Module 5-II – Trigonometric Functions
Section 5-II.1 – Definitions and Evaluation
Introduction
If, for each value of t between 0 and 90, we construct an angle of t , build a right triangle,
and measure its sine ratio, then the result will define a function of t with f (t )  sin(t ) . If
we use a different method of measuring angles then for a given value of the number t we’ll
be looking at a different angle and so get a different result. So we’ll get a different function.
For example, the functions g (t )  sin(t revolutions) , and h(t )  sin(t radians) are both
different from f (t ) .( If you put in a particular number, like 1/4, for t , they will give
different results). Of course, since an angle of t degrees is the same as an angle of
2
2
360 t radians, we have f (t )  h( 360 t ) and so the functions are related, but they are not the
same.
It turns out that the functions defined using radian measure are more convenient for many
purposes (especially in calculus) so these are the ones we’ll focus on, and it is just for these
that we use the name of the trigonometric ratio as the name of our function. (So, we’ll use
the name sin(t ) just for the case of h(t ) above.) This section is devoted to introducing these
functions and studying their basic properties.
Section Learning Outcomes
By the time you have completed this section you should be able to:


State the “unit circle” definitions of the trigonometric functions
Use the definitions to evaluate these functions (exactly if possible)
Study Plan
1.
Read the following Study Notes and Discussion, and make sure that you have
absorbed the main points by answering the questions.
2.
Read sections 5.3 and 5.4 of the text
3.
Follow the instructions regarding Further Practice.
Study Notes and Discussion
Need for Units
As pointed out in the introduction to this section, if we want to use the trigonometric ratios
to define functions of a real number variable, then it is necessary to decide what system of
units is to be used to relate angles and numbers. Because of the way it relates angles and
distances, the radian measure of angles turns out to be more convenient for many
applications (especially in Calculus), so, in Section 5.3 of the text, the trigonometric
functions of a real variable, t, are defined as the corresponding trig. ratios for an angle of t
radians.
i.e. sin(t )  sin  , for an angle,  , of t radians , etc.
Circle Picture
Another way of stating the above definition is to say that cos(t) and sin(t) are the x and y
coordinates of a point on the unit circle at a distance of t units from the positive x axis as
measured counterclockwise around the circumference of the circle.Why? (see answer #1)
The other trig functions are then defined in terms of sin and cos by the same relationships
as for the corresponding ratios.
This “unit circle definition” has the advantage that it doesn’t just work for acute angles. In
fact it makes sense for arbitrary real values of t. For example, to find cos(10) we can go 10
units around the circle. (see below)
y
Since the circumference is 2  6.28 ,
going 10 units around takes us more
than one full revolution, but that’s ok.
Another half revolution takes about
3.14 more units and we come to the
end of our 10 unit trip just a bit further
1
on as shown in the picture.
x
The x-coordinate of the endpoint is by
definition cos(10), and by measuring
t=10
on the graph we see that it’s about
-0.85 . Using your calculator (in
radian mode) you should be able to
check that this is about right.
If you estimate the y-coordinate of the terminal point in the above picture you should get
about -0.55, and if you check on your calculator you should see that this is a pretty good
estimate for sin(10).
