is the polar coordinate system. It uses a distance from the origin and a direction to locate a point.

The plane is divided into 4 quadrants. Quadrant 1 has positive x and y values. The other quadrants are reached by going counter-clockwise from the first.

Angles and their Measure

An angle in standard position has its vertex at the origin and one side along the x axis.
The angle is generated when the ray is rotated from the initial to the terminal position.
A counter-clockwise rotation is considered to be a positive rotation.

A clockwise rotation is considered to be a negative rotation.

There are two scales that are commonly used to measure the size of an angle: the degree scale and the radian scale:

Degrees

In this scale an angle is expressed in units called degrees, with a rotation of one full circle being 360 degrees (symbol ° ).

Radians

In this scale an angle is expressed in units called radians, with a full circle rotation being 2π or approximately 6.28 radians.

The radian scale is formally defined like this:

The size of angle θ expressed in radians equals s (the arc length subtended by the angle) divided by r (the radius of the circle).

For one full circle s = C (the circumference), and since C = 2 π r we get

for an angle of one full circle.

An important property of radians is that since s and r both have the dimensions of length, radians are units without dimensions.

Example: The angle θ in the picture to the right equals 2 radians since s = 6 and r = 3.

Converting between degrees and radians

To convert between degrees and radians you can use the fact that π radians = 180°. This means that:

These two fractions are called UFOOs (useful forms of one). Multiplying any angle by one of these does not change the size of the angle but it does change the units of the angle.

Example: Convert 118° to radians.

Solution:

Example: Convert 0.438 radians to degrees.

Solution:

The Angles of a Triangle

The sum of the angles in any triangle equals 180° or π radians.

Pythagoras’ Theorem and Lengths

Pythagoras’ Theorem is concerned with the lengths of the sides of a right triangle. (A right angle is a 90° angle and a right triangle is one that contains a 90° angle.) Consider the right triangle with sides of length a, b and c shown to the right. a is called the altitude, b is called the base and c, the longest side opposite the right angle, is called the hypotenuse. Pythagoras’ Theorem states that the sides are related by the formula:

c² = a² + b²

A nice animated geometric proof of Pythagoras’ Theorem is given on wikipedia.

Example: Find the length of the straight line in the Cartesian plane extending from the point (x₁, y₁) = (−1.5, 2) to the point (x₂, y₂) = (2, −1).

Solution: Extend the dotted lines horizontally and vertically to create a right triangle. This triangle has base b = 3.5 and altitude a = 3, so Pythagoras’ Theorem gives:

c² = 3² + 3.5² = 21.25

Taking the square root of both sides gives the length of the straight line as

The above example illustrates the following very useful application of Pythagoras’ Theorem. If (x₁, y₁) and (x₂, y₂) are any two points in the Cartesian plane then the distance c between them can be obtained from the formula:

The trigonometric functions

For now let us restrict ourselves to right triangles. This means that angle θ must have a value between 0°and 90°.

Notice that if the triangle is scaled up in size then the sides get longer, but the ratio of the lengths of any two sides does not change.

It is possible to define six different ratios. Let opposite and adjacent denote the lengths of the sides opposite and adjacent to angle θ and let hypotenuse denote the long side. The six ratios are named sine, cosine, tangent, cotangent, secant and cosecant and are defined as follows:

Note:

The names of the six ratios are abbreviated as sin, cos, tan, cot, sec, csc.
The six ratios depend on (are functions of) the angle θ. We can denote this using functional notation like this: sin(θ), cos(θ), tan(θ), etc.
Many textbooks on trigonometry leave out the brackets and write sin θ instead of sin(θ). This is okay, but don’t write something like sin x + y. This is ambiguous because it is not clear if you should add x and y first and then take the sine, or take the sine of x first and then add y. Also remember that computer algebra systems like the Algebra Coach require the brackets.
Three of the trigonometric functions – sine, cosine and tangent – are built into your calculator. Since their values depend on the angle θ, they are found by entering the angle into your calculator in the appropriate mode (degree or radian) and then pressing the appropriate button: sin, cos or tan.
The other three trigonometric functions – cosecant, secant, and cotangent – are not built into your calculator. To find them you must first find the sine, cosine, or tangent on your calculator and then take the reciprocal.

Examples:

sin(30°) = 0.500
tan (57°) = 1.54
cos(0.74 rads) = 0.738
tan (1.1 rads) = 1.96

Example: Find y and r in the triangle.

Solution:

To find y use the fact that .
From the calculator tan(35.8°) = 0.7212. Setting these equal to each other gives which can be solved to give y = 8.94.
To find r use the fact that .
From the calculator cos(35.8°) = 0.8111. Setting these equal to each other gives which can be solved to give r = 15.3.

Answer: The required lengths are y = 8.94 and r = 15.3.

The inverse trigonometric functions

Suppose that we wish to find the angle θ in the triangle shown.

One thing we know is that

. In the examples above we put an angle θ into the calculator and then pressed the tan button to get a value out for the tangent.

Here we want to do the inverse process - we want to put a value of tangent in and get an angle out. To do this your calculator has the inverse trigonometric functions sin⁻¹, cos⁻¹ and tan⁻¹ built in. (Note: the superscript −1 has NOTHING to do with exponents. It is merely a NOTATION meaning ‘inverse function’. To avoid this possible confusion these functions are often called arcsine, arccosine, and arctangent; abbreviated arcsin, arccos and arctan.)

For this example tan(θ) = 0.7381 so θ = tan⁻¹(0.7381). The calculator gives tan⁻¹(0.7381) = 36.4°, so θ = 36.4°.

In general the inverse trigonometric functions are defined like this:

If sin(θ) = z, then θ = sin⁻¹(z) or θ = arcsin(z).
If cos(θ) = z, then θ = cos⁻¹(z) or θ = arccos(z).
If tan(θ) = z, then θ = tan⁻¹(z) or θ = arctan(z).

How to solve right triangles

Solving a triangle means finding all the missing sides and angles of the triangle. Here are the steps involved in solving right triangles.

Draw a picture (with the angle in standard position, if appropriate).
You will be given at least (a) one side and one angle, or (b) two sides (otherwise the triangle can’t be solved).
In case (a) you can use sin, cos or tan as appropriate to find another side.
In case (b) you can use sin⁻¹, cos⁻¹ or tan⁻¹ as appropriate to find an angle.
Use Pythagoras’ Theorem to find the final side.
Use the fact that the angles sum to 180° to find the final angle.
Give your answer in a sentence.

Example: A train is traveling along a straight section of track. A pilot flying 4220 meters directly above the caboose at the back end of the train observes that the angle of depression of the locomotive at the front of the train is 68.2°. Find the length of the train.

Solution: The picture on the left describes the situation. Notice that the angle of elevation of the airplane from the locomotive is also 68.2°. The pictures on the right define the terms “angle of depression” and “angle of elevation”, which are used often in surveying work.

Using the tangent function we get:

Evaluating the tan function gives 2.500 and solving for the length of the train then gives:

The sentence answer is that to 3 significant figures the length of the train is 1690 m.

Trigonometric functions of any angle

When we first defined the trigonometric functions (click here to review that) the angle θ was between 0° and 90° and we used the terms adjacent, opposite and hypotenuse to refer to the sides of a triangle.

But we now want to allow angle θ to have values outside this range. One reason is to be able to solve oblique triangles. These triangles can have an angle that is bigger than 90°. Another reason is that we want to be able to describe the angle of a spinning object as it rotates through many revolutions. An example is an aerial skier who jumps and does 2 revolutions in the air. They are “doing a 720”, meaning that they are rotating through 720°.

To allow for angles bigger than 90° we now imagine an arrow (or vector) pointing out from the origin with length r and orientated at angle θ, and with its arrowhead located at (x, y).

We construct a triangle by drawing a line vertically from the arrowhead to the x axis and another line horizontally across to the y axis.

