### The Cartesian Plane

The cartesian plane is a plane on which an x axis is drawn horizontally and a y axis is drawn vertically. Any points in the plane are 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 in the plane is called the rectangular coordinate system (notice the dotted rectangle shown in the picture). Another coordinate system in common use, for example 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 ° ).

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 subtendend 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.

Note that since s and r have the same units, the radian is a unitless unit!

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.

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 right triangles. (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 proof of Pythagoras’ theorem can be found at http://www.math.ubc.ca/~morey/java/pyth/

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

This 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:

### Definitions of the trigonometric functions

For now let us restrict ourselves to right triangles.

Notice that if a 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. Denote this using functional notation like this: sin(θ), cos(θ), etc.

• Later when we discuss trigonometric functions of any angle it will be very useful to imagine an arrow (or vector) radiating out from the origin with length r and orientated at angle θ. Then a triangle can be constructed by drawing lines vertically and horizontally to the x and y axes. Suppose that the arrow-head is at coordinate (xy). Then the six trigonometric functions can be defined like this:
• The three trigonometric functions: sin, cos and tan 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 pressing the appropriate button, sin, cos or tan.

• The other three trigonometric functions: csc, sec and cot are not built into your calculator. To find them you must instead find the sin, cos or tan on your calculator and then take the reciprocal.

Examples:
• sin(30°) = 0.500

• tan (57°) = 1.54

• 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.

### Definitions of the inverse trigonometric functions

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

One thing we know is that . Previously we would put an angle θ into the calculator and press TAN to get a value out for the tangent.

Here we want to do the inverse process - we want to put a value of tan 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 the arcsin, arccos and arctan functions respectively.)

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).

• If cos(θ) = z, then θ = cos−1(z).

• If tan(θ) = z, then θ = tan−1(z).

### Solution of 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.

Example: A train is travelling 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 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.

 Algebra Coach Exercises

### Trigonometric functions of any angle

When we first defined the trigonometric functions 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. This was sufficient for right triangles. But we now want to allow θ to have any value. One reason is to be able to solve oblique triangles (triangles which don’t have a 90° angle). Another reason is that we want to be able to describe the angle of spinning objects as they rotate through many revolutions.

To do this we imagine an arrow (or vector) radiating out from the origin with length r and orientated at angle θ. Suppose that the arrow-head is at coordinate (xy). Then a triangle can be constructed by drawing lines vertically and horizontally from the arrowhead to the x and y axes. If θ is outside the range 0° to 90° then x or y or both can be negative. (Don’t forget, however, that r is defined to be positive.)

This means that the trigonometric functions can possibly be negative since:

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

Here are two ways to show the signs of the trigonometric functions in the various quadrants:

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

As we have seen before, if you are given any angle whatsoever, you can use the calculator to find its sin, cos or tan. The answer is unique.

The new problem is that if you are given the sin, cos or tan of angle θ, then you can use sin−1, cos−1 or tan−1 on your calculator to find one possible value of θ but you must use a picture like below to find the other possible value of θ between 0° and 360°.

Example: Find 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 two 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 two 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 sin law reduces to , namely the definition of the sin, 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 Sin 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 sin law. Otherwise you must use the cosine law.

If you are using the sin 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
A = 25.3°, c = 152 and a = 95.0

Solution: Note that the given info is not enough to decide whether the triangle is ABC or ABC'. 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 2 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°

 Algebra Coach Exercises

### Vectors

Definition: A vector is a quantity that has magnitude and direction. A scalar is a quantity that has only magnitude. An example of a vector is force; you can apply a force in any direction. An example of a scalar is temperature; it doesn’t point in any direction.

A vector is represented by an arrow. The length of the arrow represents the magnitude of the vector and the direction in which the arrow points represents the direction of the vector.

Vectors are usually drawn with their tail at the origin. This way complete information about the vector is given by the coordinates of the arrowhead.

Vector Addition: To add vectors A and B graphically, put the tail of A at the head of B or vice versa. The resulting vector, A + B, is also drawn with its tail at the origin. This is called parallelogram construction. A + B is called the resultant of adding A and B.

Vector Subtraction: The negative of a vector B, namely −B, points in the opposite direction.

The vector subtraction A − B is equivalent to the vector addition A + (−B).

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

Resolution of a vector into its components: Resolving a vector into its x and y components is done by drawing a dotted line from the arrow-head 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 )

### 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 PR or → x y ) and rectangular to polar conversion (denoted RP or → r θ ) built in. See your calculator manual. If not, you can do the following:

• If you are given the vector rθ 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 rθ where:
Note that you may have to add 180° if θ is not in the first quadrant.

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.

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 subtendend 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 have the same units, the radian is a unitless unit!

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 travelled / 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 = rE for the equator or r = rV for Vancouver. We were given rE = 3960 mi. To find rV 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 sin 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. The vertical arrows are the y components of the vector in the various positions. The graph on the right is a plot of the y component 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 cos 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. The horizontal arrows 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 tan 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 wave function:
y = A sin(ω t + φ)
where y represents the quantity of interest and t represents time. Because y is proportional to the sin function, it has the characteristic sin 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 ordinary 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 rad/sec and ω = 4 rad/sec. The wave sin(4t) oscillates 4 times as fast as the wave sin(t).

An algebraic way to see why ω = 4 compresses the wave horizontally by a factor of 4 is to note that in sin(4t), t needs to be only ¼ 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 2π 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 angle 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.

• How much is the wave shifted? That 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 to the left or right, 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 above example, y = sin(4t+1), a cycle begins when there is 0 inside the brackets, namely when t = −¼.

For the example y = sin(4t) above, a 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 angles:
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.

### 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 tB and tE , the values of the time t at the beginning and the end of the cycle. The period T is the difference tEtB. Then calculate angular velocity ω from the equation .

• Get phase angle φ by using the fact that the quantity in brackets, ω t + φ, must equal zero when t = tB, 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 labelled v. Thus the template equation is:
v = A sin(ω t + φ)
• The amplitude is A = 7 volts.

• The cycle begins at tB = −1 sec and ends at tE = 5 sec. Thus the period is T = tEtB = 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 labelling the vertical axis “y” and the horizontal axis “t”, assuming that the equation is y = A sin(ω t + φ).

• Get tB, 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 = tB.

• Get tE, 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 = tE.

• Place a box stretching from tB to tE 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.

 Algebra Coach Exercises