For negative values of t , we follow the same convention as for negative numbers on the
number line – that is we go in the opposite direction. So to find cos(-10) we go 10 units
clockwise. So making a picture like the one above for t  10 we find cos(-10)=cos(10)
and sin(10)   sin(10)  0.55 .
Try using the circle picture to evaluate each of the following as accurately as you can :a) cos  2 3 b) sin 8 3 c) csc  5 4 d) cos(1) e) sin(30) (see answer #2)
Note that with this unit circle approach there is no need to memorize a set of rules for the
signs of the trig functions. They are just the signs of the corresponding coordinates of the
point on the unit circle. So the table and mnemonic on page 384 of the text can safely be
ignored. (This doesn’t mean that you don’t need to know what they say, just that once you
understand the circle picture these facts will be obvious.)
The Basic Graphs
The graph of y  sin(t ) can be drawn by just tracing the height of the point Pt as it moves
around the unit circle with t being the arc length from the x-axis (counterclockwise). If you
imagine Pt moving around the circle in the picture below, then, as it does so, the graph on
the right will be traced out. Many important phenomena have similar infinitely repeated
regular oscillations and this is why trigonometric functions are so useful even outside of
geometry.
Pt
t
Note: We would get a similar picture if we used different angular units rather than
radians except for the fact that the horizontal distance on the t-axis would no longer
exactly match the arc length but rather would be some multiple of it. So the graph would
be stretched out unless the t-axis is re-scaled. What distinguishes the case of radian
measure is the equality of arc length and distance along the t-axis. If a point is moving
counter-clockwise around the circle, then as it passes P0  (1, 0) it is moving directly
upwards, and so, for small t , the y-coordinate is almost exactly equal to the arc length. I.e.
sin(t )  t for small t . This means that the graph of y  sin(t ) is very closely matched near
the origin by the diagonal line y  t ,(at 45 to both axes if the scales are equal).
Some Basic Identities
The unit circle picture also allows us to easily establish various identities concerning the
trig functions.
For example, the equation of the circle ( x 2  y 2  1) gives us the 'pythagorean identity',
cos2 (t )  sin 2 (t )  1 .
Other identities express relationships between trig ratios of different angles:
(i) one such is the 'negative angle relations' sin(t )   sin(t ) and cos(t )  cos(t ) which
are illustrated in Figure 12 on page 396 of the text;
(ii) another is the 'periodicity relation' sin(t  2 )  sin(t ) , cos(t  2 )  cos(t )
which corresponds to the fact that going a distance 2 around the unit circle brings us all
the way back to where we started;
(iii) and another is the pair sin(t  2 )  cos(t ) , cos(t  2 )   sin(t ) .
These three are all illustrated below (which is which?)(see answer #3)
(a)
(b)
(c)
Exercise:
Use similar pictures to show that sin(t   )   sin(t ) , cos(t   )   cos(t ) (answer #4)
and to derive formulas for sin(t  2 ) and cos(t  2 ) .( see answer #5)
Basic Trigonometric Equations
We can also use the unit circle picture to go the “other way” and determine t from the
value of sin(t) or cos(t). This amounts to solving for t in equations of the form sin(t )  c ,
and cos(t )  c . Because they are expressed in terms of trig functions, these are called
trigonometric equations.
For example, to solve sin(t )  c we need to find the numbers t for which the point Pt has
y-coordinate equal to c. These would be the t-values for intersection points of the circle
with the horizontal line y  c . For c  12 , the points are shown in the picture below.
Q
P
O
P0
y
1
2
We know that P0OP  30 