We now redefine the six trigonometric functions like this:

If θ is between 0° to 90° then both coordinates, x and y, are positive and the trigonometric functions have the same values as before. But if θ is outside the range 0° to 90° then x or y or both can be negative. (Don’t forget, however, that the arrow’s length r is defined to be positive.) This means that the trigonometric functions can now possibly be negative.

Example: The vector with length r = 2.5 and angle θ = 127° has coordinates x = −1.5 and y = 2.0. Thus:

Here are two ways to remember the quadrants in which the trigonometric functions are positive or negative. The best way, of course, is to simply learn the graphs of the sine, cosine and tangent functions. Click here to see them.

Many students use the saying All Students Take Chemistry to remember the quadrants in which the functions are positive.

Calculating the sine, cosine, or tangent now that the angle can have any value

This is easy. Your calculator is programmed to find the sine, cosine, or tangent of any angle whatsoever (minus sign and all). For example the calculator gives cos(120°) = −0.5.

The answer the calculator gives is unique but it is important to note that several angles have the same sine, cosine, or tangent. For example the calculator gives sin(60°) = 0.866 and also gives sin(120°) = 0.866.

Calculating the angle now that the sine, cosine, or tangent can have any value

Because many angles have the same sine (or cosine, or tangent) and because the calculator only finds one of them, this is not so easy. For example let’s turn the last example around: suppose we know that the sine of some unknown angle equals 0.866 and we want to find the angle. We use the sin⁻¹ button on the calculator and it gives sin⁻¹(0.866) = 60°. It does not give sin⁻¹(0.866) = 120°.

How do we find all the angles whose sine is 0.866? One solution is to use a picture like one of these:

Example: Find all the values of θ between 0° and 360° for which sin(θ) = +0.6293.

Solution: Since

, this means that

.

Recall that r is defined positive. If we let r = 1, for example, then y = +0.6293. This means that the angles must be in quadrants 1 and 2.

The calculator gives sin⁻¹(0.6293) = 39.0°. The picture shows that this means that the angles are θ = 39.0° and θ = 180° − 39.0° = 141.0°

Example: Find all the values of θ between 0° and 360° for which tan(θ) = −1.24.

Solution: Since

, this means that

.

The picture shows that there are two ways that this ratio can happen.

If we let x = 1 then y = −1.24. This means that one angle is in quadrant 4.

And if we let x = −1 then y = + 1.24. This means that the other angle is in quadrant 2.

The calculator gives tan⁻¹(−1.24) = −51.1°. The picture shows that this means that the angles are θ = 128.9° and θ = 308.9° (These values are gotten by taking −51.1° and adding 180° once and once again.)

Example: Find all the values of θ between 0° and 360° for which sec(θ) = −4.21

Solution: Since

, this is exactly the same as saying that

. Since

this means that

.

If we let r = 1 then x = −0.2375. This means that the angles are in quadrants 2 and 3.

The calculator gives cos⁻¹(−0.2375) = 103.7°
The picture shows that this means that the angles are θ = 103.7° and θ = 360° − 103.7° = 256.3°.

Oblique triangles

An oblique triangle is one that doesn’t contain a right angle.

The naming convention for the angles and sides is that the angle and the side opposite it have the same letter, as shown in the picture to the right.

The sine law and cosine law (derived below) are used to solve oblique triangles

the sine law:
the cosine law:

Note the following points regarding the sine and cosine laws:

If C = 90° then the sine law reduces to , namely the definition of the sine, and the cosine law reduces to c² = a² + b², namely Pythagoras’ Theorem.
There is nothing special about angles A, B or C. We could just as well write the sine and cosine laws as and

Derivation of The Sine Law

Break the triangle into 2 right triangles by dropping the perpendicular with length h. Then:

and

Dividing the first equation by the second gives

which is another way to write the sine law.

Derivation of The Cosine Law

Again break the triangle into 2 right triangles. Applying Pythagoras’ Theorem to the two right triangles gives:

Subtracting the second one from the first gives:

This is the cosine law:

How to use the Sine and Cosine Laws

To solve an oblique triangle, a combination of at least 3 angles and sides must be given. The cases are classified SAS, ASA, ASS, etc. If a side and the angle opposite it are given then you can use the sine law. Otherwise you must use the cosine law.

If you are using the sine law to find an angle you will eventually need to evaluate a sin⁻¹. If the angle you are looking for is acute then the calculator returns the correct value. But if the angle is obtuse then the angle given by the calculator is not the correct one. You need the one in the second quadrant (which can be gotten by subtracting the calculator angle from 180°.) (Recall that an acute angle is an angle between 0° and 90°. An obtuse angle is an angle between 90° and 180°.)

Example: Solve the oblique triangle.

Solution: Since the angles add to 180°:

C = 180° − 32.5° − 49.7° = 97.8°

Now use the sine law to get sides b and c:

Solving for b gives
Solving for c gives

The unknown sides and angles are C = 97.8°, b = 321 and c = 417.

Example. Ambiguous case: Solve the oblique triangle with given information about an angle and two sides:

A = 25.3°, c = 152 and a = 95.0

Solution: Note that the given information is not enough to decide whether the triangle is ABC or ABC' shown here:

This is called the ASS ambiguity. If we use the sine law to get angle C, we get:

Notice the two values of sin⁻¹ give the two values of C in the two possible triangles.

First possibility: C = 43.1° Then B = 180° − 25.3° − 43.1° = 111.6° and b comes from the sine law:
Second possibility: C = 136.9° Then B = 180° − 25.3° − 136.9° = 17.8° and b comes from the sine law:

In summary there are two possibilities: either C = 43.1°, B = 111.6° and b = 207
or C = 136.9°, B = 17.8° and b = 68.1

Example: Solve the oblique triangle shown.

Solution: Since we don’t know any side - opposite angle pair we must begin with the cosine law:

Now find angle B using the sine law:

From the picture it is clear that B is acute and that the only possibility is that B = 30.2° And therefore A = 127.6°

In summary, c = 91.5, B = 30.2° and A = 127.6°

Vectors

Definition: A vector is a quantity that has a direction as well as a magnitude. Compare this with a scalar, which is a quantity that has only magnitude. An example of a vector is force. To accelerate an object in a certain direction you need to apply a force in that direction. An example of a scalar is temperature; temperature doesn’t point in any direction.

An arrow can be used to represent a vector. The length of the arrow represents the magnitude of the vector and the direction of the arrow is the direction of the vector.

Vector Addition: To add two vectors A and B geometrically you put the tail of B at the head of A or vice versa. The resulting vector, A + B, shown below in red, points from s (start) to f (finish). This is called parallelogram construction. The vector A + B is called the resultant of adding vectors A and B.

(This tail-to-head addition is an extension of the way that ordinary numbers are added. Click here to see the numbers 2 and 3 and their sum 5.)

The definition of the vector does not say anything about where the vector is located. Vectors are usually drawn with their tail at the origin. This is called the standard position. One advantage of standard position is that complete information about the vector is given by the coordinates of the arrowhead. Here is the same vector addition done with the vectors in standard position.

Vector Subtraction: Vector subtraction can be understood in two ways.

First, the subtraction B − A is equivalent to the addition B + (−A). (The negative of a vector A, namely −A, points in the opposite direction of A.)

The second way is to think of B − A as the difference between B and A. That is, the difference between their arrowheads. Thus it is the vector that points from the arrowhead of A to the arrowhead of B, which is shown in gray, and is then redrawn in standard position in red.

Resolving a vector into its components: Resolving a vector into its x and y components is done by drawing a dotted line from the arrowhead parallel to the y axis until it hits the x axis and another dotted line parallel to the x axis until it hits the y axis, and then drawing vectors along both axes until they touch these lines.

Coordinate Systems for Vectors: Vectors can be expressed in polar coordinates or rectangular coordinates.

