Download 3. TRIGONOMETRIC FUNCTIONS

3. TRIGONOMETRIC FUNCTIONS
§3.1. Elementary Trigonometry
In this section we revise the elementary trigonometry that we learn in High School. It
concerns angles and triangles, especially right-angled ones.
Angles are traditionally measured in degrees, where 360 degrees constitute a complete
revolution. This is a legacy from the Babylonians who used a number system based on base
60. The other measure that comes from the Babylonians is time where there are 60 minutes
in an hour and 60 seconds in a minute.
A perpendicular angle, or a right angle, has 90 degrees. We write this as 90. A
right-angled triangle is one where one angle is a right angle. The other two angles add up to
90 and are called complementary angles.
The longest side of a right-angled triangle, the side opposite the right angle, is called
the hypotenuse. The fundamental fact about right-angled triangles is the theorem that is
attributed to Pythagoras: The square on the hypotenuse is the sum of the squares on the other
two sides.
Theorem 1 (Pythagoras): If the lengths of the sides of a right-angled triangle are a, b, c and
c is the length of the hypotenuse then a2 + b2 = c2.
Proof:
a
b
a
b
c
c
c
c
b
a
a
b
The areas of the larger square is (a + b)2 = c2 + 4. ½ .ab = c2 + 2ab.
Hence a2 + b2 = c2.
Suppose we have a right-angled triangle where the angles are 90,  and 90  .
hypotenuse
opposite

adjacent
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If we focus on the angle  then the opposite side is, as the name suggests, the side
opposite that angle. The hypotenuse is the side opposite the right-angle and the remaining
side is called the adjacent side.
The sides of the triangle will change as the triangle gets bigger or smaller, with the
same angle . But, since corresponding sides of similar triangles are proportional the ratio of
corresponding sides will depend only on the angle.
opposite
We define the sine of  as sin =
,
hypotenuse
adjacent
the cosine of  is defined as cos =
and
hypotenuse
sin opposite
the tangent of  is defined as tan =
=
.
cos adjacent
A consequence of Pythagoras’ Theorem is that:
sin2 + cos2 = tan2.
[Here we write sin2 when we mean (sin)2. We could have meant sin(sin). Similar
comments apply to cos2 and tan2.]
The three other rations are the reciprocals of the above.
adjacent
The secant of  is sec = hypotenuse ,
hypotenuse
the cosecant of  is cosec = opposite and
adjacent
the cotangent of  is cot = opposite .
Rather than burden our brains with these definitions we remember that
1
1
1
sec =
, cosec =
and cot =
.
cos
sin
tan
Simple consequences of these definitions are the trigonometric identities:
sec2 = 1 + tan2 and cosec2 = 1 + cot2.
The first of these is important enough to be memorised but the second is not. It can always
be worked out when needed.
A general principle with an area of mathematics, like trigonometry, with a lot of
formulae is to remember only a key few and to be able to work out any others that one might
need as the occasion arises.
§3.2. Special Angles
There are two right-angled triangles that are rather special: the 45 degree triangle and
the 60-30 triangle. Indeed these are the two standard shapes for set squares.
2
2
1
1
1
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3
The dimensions can be obtained from the fact that a 45 degree right-angled triangle is
isosceles, a 60-30 triangle is half an equilateral triangle and, of course, Pythagoras’ Theorem.
Although a right-angled triangle with two right angles and one zero angle is not a proper
triangle we can infer the trigonometric functions for 0 and 90. These are included in the
following table.

