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An Introduction to Trigonometry
P.Maidorn I. Basic Concepts
The trigonometric functions are based on the unit circle, that is a circle with radius r=. Since the circumference of a circle with radius r is C=2Br , the unit circle has circumference 2B.
For any point (x,y) on the unit circle, the associated angle 2 can be measured in two different ways:
- degree measure: in this case the circumference is divided into 360 equal parts, each part has measure one degree (written 1 ). A right angle, for example, is a 90o^ o angle. Positive angles are measured in the counter-clockwise direction.
- radian measure: radian measure is defined as the actual length of the arc between the points (1,0) and (x,y). One entire revolution (i.e. 360 ) hence has a radian measureo of 2B. A right angle (that is a quarter of one revolution) would have radian measure B/2. Note that the angle is simply denoted “B/2”, not “B/2 radians”.
One can easily convert between these two measures by keeping in mind that a 180 angleo (in degrees) is equivalent to a B angle (in radians). Note that angles in Calculus-related problems are usually denoted in radian measure, hence it is important to be comfortable with this measurement.
Examples:
- A 270 angle is 3/2 times a 180 angle, hence in radian measure the angle would beo^ o denoted 3B/2.
- A 7B/5 angle would simply have degree measure 7/5 times 180 , i.e. 252 .o^ o
Exercises:
Convert each angle to radians. a) 120 o^ b) 315 o^ c) -420o Convert each angle to degrees, to the nearest tenth of a degree. d) -2B/3 e) 3B f) 4.
Turning the above discussion around, each time we choose an angle 2 , we find a unique point (x,y) on the unit circle. Hence both “x” and “y” can be considered functions of 2. Since these particular functions are of great importance to both pure and applied mathematics, they are given special names and symbols, and are called the trignometric functions.
Specifically:
The length “y” is called the sine of the angle 2 , and is denoted by y=sin( 2 ). The length “x” is called the cosine of the angle 2 , and is denoted by x=cos( 2 ).
Other trigonometric functions can be calculated from the sine and cosine functions: the tangent of 2 is defined as tan( 2 )=sin( 2 )/cos( 2 ) (or y/x), the secant of 2 is defined as sec( 2 )=1/cos( 2 ), the cosecant of 2 is defined as csc( 2 )=1/sin( 2 ), and the cotangent of 2 is defined as cot( 2 )=cos( 2 )/sin( 2 ).
One can also find trigonometric values for the angles 2 =B/6 (or 30 ) ando^2 =B/3 (or 60 ).o This set of angles is sometimes called the “special” angles, and their associated sine and cosine values are listed in the table below:
2 sin( 2 ) cos( 2 )
0 0 1
B/6 ½ /3/
B/4 /2/2 /2/
B/3 /3/2 ½
B/2 1 0
We can immediately use these values to calculate other trigonometric functions of these special angles.
Examples:
- Since tan( 2 )=y/x, i.e. tan( 2 )=sin( 2 )/cos( 2 ), we simply divide sin(B/3) by cos(B/3) to find that tan(B/3) = /3.
- Similarly csc(B/6) = 1 / sin(B/6), that is csc(B/6) = 2.
Note that not all trigonometric functions are defined for all angles. For example, the tangent of 2 =B/2 does not exist, since here the denominator is equal to zero.
Exercises:
Calculate: a) sec(B/3) b) csc(45 )o c) cot(0)
III. Other Angles
Once you know how to find the trigonometric functions for the above special angles, it is important to learn how to extend your knowledge to any angle that is based on one of 0, B/6, B/4, B/3 or B/2, such as for example 2B/3, -B/6, 7B/4, -5B/2, and others.
Let’s examine the angle 2 =-B/3. Clearly, it is somehow related to the angle B/3. Draw a unit circle for both angles side-by-side, and indicate the sine and cosine on them:
Clearly, the right-angle triangles that are formed are identical, except that they are mirror images of each other. The cosine (“x”) in both cases is the same, hence we know that cos(-B/3) is identical to cos(B/3), that is cos(-B/3)=½. The sine (“y”) is the same length, but has opposite sign (it is negative). Since sin(B/3)=/3/2, then sin(-B/3)=-/3/2.
In each of the examples below, proceed with the same method. It is imperative that you draw the unit circle each time until you become comfortable with these types of questions.
Examples:
- Find sin(5B/6). The triangle formed by 2 =5B/6 is identical to that formed by 2 =B/6, except that it is reflected across the y-axis. The sine of 5B/6 is hence the same length, and has the same sign, as the sine of B/6. Therefore sin(5B/6)=½.
Exercises:
Use your calculator to compute, rounding to four decimal places: a) sin(53.7 )o^ b) cos(11B/7) c) csc(-32 )o
V. Right-Triangle Applications
In order to apply the trigonometric functions based on the unit circle to right triangles of any size, it is important to understand the concept of similar triangles. Two triangles are said to be similar if the ratio of any two sides of one triangle is the same as the ratio of the equivalent two sides in the other triangle. As a result, similar triangles have the same “shape”, but might differ in size. For example, the sides in the triangles below have the same ratios to each other.
Consider the right triangle inscribed in the unit circle associated with an angle 2. We can calculate the length of the side adjacent to the angle 2 (i.e. cos( 2 )) and the length of the side opposite the angle 2 (i.e. sin( 2 )). Since the unit circle has radius one, the hypotenuse of these triangles is always equal to one.
If we were given a triangle with identical angle 2 but with a hypotenuse twice the length, each of the other sides would be twice the length as well, as the triangles are similar. We can use this fact to now compute side-lengths of any right triangle, if the angle 2 and one side-length are known. In general, we have
sin( 2 ) = opposite / hypotenuse and cos( 2 ) = adjacent / hypotenuse
and since tan( 2 )=sin( 2 )/cos( 2 ), we have
tan( 2 ) = opposite / adjacent.