In Polar Coordinates the vector’s magnitude and angle are given directly. Vector V in the picture is denoted like this:
V = 2.5 ∠ 53°
∠ is called the angle symbol. The number in front of the angle symbol is the magnitude or length of the vector. The number after the angle symbol is the direction of the vector, expressed as an angle measured counterclockwise from the positive part of the real axis.

In Rectangular Coordinates the x and y coordinates of the vector’s arrowhead are given. Now vector V in the picture is denoted like this:
V = (1.5, 2)

Resolving a vector expressed in rectangular coordinates into its x and y components is easy. For example:

Adding two vectors expressed in rectangular coordinates is also easy. You add the two x components to get the resultant’s x component and add the two y components to get the resultant’s y component. For example, if A = ( 2, 4 ) and B = ( 3, 1 ), then:

A + B = ( 2, 4 ) + ( 3, 1 ) = ( 5, 5 )

The picture shows that this works because:

A + B = ( 2, 4 ) + ( 3, 1 )
= ( 2, 0 ) + ( 0, 4 ) + ( 3, 0 ) + ( 0, 1 )
= ( 2, 0 ) + ( 3, 0 ) + ( 0, 4 ) + ( 0, 1 )
= ( 5, 0 ) + ( 0, 5 )
= ( 5, 5 )

We only describe 2-dimensional vectors here, but mathematicians use vectors in any number of dimensions.

Converting between Polar and Rectangular Coordinates

Often you must convert a vector expressed in polar coordinates to rectangular coordinates or vice versa.

Your calculator may have polar to rectangular (denoted P → R or → x y ) and rectangular to polar conversion (denoted R → P or → r θ ) built in. See your calculator manual. If not, you can do the following:

If you are given the vector r∠θ in polar, then in rectangular it is (x, y) where:
x = r cos(θ) and y = r sin(θ).
If you are given the vector (x, y) in rectangular, then in polar it is r∠θ where:
Note that you may have to add 180° if θ is not in the first quadrant.

Addition of Polar Vectors

In general there is no simple, direct way to add two polar vectors. For example it turns out that (see example below):

8 ∠ 30° + 10 ∠ −60° = 12.8 ∠ 21.3°

The only exception is if the two vectors lie along the same line. For example,

1 ∠ 50° + 2 ∠ 50° = 3 ∠ 50°

In general, to add two polar vectors you must:

convert both to rectangular
do the addition while in rectangular
convert the resultant back to polar

Example: Add the polar vectors 8 ∠ 30° and 10 ∠ −60°.

Solution:

Applications of Vectors

Statics If an object is in static equilibrium then the forces on it must sum to zero. In other words the sum of the forces to the right equals the sum of the forces to the left and the forces up equal the forces down.

Example: A weight of W = 510 N is suspended by two ropes at the angles shown. (N stands for newtons which are the units of force in the metric system.)

(a) What are the tensions T₁ and T₂ in the ropes?

(b) If the ropes have a tensile strength of 1000 newtons, what is the maximum weight that can be supported?

Solution: The forces T₁ , T₂ and W must be in balance:

Substituting 1.049 T₁ in for T₂ in the second equation gives:

Thus the tensions are T₁ = 678 N and T₂ = 711 N.

Part (b): Notice that T₂ is the bigger tension. The tensions are proportional to the weight and when W = 717 N then T₂ = 1000 N. Thus the maximum weight that can be supported is 717 N.

Example: A cart with a weight of W = 1250 newtons is on a plane inclined at an angle θ = 13°. What minimum force F must be applied to keep the cart from rolling downhill, if there is no friction?

Solution: The weight W is due to gravity and so acts vertically downward. The key is to resolve W into components parallel to and perpendicular to the inclined plane. The component of W parallel to the plane is W sin(θ). F must equal at least this much or the cart will roll downhill. Plugging in the numbers we get:

F = W sin(θ) = (1250 N) sin(13°) = 281 N

Thus the minimum force is 281 newtons.

Applications of Radian Measure

We saw above that angles can be measured in radians, with a full circle rotation being 2π or approximately 6.28 radians.

The size of an angle θ in radians equals s (the arc length subtended by the angle) divided by r (the radius of the circle).

For an angle of one full circle the arc length s becomes the circumference C. Since C = 2 π r we get

for an angle of one full circle.

Note that since s and r both have dimensions of length the radian is a unit with no dimensions. This is an important feature of radians.

Example: Express angle θ in the picture in radians.

The equilateral triangle has three angles equal to 60°. Bowing out one side into an arc produces an angle of 1 radian. We see that 1 radian is slightly less than 60°.

It is often useful to express a radian angle as a multiple of π. To do so, multiply by the symbol π and divide by the value of π, namely by 3.14159.

Example: Express 4.22 rads as a multiple of π.

The picture defines the sector, chord and segment.

The area of a circle is:

A = π r²

The area of a sector is a fraction s/C of the area of a circle, where C is the circumference. Thus:

Since C = 2 π r the area of the sector is:

Note the similarity to the formula for the area of a triangle:

Example: What is the distance between a point at latitude North 43.6° and a point due south of it on the equator? (Use the fact that the earth has a diameter of 7920 mi.)

Solution: Solving the equation

for the arc length gives s = r θ. This formula requires the radius,

, and the angle θ expressed in radians:

Thus the arc length is:

s = r θ = (3960 mi)(0.7610 rads) = 3010 mi,

to 3 sig figs. Notice that the units of s are the same as those of r (miles), since the units of θ (radians) are actually unitless.

Example: What distance does the rack gear move if the pinion gear rotates through an angle of 300°?

Solution: Because the rack and pinion gear mesh, the distance that the rack moves is equal to the arc length s subtended by the angle 300° on the pinion gear, which is given by s = r θ. The radius of the pinion gear is r = 11.25 mm and the angle θ through which it turns expressed in radians is:

Thus:

s = r θ = (11.25 mm)(5.236 rads) = 58.9 mm.

Notice that the units of s are the same as those of r, namely millimeters.

Uniform circular motion

Suppose that an object is moving at a constant velocity v in a circle of radius r and that it moves from A to B, covering a distance s, in time t. Then

since rate = distance traveled / time required.

As this happens angle θ increases. The rate at which θ increases is angle swept out / time required, namely θ / t. This quantity is called the angular velocity ω (the Greek letter omega). Thus:

Angular velocity ω can have any units of angle divided by time but the most useful units are radians / sec.

Example: An object rotates at an angular velocity of ω = 48 π rads / sec. Through what angle does it rotate in 0.25 sec?

Solution: Solving

for θ gives:

θ = ω t = (48 π rads / sec)(0.25 sec) = 12 π rads.

(Note that 48 π rads / sec = 24 cycles / sec and that 12 π rads = 6 cycles)

Example: What is the speed in miles per hour of a point on the equator and of a point in Vancouver due to the rotation of the earth? Use the fact that the radius of the earth is 3960 miles, that it rotates once every 24 hours and that Vancouver is at latitude 49.2° north.

Solution: The first picture on the right looks down on the north pole and shows how far a point in Vancouver and a point on the equator move in 1 hr. Both have the same angular velocity ω, namely:

but they have different so-called linear velocities v because they have different radii from the earth’s axis.

To find v use the equation v = r ω, where either r = r_E for the equator or r = r_V for Vancouver. We were given r_E = 3960 mi. To find r_V use the side-view picture of the Earth to the right. Trigonometry gives:

Thus the velocities are:

Thus the speed of a point on the equator is 1037 mi / hr and of a point in Vancouver is 677 mi / hr due to the rotation of the earth. (This extra speed is the reason launch sites for orbital rockets are chosen near the equator.)

Graphs of the sine, cosine and tangent functions

Recall that

, and ,

where r is the length of a vector and x and y are its x and y components. Let r = 1. Then sin(θ) = y and cos(θ) = x. In other words sin(θ) is just the y component and cos(θ) is the x component of a vector of length 1.