6
radians .
How? (see answer #6)

So P  P and t  is one solution.
6
6
By symmetry we see that Q  P


6
, so t 
5
is
6
another solution.
And adding any multiple of 2 brings us back to the same point so we find that the

5
 2n , where n can be any whole number.
general solution is t   2n , or
6
6
Even when we can’t determine an exact value, we can use the circle picture to estimate
the solutions. A bigger picture is easier to measure on, and it helps to have the arc lengths
labelled as in the text’s illustration for Exercises 69-72 on page 406. Try using that one to
estimate the solutions of sin(t )  0.8 . (see answer # 7)
Note that the calculator’s sin 1 button can be used for such problems to get a better
approximation. It only gives one of the possible solutions, but the others can be found by
reflecting across the y-axis and adding multiples of 2 .
Similarly, to solve cos(t )  c we look for intersections with the vertical line x  c . Again
there are two points of intersection, but this time placed symmetrically across the x-axis,
and each corresponds to an infinite family of t values by adding multiples of 2 .
For the case of tan(t )  c , it is neither x,nor y that we must match but rather their ratio
y / x , which is just the slope of the ray from the origin to the required point. So to solve
tan(t )  2 , for example, we just draw the line y  2 x , and check where it crosses the
circle. By eye we can read off that one point is at t  1.1 (calculator gives 1.107..), and
the other one is exactly halfway around the circle - which we can reach by adding  .
Further Reading
In section 5.4, the text uses some of the basic identities discussed above (and in the text’s
section 5.3) to express the trig ratios for any angle in terms of a “reference angle”
between 0 and 90 (i.e. between 0 and 2 ).
You may find that by trying to explain it in words the book makes the “reference angle”
approach seem more complicated than it really is. For example on page 409, the pictures
in Figure 2 may be easier to understand if you don’t read the text beside them. In fact you
can evaluate trig ratios for general angles just as easily by using the unit circle, measuring
off the angle (equal in radians to the distance around the circumference) and using basic
trigonometry on the right triangles joining the terminal point to the coordinate axes to
determine the x and y coordinates (of the point corresponding to the given angle.)
Try using this approach to solve the text’s Example 2 on page 410. (see answer #8)
The text material about finding angles from trig ratios on pages 411-414 can also be
understood most easily in terms of the unit circle picture as we have described above.
Try using the circle approach to solve the text’s Example3 on page 413. (see answer #9)
Further Practice
Try enough of the odd numbered questions in the text sections 5.3 and 5.4, using the
solutions guide if you need to, until you feel able to do similar questions on your own.
From section 5.3 you should do at least #3,13,23,33,53,69, and 73, and from section 5.4
at least #7,27, and 39.
Answers to Questions
1. Because an angle of t radians corresponds to an arc length of s  rt around a circle of
radius r, and so if r  1 we get s  t and the angle of t radians corresponds to going t units
around the circumference.
2.
a) In the picture below, the triangle has angle 60 (  
2 
= radians ) .
3
3
So it is equilateral and the altitude (vertical line) splits the base in half .
P2 3   cos  2 3 , sin  2 3  
P0  (cos(0), sin(0))  (1,0)
So cos  2 3   1
2
b) Since 8 3  2 3  2 , we reach the point P8 3 by going past P2 3 for one
full revolution – which brings us back to the same point.
So,
sin 8 3  sin  2 3  12   12  
2
3
3

4
2
c) Since 5 4     4 the triangle in the picture has 45 angles and since the
1
hypotenuse is 1 each of the other two sides is of length
. So sin  5 4    1
and
2
2
1
1
csc  5 4  

 2
1
sin  5 4  
2
P0  (cos(0), sin(0))  (1,0)
P5 4   cos  5 4  , sin  5 4  
d) Since 3  1.05 , 1 radian is just a bit less than 60 (actually about 57.3 ).
So cos(1) is just a bit more than .5 (as can be seen from the left picture below).
P1   cos 1 , sin 1 
P10  P0  (1, 0)
P0   cos  0  , sin  0    (1,0)
P30   cos  30  , sin  30  
e) Note that one revolution is about 6 radians, so 30 radians corresponds to about 5
revolutions. But 5 full revolutions is actually 10 radians, and 10  31.4 , so by going all
the way to P10  P0  (1, 0) we will have gone too far. Since, 30  10 1.4 . So P30 is
found by going back (clockwise) a distance of about 1.4 units ; and since 1.4 is a bit less
than 2 , P30 is just a bit to the right of P10    P3  (0, 1) . So sin(30)  1 (see right
2
2
picture above).
3.
4.
(i)(c), (ii)(a), (iii)(b)
5.
cos(t  2 )  sin(t )
sin(t  2 )   cos(t )
6.
This is one of our geometric special cases, sin( 6 )  sin(30)  12 .(see Module 5.3)
7.
Placing a ruler horizontally at y  0.8 on the picture (on page 396 of the text),
we see that it cuts the circle at t  0.95 and 2.2 . So the general solution will be
t  0.95  2n and 2.2  2n . (Using the calculator, we get more decimal places. The
first solution is t  0.9273 , and the second is  minus that, or approximately 2.2143. But
often just the quick “by eye” estimate is good enough for what we need.)
8.
a)
P5
6
b)
P 
6

3
2
, 12

P 

1
2
,
1
2

4
P9   cos(315),sin(315) 
4
( 315  94 radians )
9.
Points corrresponding to equal tangent ratios are on opposite sides of the circle.
So if the calculator gives one answer, others can be obtained by adding or subtracting
integer multiples of 180 .