Examples:
- Find all missing sides and angles in the given triangle.
First, given that the sum of all angles in any triangle must equal 180 (oro B), the missing angle measures 30. We know the length of the opposite side of angleo 2 , hence we can use the sine to find the length of the hypotenuse. Since
sin(60 ) =o /3/2 and sin( 2 ) = opposite / hypotenuse
we have /3/2 = 4 / c, which we can solve for c = 8//3 or c = 8/3 / 3.
Finally use, for example, the Pythagorean theorem to find that b = 4/3 / 3.
- To find the height of a building, a person walks to a spot 130m away from the base of the building and measures the angle between the base and the top of the building. The angle is found to be 51.2. How tall is the building?o
We know the length of the adjacent side of the right triangle that is formed, and wish to find the length of the opposite side. The quickest method to do so is to use the tangent, since tan 2 = opposite / adjacent. Using the calculator, tan(51.2 )o.1.2437. Hence the height of the building is h = 130 tan(51.2 )o. 161.7 metres.
Are there any other such angles? By remembering that we can always add or subtract one complete revolution (2B) from an angle to end up in the same position, we can in fact generate infinitely many such angles. For example, 23 = B/6 + 2B = 13B/6 is another such angle, as is 24 = 5B/6 - 4B = -19B/6.
Hence all angles 2 = B/6 + 2kB and 2 = 5B/6 + 2kB for any k 0 I (that is k can be any integer ...-3,-2,-1,0,1,2,3,... ) are solutions to the equation sin(x)=½.
Exercises:
Find all “x” such that a) cos(x) = /3/2 b) cos(x) = -/3/2 c) sin(x) = - d) sin(x) = - ½ e) cos(x) = - ½ f) tan(x) = 1
VII. Trigonometric Identities
There are several trigonometric identities, that is equations which are valid for any angle 2 , which are used in the study of trigonometry.
From the unit circle, we have already seen that the cosine of an angle is identical to the cosine of the associated negative angle, that is
cos(- 2 )=cos( 2 ) for any angle 2. (1) Similarly, sin(- 2 )=-sin( 2 ) for any angle 2. (2)
For example, sin(B/6) = ½ and sin(-B/6) = -½.
Also from the unit circle, which has equation x +y =1, we can substitute x=cos(^2 22 ) and y=sin( 2 ) to obtain the identity
sin (^2 2 )+cos (^22 )=1, for any angle 2. (3)
Another useful trigonometric identity concerns the sum of two angles A and B. We have:
sin(A+B) = sin(A) cos(B) + sin(B) cos(A) (4) and cos(A+B) = cos(A) cos(B) - sin(A) sin(B) (5) for any angles A and B.
Note that you cannot simply “distribute” the sine through a sum. It is false to state that, for example, sin(A+B)=sin(A) + sin(B).
The above five identities can be used to derive many other useful identities, which then no longer need to be memorized.
Examples:
- Dividing equation (3) by cos (^22 ), we obtain the identity 1 + tan (^2 2 ) = sec (^22 ).
- Identities involving the difference of two angles A-B can be obtained by combining equation (4) or (5) with equations (1) and (2). As an example,
sin(A-B) = sin(A+(-B)) = sin(A) cos(-B) + sin(-B) cos(A) using (4) = sin(A) cos(B) - sin(B) cos(A) using (1) and (2) on the left and right term respectively.
- The double-angle formula cos(2 2 ) = 2 cos (^22 ) -1 can be derived using equation (5) and setting both A= 2 and B= 2 :
cos(2 2 ) = cos ( 2 + 2 ) = cos( 2 )cos( 2 ) - sin( 2 )sin( 2 ) using (5) = cos (^2 2 ) - sin (^22 )
Now, re-arrange (3) to obtain sin (^2 2 )=1-cos (^22 ) and substitute:
cos(2 2 ) = cos (^2 2 ) - (1-cos (^22 )) = 2cos (^22 ) - 1.
- Find all “x” such that 2sin (x) + cos(x) = 1.^2 To solve this trigonometric equation, we will first need to simplify it using the trigonometric identities. Using (3) we can substitute sin (x)=1-cos (x) and obtain an equation that^2 only involves cosines: 2 - 2cos (x) + cos(x) = 1^2 or 2cos (x) - cos(x) -1 = 0.^2
The above is a quadratic equation in cos(x), and can be factored (2cos(x)+1) (cos(x)- 1) = 0.
As a result, we now need to find all “x” such that either 2cos(x) + 1 = 0, i.e. cos(x)=-½, or cos(x)-1 =0, i.e. cos(x)=1.
From the unit circle, cos(x)=1 when x= 0 +2kB, k 0 I, i.e. x=2kB, k 0 I Further, cos(x)=-½ when x=2B/3+2kB or x=4B/3+2kB , k 0 I. The solution to the given equation is hence the set of all x-values listed above.
The graph of y = cos(x) is obtained in a similar fashion.
To obtain the graph of y = tan(x), divide sin(x)/cos(x) as before. Note again that tan(x) does not exist for values of x=B/2 + kB, k 0 I.
You can now use the techniques for shifting and scaling function graphs to obtain the graphs for any trigonometric function.
Examples:
- Graph y = 2 sin(x+B).
This graph is identical to that of y=sin(x), except it is shifted to the left by B units, and scaled vertically by a factor of 2.
- Graph y = - cos(x-B/6).
This graph is obtained by shifting the graph of y=cos(x) to the right by B/6 units, then flipping it across the x-axis.
Exercises:
Sketch graphs for the given functions: a) y = cos(x+B/6) - ½ b) y = ½ sin(x-B/2) + 1