The sine function: Let the angle θ vary from 0 to 2π radians in 8 steps. The vector of length 1 rotates to positions a through h in the circle below (indicated by the red arrows). The vertical arrows (colored blue) are the y components of the vector in the various positions. The graph on the right is a plot of the y components vs θ, hence a graph of sin(θ) as a function of θ.

The circle produced by the vector of length 1 as it rotates is called the unit circle and we say that the sine function is generated by the y component of this rotating vector.
The cosine function: Again let the angle θ vary from 0 to 2π radians in 8 steps. Again the vector of length 1 rotates to positions a through h in the circle below (red arrows). The horizontal arrows (blue) are the x components of the vector of length 1 in the various positions. The graph on the right is a plot of the x component vs θ, hence a graph of cos(θ) as a function of θ.

We say that the cosine function is generated by the x component of a rotating vector of length 1.
The tangent function: This graph was drawn by calculating the ratio y/x for various values of θ.

Note that tan(π/2) = ± ∞ and tan(3π/2) = ± ∞ since x equals 0 (causing a division by 0) at those angles.

Note the following features of all three curves:

The θ axis for all three of the above graphs could use degrees. Just replace π radians with 180°, etc.
sin(θ), cos(θ) and tan(θ) are periodic functions - this means that they repeat their patterns for θ < 0 and θ > 2π radians. The functions sin(θ) and cos(θ) have period 2π radians since they repeat every 2π radians. The function tan(θ) has period π radians since it repeats every π radians.

The Sinusoidal Function y = A sin (ω t + φ)

Waves on the water, vibrations on a string, sound, light, animal population cycles, economic cycles and alternating current can all be described by the sinusoidal function or sinusoidal waveform:

y = A sin (ω t + φ)

where y represents the quantity of interest and t represents time. Because y is proportional to the sine function, it has the characteristic sine wave shape. By changing A, ω and φ we can change the height or width of the wave or shift it left or right to suit the application. Let’s look at each of these parameters, A, ω and φ, in turn:

In the function y = A sin (ω t + φ), the parameter A is called the amplitude of the wave. The sin function itself ranges from −1 to +1 without units. Multiplying the sin wave by A causes y to stretch vertically from −A to +A with the units of A.
In the function y = A sin (ω t + φ), the parameter ω is called the angular velocity of the wave.

(Note: This is the greek letter “omega”, not the latin letter w.)

ω refers to the rate of rotation of the rotating vector which generates the wave. It has units of radians/sec.

The figure compares ω = 1, 2, and 3 rad/sec. The wave sin(3t) oscillates 3 times as fast as the wave sin(t).

An algebraic way to see why ω = 3 compresses the wave horizontally by a factor of 3 is to note that in sin(3t), t needs to be only ⅓ as big as it needs to be in sin(t) to yield the same value in the brackets. When the value in the brackets equals 2 π radians, the sin function completes one cycle.

Related to angular velocity are the frequency f and the period T.
- The frequency is the number of cycles of the wave that occur per second. Because there are 2π radians in 1 cycle, angular velocity and frequency are related by the formula:
  ω = 2 π f.
  The units of f are cycles/sec or Hertz.
- The period is the time in seconds required to complete one cycle. It is the reciprocal of the frequency:
  The units of T are seconds. Because ω = 2 π f this can also be written as:
In the function y = A sin (ω t + φ), the parameter φ is called the phase shift of the wave.

(Note: This is the greek letter “phi”, which is pronounced “fi” and which rhymes with “cry”.)

If φ is positive then the wave is shifted to the left. If φ is negative then the wave is shifted to the right. φ may be given in radians or degrees.
To understand how much the wave is shifted note that φ measures the shift in phase, 2π being a shift by an entire cycle, π a shift by half a cycle, etc.

The shift in time depends on both ω and φ. The example to the right shows that the starting point of the wave on the t axis is not simply the value of the phase angle φ, which many beginning students believe to be true.

To understand how ω and φ act together to cause the shift in time, think of sin ( ) as a function machine. An angle goes into the brackets ( ) and a point on the sin curve comes out. When 0 goes into the brackets the beginning of a cycle comes out and when 2π radians goes into the brackets the end of the cycle comes out.

For the red curve, y = sin(4t+1), the cycle begins when there is 0 inside the brackets, namely when t = −¼.

For the black curve, y = sin(4t), the cycle ends when there is 2π inside the brackets, namely when 4t = 2π or (solving for t) when t = π/2.

If we imagine that the sinusoidal curve is generated by a rotating vector then φ can be though of as the angle at which the rotating vector points initially, that is, at time t = 0.

Here are some special phase shifts:
φ = 2π rads or 360°. The wave is shifted left by one complete cycle, so the wave appears unchanged.

φ = π rads or 180°. The wave is shifted left by half a cycle, so the wave appears to be turned upside-down.

φ = π/2 rads or 90°. The wave is shifted left by one quarter of a cycle, so that, for example, a sin wave becomes a cosine wave.

Sinusoidal Waveform Plus DC Component

In some applications the sinusoidal function discussed in the previous section must be shifted up or down by some amount. To do this we add a constant y_DC to the right side of the formula, like this:

y = A sin (ω t + φ) + y_DC

The graph is shifted upward by the amount y_DC (or downward if y_DC is negative).
y_DC is the average value (average height) about which y oscillates.
In electricity the term y_DC is called the DC (direct current) component of y.
The term A sin (ω t + φ) is called the AC (alternating current) component.

The graph compares DC components of 0, 1 and 2.

Finding the Equation of a Sinusoidal Curve

To find the equation of a sinusoidal curve follow these steps:

Begin by writing down the “template” equation y = A sin(ω t + φ) (assuming that the label on the vertical axis is y and the label on the horizontal axis is t). Then use the next three steps to get A, ω and φ.
Get the amplitude A by reading the peak value of the curve from the vertical axis. Get the units of A from the label on the vertical axis.
Read t_B and t_E , the values of the time t at the beginning and the end of the cycle. The period T is the difference t_E − t_B. Then calculate angular velocity ω from the equation .
Get phase shift φ by using the fact that the quantity in brackets, ω t + φ, must equal zero when t = t_B, at the beginning of a cycle.

Example: Find the equation of the sinusoidal wave shown to the right. This represents an alternating voltage. v is the voltage in volts and t is the time in seconds.

Solution:

The vertical axis is labeled v. Thus the template equation is:
v = A sin(ω t + φ)
The amplitude is A = 7 volts.
The cycle begins at t_B = −1 sec and ends at t_E = 5 sec. Thus the period is T = t_E − t_B = 6 sec and the angular velocity is:
The equation at this point is . To find the only remaining unknown parameter, namely the phase angle φ, use the fact that the brackets must contain 0 at time t = −1:
Thus the equation is .

Note: Another way to find the phase angle φ is to notice that the wave is shifted left by 1/6 cycle. Since there are 360° or 2π radians in a cycle, this means that .

Drawing the Graph of a Sinusoidal Function

To draw the graph of a sinusoidal function follow these steps:

Begin by labeling the vertical axis “y” and the horizontal axis “t”, assuming that the equation is y = A sin(ω t + φ).
Get t_B, the time of the beginning of the cycle, by using the fact that the quantity in brackets, ω t + φ, must equal 0 when t = t_B.
Get t_E, the time of the end of the cycle, by using the fact that the quantity in brackets, ω t + φ, must equal 2π when t = t_E.
Place a box stretching from t_B to t_E on the horizontal axis and from −A to +A on the vertical axis. Label the edges of the box.
Draw one cycle of the sin curve to fill the box.

Example: Graph the function

, where v is in volts and t is in seconds.

Solution: Since the first term inside the brackets, namely

, is in radians we must change the second term, 60°, to π/3 radians to make the units agree. Thus we will graph

Place a box stretching from −3 to +15 in the horizontal direction and from −12 to +12 in the vertical direction. Label the edges of the box.

Break the box into four sub-boxes and use them to outline the sinusoidal curve (the red dots).

Draw the sinusoidal curve by smoothly connecting the dots.

14.2 - Trigonometric Identities

We begin this section by stating about 20 basic trigonometric identites. You can refer to books such as the “Handbook of Mathematical Functions”, by Abramowitz and Stegun for many more. To understand them we will organize them into 9 groups and discuss each group.

Then we finish this section with 7 examples.

Identities and equations compared: An identity is a statement that is always true, whereas an equation is only true under certain conditions. For example

3x + 2x = 5x

is an identity that is always true, no matter what the value of x, whereas

3x = 15

is an equation (or more precisely, a conditional equation) that is only true if x = 5.

A Trigonometric identity is an identity that contains the trigonometric functions sin, cos, tan, cot, sec or csc. Trigonometric identities can be used to:

Simplify trigonometric expressions.
Solve trigonometric equations.
Prove that one trigonometric expression is equivalent to another, so that we can replace the first expression by the second expression. The second expression can give us new insights into some application that the first one doesn’t show.

The reciprocal identities

1a.

1b.

1c.

This group of identities states that csc and sin are reciprocals, that sec and cos are reciprocals, and that cot and tan are reciprocals. They follow directly from the definitions of the trigonometric functions. (Click here to see them again.) Many mathematicians consider them to be merely basic simplifications to be used to get rid of csc, sec and cot in favor of sin, cos and tan.

The negative angle identities

2a.

cos (−θ ) = cos ( θ )

2b.

sin (−θ ) = − sin ( θ )

2c. tan (−θ ) = − tan ( θ )

These identities describe the left-right symmetry of the cos, sin and tan curves. Many mathematicians consider these identities to be just basic simplifications to be used to get rid of negative angles inside a cos, sin or tan.

Graphically, identity (2a) says that the height of the cos curve for a negative angle is the same as the height of the cos curve for the corresponding positive angle. Any curve having this property is said to have even symmetry.

Identity (2b) says that the height of the sin curve for a negative angle is the negative of the height for the corresponding positive angle. Any curve having this property is said to have odd symmetry.

Identity (2c) says that the tan curve also has odd symmetry.

The “left shift by 90° ” identities

3a.

3b.

3c.

The graph on the left explains identity (3a). It shows that shifting a sin curve to the left by 90° (or by π/2 radians or ¼ cycle) produces a cos curve. The graph on the right explains identity (3b). It shows that shifting a cos curve to the left by 90° (or by π/2 radians or ¼ cycle) produces an upside-down sin curve.

Identity (3c) can be proven by using identities (3a) and (3b) and a bit of algebra.

These identities are also considered to be basic simplifications and are used to get rid of the 90° shift. By using them repeatedly they can be used to get rid of shifts of 180°, 270°, 360°, etc.

Pythagoras’ theorem

4. sin²(θ) + cos²(θ) = 1

This is the trigonometric form of Pythagoras’ theorem.

Note that the notation sin²(θ) means (sin(θ))², and the same goes for cos. In other words you must take the sin first and then square. The reason for not placing the exponent at the end is to make it clear that it is not angle θ that is squared, rather it is the sin of θ that is squared.

Identity (4) is considered to be a basic simplification. We can also rearrange it as

1 − sin²(θ) = cos²(θ)

or as

1 − cos²(θ) = sin²(θ)

and these are also considered to be basic simplifications when they are used to replace two terms on the left with the one term on the right.

To prove identity (4) simply construct a right triangle with hypotenuse 1 and angle θ. Then the base and altitude are given by cos(θ) and sin(θ), and the original form of Pythagoras’ theorem, namely a² + b² = c², turns into identity (4).

The graph to the right illustrates Pythagoras’ theorem by showing how the height of the sin²(θ) curve (red) and the height of the cos²(θ) curve (blue) add to always equal 1 (black dashed line).

We can also write Pythagoras’ theorem in two other forms. If we divide both sides of identity (4) by cos²(θ) and simplify then it reads

1 + tan²(θ) = sec²(θ)

Alternatively, if we divide both sides of identity (4) by sin²(θ) and simplify then it reads

1 + cot²(θ) = csc²(θ)

The identities that we have discussed so far (identities 1 to 4) are really just simplifications that are always applied from left to right (i.e. by replacing the left-hand-side expression anywhere it appears by the right-hand-side expression.)

The rest of the identities (identities 5 to 9, discussed next) are stated so that the left-hand-side is the combined form of some expression and the right-hand-side is the broken apart form. They can be applied in either direction depending on what we want to accomplish:

From left to right: The purpose is to break apart a combined angle into two separate angles, or break one trigonometric function into two other ones, or break up any other trigonometric functions into sines and cosines. Going in this direction is easy. This technique is used to prove new trigonometric identities.
From right to left: The purpose is to combine two separate angles into one combined angle or to replace two terms or functions with one. Going in this direction is usually hard. This technique is used to create more compact expressions.

The tangent identity

5.

This identity follows from the definitions of sin, cos and tan. To derive it simply divide the ratio for sin by the ratio for cos and simplify. The result is the ratio for tan:

The sum of angles identities

6a.

sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)

6b.

6c.

These identities describe how to break apart the trigonometric function of a sum or difference of angles α and β into the trigonometric functions of the separate angles α and β.

These are actually 6 identities, 3 come from using the upper signs and 3 come from using the lower signs. For example identity (6c) with the lower signs reads:

A common mistake is to ignore these identities and believe that, for example

sin(α + β) = sin(α)+ sin(β) WRONG!

This is not correct! It could only be correct if the sin curve was a straight line through the origin, instead of a wavy curve.

Here is an example that uses identity (6a) to describe what happens when a sin wave is shifted left by 30°.

We see that it becomes part sin wave and part cos wave. It is interesting to compare this result with identity (3a), which shows that shifting a sin wave left by 90° turns it completely into a cos wave.

It is possible (but more difficult) to use identity (6a) in the opposite direction and show how the sum of a sin and a cos can be expressed as a sin with a phase shift. Click here to see an example of that.

We will now prove that sin(α + β) = sin(α)·cos(β) + cos(α)·sin(β).

Proof: The picture shows a big yellow oblique triangle with two small right triangles inside it. Apply the sine law to the big triangle.

Use the fact that sin(γ) = h/c and then multiply (a + b) across to the other side.

Expand the right-hand-side and then use the two small triangles in the picture to replace each fraction by the appropriate trigonometric function.

We are done.

Identity (6a) with the upper sign, namely:

sin(α + β) = sin(α)·cos(β) + cos(α)·sin(β)

which we just proved, is probably the most important identity of the group. Every other identity in the group (and in fact almost any trigonometric identity) can be derived from it. For example:

The identity for sin(α − β) can be derived by writing it as sin(α + (−β)), applying identity (6a) and then using the negative angle identities, (2a) and (2b), to clean up.
The identity for cos(α + β) can be derived by using identity (3a) to write it as a sin with a phase shift:
sin((α + 90°) + β),
then applying identity (6a) and then using (3a) again to clean up.
The identity for tan(α + β) can be derived by dividing identity (6a) by (6b).

The double angle identities

7a.

sin(2α) = 2·sin(α)·cos(α)

7b.

cos(2α) = cos²(α) − sin²(α)

= 1 − 2 sin²(α)

= 2 cos²(α) − 1

7c.

These identities are special cases of identities (6a), (6b) and (6c) with the upper signs, which read:

Let β equal α and identities (7a), (7b), (7c) immediately pop out. Identity (7b) is written in 3 different forms. The second and third forms result from using Pythagoras’ theorem on the first form. They are the preferred forms because they only involve sin or only cos, and not both.

These identities are used to convert a trigonometric function of twice an angle into a trigonometric function of the angle itself.

The half angle identities

If we solve the second form of identity (7b) for sin²(α) and the third form for cos²(α) then we get these two identities:

If we now change the name of the angle α to be A/2 then these become the so-called half angle identities:

8a.

8b.

These identities are used to convert a trigonometric function of half an angle into a trigonometric function of the angle itself.

The product identities

If we add or subtract identities (6a) and (6b) in various combinations then we get the so-called product identites:

9a.

sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]

9b.

sin(α)·cos(β) = ½[sin(α + β) + sin(α − β)]

9c. cos(α)·cos(β) = ½[cos(α + β) + cos(α − β)]

These are useful to change products of trigonometric functions into sums of trigonometric functions or vice versa.

Examples

We finish this section with seven examples. The first three examples show how identites 1 to 9 can be used to prove new trigonometric identities.

The last four examples show how converting a trigonometric expression to another form leads to new insights that were not previously evident. The examples come from electrical technology where AC (alternating current) has the form of a sin wave.

Trigonometric identities are also used to help solve trigonometric equations. That topic is covered in the next section.

How to prove a trigonometric identity:

Proving an identity is different than solving an equation. Even though the identity may contain an = sign you must not transpose a quantity from one side to the other because doing so changes the value of both sides. Instead you must use trigonometric identities to modify the left side or the right side or both sides until they are identical.
The first step is to use the reciprocal identities and the tangent identity to replace tan, cot, sec, csc wherever they occur by sin and cos.
These identities produce fractions. Combine (add) the fractions.
Use the sum of angles identities or double angle identities to break apart any sums of angles or to replace double angles.
If you end up with a fraction on one side of the identity but not the other then multiply the non-fraction side by a UFOO to convert into a fraction. A UFOO is a fraction that equals 1 because it has equal numerator and denominator. Example 3 requires one.
When the two sides are identical the identity is proven.

Example 1. Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Use the tangent identity to replace the tan function with sin/cos on the LHS (left-hand-side).
Do the same on the right-hand-side. Also since the tan and cot functions are reciprocals replace the cot function with cos/sin.
The right side is now a compound fraction (a fraction that contains more fractions). Combine the fractions in the numerator and the fractions in the denominator.
Use the invert and multiply rule to divide the fractions and simplify. The two sides are now identical so the identity is proven.

Example 2. There are two new features in this example:

There are several different angles involved.
We need to compare the two sides for guidance on what to do.

Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Use the reciprocal identities to replace the sec and csc functions with cos and sin.
Comparing the two sides we notice that the angles on the RHS are 4x and x. We can have the same angles on the LHS if we replace angle 5x by the sum of angles 4x + x.
Apply the sum of angles identity to sin(4x + x) in the numerator.
Break the single fraction into two fractions. The two sides are now identical so the identity is proven.

Example 3. The new feature here is that the right side is a fraction and the left side is not. We will have to make it a fraction.

Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Since the angle on the RHS is double the angle on the LHS, replace angle 2α on the RHS using the double angle identity for sin.
The RHS has a denominator sin(α) + cos(α). To get the same denominator on the LHS we apply a UFOO to it; that is, we multiply and divide the LHS by the same quantity, namely sin(α) + cos(α).
Expand the numerator.
Use Pythagoras’ theorem. The two sides are now identical so the identity is proven.

Example 4. Adding a sin wave and a cos wave of the same frequency produces another wave of the same frequency

(a) Prove the identity

where on the right-hand-side C and φ are given by

(b) Use the identity to express the sum of the two sinusoidal waveforms,

y₁ = 3 sin(θ) and y₂ = 4 cos(θ),

as a single sin wave with a phase shift. Then plot all three waveforms in a graph.

Solution part (a): Because it is easier to break angles apart than it is to combine them we will work on the right-hand-side of this identity and convert it into the left-hand-side.

First notice that the pair of equations for C and φ describe the conversion of a vector from rectangular to polar coordinates. Referring to the figure to the right, the vector is (A, B) in rectangular coordinates and C∠φ in polar coordinates. For use in step 3 of the proof notice that the diagram also shows that

A = C cos(φ) and B = C sin(φ).

Here are the steps in the proof:

Details of the steps:

Use the sum of angles identity for sin to break apart angles θ and φ.
Expand.
Use the fact (mentioned above) that A = C cos(φ) and B = C sin(φ). The two sides are now identical so the identity is proven.

Solution part (b): We will now use the identity proven in part (a) to express the sum y = 3 sin(θ) + 4 cos(θ) as a single sin wave with a phase shift. Let A = 3 and B = 4 in the identity. This gives

so the sum of the waveforms y₁ = 3 sin(θ) and y₂ = 4 cos(θ) is

y = 5 sin(θ + 53.13°)

The graph to the right shows the separate waveforms in green and blue and the resultant waveform in red. (Click here to see how the red waveform can be graphed.)
Notes:

The significance of this example is that it shows that adding two sinusoidal waveforms of the same frequency results in another sinusoidal waveform of the same frequency but with a phase shift. Compare this result with the result found in the next example.
We could add any number of out-of-phase sin or cos waves of the same frequency to produce a single sin with a phase shift by following these steps:
1. Break each waveform into a sin and a cos by applying the identity from right to left (in the Algebra Coach click the Break apart trig button),
2. Add all the sines and add all the cosines,
3. Combine the remaining sin and cos by applying the identity from left to right (in the Algebra Coach click the Combine trig button).
One of the most important situations where one is concerned with out-of-phase waves is in an alternating current (AC) network. An example is the continent-wide electrical energy grid. Every generator that is connected to the grid must produce voltage that is in phase with all the other generators on the grid. Otherwise the voltages will tend to cancel.

Example 5. Waves with slightly different frequencies produce beats

(a) Prove the identity

(b) The left side of the identity can be considered to be the sum of two sinusoidal waveforms

with slightly different frequencies. Plot waveforms y₁ and y₂ on the same graph and compare them.

(c) Now plot their sum in another graph. This task would be very difficult were it not for the identity proven in part (a). The identity can be used to convert the sum of the waveforms into a single term. Show that this single term can be interpreted as a sin wave with a slowly varying amplitude. Use the single term side of the identity to make the plot.

Solution part (a): We are essentially proving the product identity (9b). The key idea in the proof is noticing that the angles 21t and 19t can be thought of a 20t ± 1t and then breaking the angles apart with a sum of angles identity.

Here are the steps in the proof:

Details of the steps:

The RHS has angles t and 20t. To get the same angles on the LHS write 21t as 20t + 1t and 19t as 20t − 1t.
Apply the sum of angles identity for sin to both terms. That identity says
sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)
In this case let α = 20t and β = 1t.
Simplify.

Solution part (b):Here is the graph of the two sinusoidal waveforms:

Notice that the waveform with angular velocity ω = 21 (the blue one) oscillates slightly faster than the waveform with ω = 19 (the green one). The blue one has 10.5 cycles in the time shown while the green one has 9.5 cycles; one cycle less. The result is that at certain times, for example at time t = 0 and t = 3.14, the waves are in phase (that is, peaking together), and at other times, for example at time t = 1.57, the waves are 180° out of phase, (that is, one is at a crest when the other is in a trough).

Solution part (c): It would be very difficult to add the waveforms graphically. All we could say for sure is that when they are in phase they add to give an amplitude of 1 and when they are 180° out of phase they cancel to give an amplitude of 0.

What is more informative is to make a plot of the other side of the identity, namely of

y = cos(t) · sin(20t).

This graph is plotted below in red. We see that the factor cos(t) in this product plays the role of a slowly varying amplitude for the rapidly varying oscillation sin(20t). That is, we have a sinusoidal wave sin(20t), but that at certain times it has a small amplitude and at other times it has a large amplitude.

To make this graph we first draw one cycle of the curve y = cos(t) (and its negative), both shown in gray. This is called the envelope and is used as a guide. Then we draw the curve y = sin(20t) inside the envelope. We draw 20 oscillations and stretch and squeeze them to fit inside the envelope.

Notes:

In acoustics the periodic constructive and destructive interference is known as beats. Beats can be quite annoying if they are due to, say, the twin engines of an airplane running at slightly different speeds. But they can be useful, for example, for checking if two guitar strings are in tune.
In electrical energy generation every generator that is connected to the electrical grid must produce voltage at the same frequency as all the other generators on the grid. Otherwise the voltages will tend to periodically cancel.

Example 6. Electric power dissipated by a resistor

There are two types of electric current in common use: direct current (DC) and alternating current (AC).

Direct current is current that is constant in time. It is typically produced by a battery. For this example we will consider a direct current of 0.7071 amperes:
i = 0.7071 amperes
Alternating current is current that flows back and forth in an electric circuit and is described by a sinusoidal function of time. It is typically produced by an electric generator. For this example we will use an alternating current with an amplitude of 1.0 amperes:
i = 1.0 sin(t) amperes

The graph on the left shows the direct current and the graph on the right shows the alternating current:

When an electric current flows through a resistor (such as a toaster or light bulb) it dissipates heat or light energy. The power (i.e. the rate at which the energy is dissipated) is given by the formula

p = i² R,

where p is the power in watts, i is the current in amperes, and R is the resistance of the resistor in ohms. If the current changes with time then the power must be calculated at each instant in time with this formula. We say that the formula gives the instantaneous power.

Problem: Show that a direct current of 0.7071 amperes and an alternating current with amplitude 1.0 amperes both dissipate the same amount of power in a 1000 Ω resistor, namely 500 W, when averaged over time.

Solution: The basic idea is shown in the graphs below where we have plotted the quantity i² corresponding to the graphs of i shown above:

On the left we see that squaring a constant current i = 0.7071 gives a constant i² = 0.5. When this i² is multiplied by R = 1000 Ω it gives a constant power p = 500 watts.

On the right we see that squaring a sinusoidal current i = 1.0 sin(t) gives another wave shape, i² = 1.0 sin²(t), with these features:

Both when i = + 1 and when i = −1 then i² = 1, and whenever i = 0 then i² = 0 also.
The wave shape looks like a cosine curve turned upside-down and raised to an average height of 0.5.

When this wave shape i² is multiplied by R = 1000 Ω it gives exactly the same upside-down cosine wave shape for the instantaneous power. But when the average height of i², namely 0.5, is multiplied by R = 1000 Ω then this gives an average power p = 500 watts, as was to be shown.

Here is the proof that the average height of the i² curve is ½.

Start with: i = 1.0 sin(t) and square it: i² = 1.0 sin²(t).
Use half-angle identity (8a), modified to read .
Interpret the RHS of the identity. It is a sinusoidal waveform plus a DC component.
The last term is the DC component. It causes i² to have an average value of ½.

Notes:

This example showed that an alternating current with an amplitude of 1 ampere is as effective as a direct current of 0.7071 amperes in powering a resistor such as a lightbulb or toaster. This result can be generalized. The effective value (also called the RMS value) of any AC current is found by multiplying its amplitude by 0.7071.
In cases where the frequency of the alternating current is high the average value of the power is more relevant than the instantaneous power. An example is when a 60 Hz AC current goes through a kitchen toaster. The toaster element doesn’t have time to heat up and cool down as the current changes and just stays at one constant temperature. Another example is the same 60 Hz current flowing through a fluorescent light. In this case the light actually goes on and off as the current alternates, but the human eye can’t respond fast enough to see it. In both situations we are more interested in the average power than the instantaneous power.

Example 7. Electric power generated by a generator

In the electric circuit shown to the right the generator produces energy and the load consumes that energy. The power consumed by the load equals the power produced by the generator. This is the law of conservation of energy.

If the load is a resistor then the previous example shows how to calculate the power consumed by the resistor. In this example we make two changes:

We study the power that is produced by the generator.
We let the load contain an inductor or capacitor. An inductor is any device that creates a magnetic field when a current flows through it. An example is an electric motor. A capacitor is any device that creates an electric field when electric charge is stored in it. In reality every circuit has some capacitance and some inductance either incidentally or by design.

The main consequence of the load having inductance or capacitance is that the alternating current and voltage produced by the generator can be out-of-phase by up to 90°.

For example the picture shows a voltage out of phase with the current, with the voltage leading the current by 60°.

The power p delivered by the generator at any instant in time is given by the formula

p = i v,

where i is the current in amperes at that instant and v is the voltage of the generator in volts. Suppose that the generator produces the AC current

i = I₀ sin (ωt)

and that the load is such that the voltage is out of phase with the current by angle φ. (The phase shift φ could be positive or negative.)

v = V₀ sin (ωt+φ)

Since i and v change with time, so does p. But as in the previous example, rather than the instantaneous power, we are interested in the average power.

Problem: Find the average power delivered by the generator.

Solution: We follow the same steps as in the previous example. Substituting the given i and v into the power formula gives

We can change the product of trigonometric functions into a sum of trigonometric functions by using identity (9a) which says that sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]. If we make the replacements α→ωt+φ and β→ωt then identity (9a) reads

sin(ωt+φ)·sin(ωt) = ½[cos(φ) − cos(2ωt+φ)]

Substituting this into the formula for p gives

or expanding,

Because φ is a constant (remember it is the phase shift between v and i) the instantaneous power p is again a sinusoidal waveform with a DC component. The DC component gives the average power,

Some special loads.

Perfect resistors: i and v are in phase (φ = 0) so the above formula gives . Since V₀ = I₀ R this can also be written as
This agrees with the result we derived in the previous example, which was that p_avg = ( I_effective )² R where I_effective = 0.7071 I₀
Perfect inductors: φ = +90^o and perfect capacitors: φ = −90^o. In both cases the above formula gives
p_avg = 0.
How can this be? The explanation is that inductors and capacitors absorb energy for one half of each cycle as they build up their magnetic and electric fields and then release the energy back to the generator in the other half of the cycle when the fields collapse.

The following graph shows the instantaneous power (solid red curve) and the average power (dashed red line) that result from out-of-phase current i (blue curve) and voltage v (gray curve).

Note that at points a and b the instantaneous power is actually negative. The reason is that at those moments i and v have opposite signs and the explanation, as mentioned above, is that power is going back to the generator. Also note that if φ = 0 then the red curve reduces to the i² curve in the previous example.

14.3 - Trigonometric Equations

Before reading this section you may find it helpful to review the following topics:

Basic procedures for solving equations.
Trigonometric functions of any angle.
Trigonometric identities.

A trigonometric equation is one in which the unknown to be solved for is an angle (call it θ) and that angle is in the argument of a trigonometric function such as sin, cos or tan. A trigonometric equation always has an infinite number of solutions, but it is customary to list only those angles between 0° and 360°.

Only certain types of trigonometric equations can be solved using algebra. We will study the following types. (Hover over the links to see examples.)

• θ occurs only once.
Learn about this type sin(θ−40°) = −3

• θ occurs several times but always as the argument of the same trigonometric function.
Learn about this type 6 sin²(θ) − sin(θ) − 2 = 0

• the equation has different trigonometric functions but they all have the same argument, θ.
Learn about this type 2 cos(θ) + tan(θ) = 4 sec(θ)

• the equation contains trigonometric functions with different arguments.
Learn about this type 4 sin(θ) + 3 cos(θ + 40°) = 2
or
3 sin(θ) + 5 sin(2 θ) = 1

Trigonometric equations in which θ occurs only once

If the unknown angle appears only once, then follow the same procedure as for any other equation where the unknown occurs just once, and invert the operations that were applied to it in the reverse order in which they were applied.

Example: Solve the equation sin(θ) = ½ for all angles θ between 0° and 360°.

Note: You can use the CAST method to derive the second angle, 150°.

Example: Solve the equation 5 sin(θ − 40°) − 2 = 0 for all angles θ between 0° and 360°.

Note: You can use the CAST method in the second-to-last step to derive the “other” angle, 156.4°. Make sure that you add the 40° last.

Trigonometric equations in which θ occurs several times but always as the argument of the same trigonometric function

If the unknown angle appears more than once, then do the following:

Transpose everything to the left side of the equation.
Factor the left side of the equation so the equation is now of the form a · b = 0.
Use the fact that if a·b = 0 then either a = 0 or b = 0, so the equation is replaced by a pair of simpler equations a = 0 and b = 0, each of which must now be solved.

A useful trick is to use an alias. Suppose that sin (θ) appears several times in the equation. Then replace all occurrences of sin (θ) with, say Q. This may turn the equation into a linear or quadratic or some other simpler type of equation in the variable Q. Solve this equation first for Q and then substitute back sin (θ) for Q.

Example: Solve the equation 2 sin²(θ) = sin(θ) for all angles θ between 0° and 360°.

Replace this equation by two simpler equations which are solved separately:

Note: You could use an alias to solve this equation. If you replace all occurrences of sin (θ) with Q then this becomes a quadratic equation in the variable Q:

2 Q² = Q.

Solve this equation first for Q:

Q (2 Q − 1) = 0 → Q = 0 and Q = ½.

Then substitute back sin (θ) for Q:

sin(θ) = 0 and sin(θ) = ½.

Now solve these quations for θ as above.

Example: Solve the equation 2 sin²(3 θ) = sin(3 θ) for all angles θ between 0° and 360°. Note that this problem is almost identical to the previous one – the only difference is that the argument of the sin function is 3 θ rather than θ.

Replace this equation by two simpler equations which are solved separately:

Note: You can use the CAST method to derive the angles 180°, 360°, and 150°. The other angles are gotten by going around the unit circle multiple times.

Trigonometric equations having different trigonometric functions but all with the same argument

These strategies may help:

Replace any other trigonometric functions with sines and cosines.
Often you arrive at the equation a sin(θ) = b cos(θ). Then you can divide both sides by a cos(θ) then according to the tangent identity this becomes tan(θ) = b/a.
Otherwise try transposing everything to one side and factoring so that each factor contains a single trigonometric function.

Example: Solve the equation 2 csc(θ) − cot(θ) = tan(θ) for all angles θ between 0° and 360°.

Equations containing trigonometric functions with different arguments

If the equation contains trigonometric functions with different arguments (angles) then use the appropriate trigonometric identity to express everything in terms of a single angle. (For example if one angle is twice another angle then use a double angle identity to get rid of the double angle. Or if one angle is another angle plus some difference then use a sum of angles identity to get rid of the first angle.) Then proceed as before.

Example: The two vectors V₁ and V₂ point in directions that are 65° apart. Find the value of the angle θ that causes the vertical components of the two vectors (the dotted lines) to be equal in magnitude.

Solution: The length of the vertical component of vector V₂ is 5 cos(θ). The length of the vertical component of vector V₁ is 3 cos(115° − θ). Using the fact that they are supposed to be equal gives us our starting equation,

5 cos(θ) = 3 cos(115° − θ).

Use trigonometric identity (6b) with the lower sign to separate angle 115° from angle θ on the RHS of the equation.

5 cos(θ) = 3 cos(115°) · cos(θ) + 3 sin(115°) · sin(θ)

Evaluate the sin and cos of 115°. Notice that now the only angle remaining is θ.

5 cos(θ) = −1.268 cos(θ) + 2.719 sin(θ)

Collect terms.

6.268 cos(θ) = 2.719 sin(θ)

Divide both sides by 2.719 cos(θ) and use the tangent identity to turn sin/cos into tan.

tan(θ) = 2.305

θ = tan⁻¹(2.305) = 66.5° or 246.5°

From the diagram above we see that the angle we want is θ = 66.5°. The other solution corresponds to having the vectors rotated by 180° and being in quadrants 3 and 4.

Example: The following type of problem arises often in electrical technology. Given two following two sinusoidal waveforms:

y₁ = 4 sin(θ) and y₂ = 3 cos(θ + 40°),

find all angles θ between 0 and 2π radians for which their sum equals 2.

Solution: We must solve the trigonometric equation

4 sin(θ) + 3 cos(θ + 40°) = 2.

We can use the sum of angles identity, (6b) to break apart the two angles, θ and 40°, in the cos. It states that

cos(θ + 40°) = cos(θ) · cos(40°) − sin(θ) · sin(40°),

or evaluating the sin and cos of 40°,

cos(θ + 40°) = −0.643 sin(θ) + 0.766 cos(θ).

When we substitute this into the equation we get

4 sin(θ) + 3 ( −0.643 sin(θ) + 0.766 cos(θ) ) = 2,

or expanding and simplifying,

2.072 sin(θ) + 2.298 cos(θ) = 2.

This looks similar to the previous example but we have a “2” on the right-hand-side so we must use a different method to solve it. In an example in the page on trigonometric identities we proved the following identity, which states that a sin waveform and cos waveform can always be added to give a sin waveform with a phase shift:

, where

We can use this identity to replace the two occurrences of θ in our equation by one. In our case A = 2.072 and B = 2.298, so we get C = 3.094 and φ = 47.97° = 0.8372 radians, and our equation becomes

3.094 sin(θ + 0.8372) = 2.

Now θ occurs just once and we simply isolate it:

sin(θ + 0.8372) = 0.6464

Of the three angles listed in black, red and blue in the last step,

the first angle comes from the calculator (but in the next step it will turn out to be too small to result in a positive θ which this question requires),
the second angle is gotten using the unit circle or CAST method, and
the third angle is just the first angle plus 2π (i.e. plus one full circle).

The last step to isolate θ is to subtract 0.8372 from both sides of the equation:

The graph to the right shows the two waveforms y₁ and y₂ (the green and blue curves) as well as their sum (the red curve). The red dots in the graph are the locations where the sum equals 2. To answer the question we drop the first solution because it is not between 0 and 2π and keep the other two.

Example: Voltage v (the dark line) is the sum of two sinusoidal waves (the two light lines) and is given by the equation:

v = 100 sin(1 t) + 50 cos(2 t).

Determine the times t when v = 0 in the cycle shown (indicated by the black dots).

(Note that the angular velocities of the sinusoidal waves are 1 radian/sec and 2 radians/sec and that time t is in seconds, so the angles 1 t and 2 t are in radians.)

Solution: Because one angle is twice the other, use the double angle identity in the form:

cos(2 t) = 1 − 2 sin²(t)

to get rid of the angle 2 t in favor of the angle t:

0 = 100 sin(t) + 50·{1 − 2 sin²(t)}

Distribute, transpose everything to the left side of the equation and divide by 50:

100 sin²(t) − 100 sin(t) − 50 = 0

2 sin²(t) − 2 sin(t) − 1 = 0

This is a quadratic equation in sin(t). (In fact replacing sin(t) with the alias Q would turn it into the quadratic equation 2 Q² − 2 Q − 1 = 0.) Plugging a = 2, b = −2 and c = −1 into the quadratic formula we get:

It is impossible for the sin of anything to equal 1.366. Thus we are left with:

sin(t) = −0.366

t = sin⁻¹(−0.366)

Putting the calculator in radian mode gives t = −0.3747. However the unit circle picture to the right shows that the values we require to answer this question are t₁ = 3.52 and t₂ = 5.91 seconds.

• θ occurs only once.	Learn about this type sin(θ−40°) = −3
• θ occurs several times but always as the argument of the same trigonometric function.	Learn about this type 6 sin²(θ) − sin(θ) − 2 = 0
• the equation has different trigonometric functions but they all have the same argument, θ.	Learn about this type 2 cos(θ) + tan(θ) = 4 sec(θ)
• the equation contains trigonometric functions with different arguments.	Learn about this type 4 sin(θ) + 3 cos(θ + 40°) = 2 or 3 sin(θ) + 5 sin(2 θ) = 1