0
30
45
60
90
1
1
0
1
sin
3
2
2
2
1
1
1
0
cos
3
2
2
2
1
1

3
3
The last value is a bit of a cheat because there is no such number as infinity. It
reflects the fact that if A is very close to 90 the adjacent side is very short and so tan(A) is
very large. In fact we can make tan(A) as large as we like by making A sufficiently close to
90. Usually we capture this property by saying that as A approaches 90, sin(A) approaches
infinity.
0
tan
§3.3. Sum And Difference of Angles
Theorem 2: sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
cos(A + B) = cos(A)cos(B)  cos(A)cos(B)
tan(A) + tan(B)
tan(A + B) =
.
1  tan(A)tan(B)
Proof: In the following diagram we have three right-angled triangles with angles A, B and
A + B.
A
k
y
d
b
h
B
z
z
A
a
b
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z+y z h y k
= . + . = sin(A).cos(B) + cos(A).sin(B).
d
h d k d
a (a + b)  b (a + b) h b k
cos(A + B) = d =
= h . d  k . d = cos(A).cos(B)  sin(A).sin(B).
d
sin(A + B)
sin(A).cos(B) + cos(A).sin(B)
tan(A + B) = cos(A + B) =
cos(A).cos(B)  sin(A).sin(B)
tan(A) + tan(B)
=
after dividing top and bottom by cos(A).cos(B).
1  tan(A).tan(B)
Corollary: sin(2A) = 2sin(A).cos(A),
cos(2A) = cos2(A)  sin2(A) and
2tan(A)
tan(2A) =
.
1  tan2(A)
sin(A + B) =
The above identities are usually referred to as the Double Angle Formulae. The
following are the Half Angle Formula where sin(A), cos(A) and tan(A) are all expressed in
terms of t = tan(A/2).
Theorem 3: If t = tan(A/2) then
2t
sin(A) = 1 + t2 ,
1  t2
cos(A) = 1 + t2 and
2t
tan(A) =
.
1  t2
Proof: The third of these is simply the Double Angle Formula, using A/2 in place of A.
By drawing a suitable right-angled triangle, with sides 2t and 1  t2 we can deduce that the
hypotenuse is 1 + t2 and hence read off sin(A) and cos(A).
1 + t2
2t
1  t2
Theorem 4: sin(A  B) = sin(A)cos(B)  cos(A)sin(B),
cos(A  B) = cos(A)cos(B) + cos(A)cos(B) and
tan(A)  tan(B)
tan(A  B) = 1 + tan(A)tan(B) .
Proof: This is proved similarly using the diagram from Theorem 2. Let C = A + B. Then by
similar methods to those used in the proof of Theorem 2 we get:
sin(C  B) = sin(C).cos(B)  cos(C).sin(B) which is what we have to prove but with different
notation. The proof of the other identities is left as an exercise.
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§3.4. Triangle Identities
We now move away from right-angled triangles to consider general triangles. We
shall label the three vertices A, B, C and also use the symbols A, B, C for the values of the
corresponding angles. The sides opposite these sides will be labelled a, b, c respectively.
B
a
c
C
A
b
sin(A) sin(B) sin(C)
= c .
a = b
Proof: Drop a perpendicular from A to the side BC.
Theorem 5 (Sine Rule):
B
a
c
h
C
A
z
bz
Then h = c.sin(B) = b.sin(C). By drawing perpendiculars to the other two sides the result
follows.
Theorem 6 (Cosine Rule): c2 = a2 + b2  2bc.cos(A).
Proof: z = c.cos(A) and h = c.sin(A).
Hence, by Pythagoras in the right triangle, (b  c.cos(A))2 + c2sin2(A) = a2.
Hence b2 + c2cos2(A)  2bc.cos(A) + c2sin2(A) = a2
and so a2 = b2 + c2  2bc.cos(B).
These formulae are useful, in many cases, for finding the remaining sides and angles
given three pieces of information. If we are given two angles and a side we know the third
angle and we can use the Sine Rule to obtain the other two sides.
If we are given two sides and the included angle we can use the Cosine rule to find the
other side, and then the Sine Rule to find the other angles.
If we are given two sides and a non-included angle or three angles the triangle is not
uniquely defined. If we are given three sides we can use the Cosine Rule to find the three
angles.
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§3.5. Rotations In The x-y Plane
So far angles have been restricted to the range 0 to 90. We now extend the
definition of sin, etc for all . We do this by considering rotations in the x-y plane.
We begin with the positive x-axis and measure rotations as positive in the anticlockwise direction and negative in the clockwise direction. The rotation indicated above is
positive and its magnitude is about 225. If this rotation were to be carried out clockwise it
would have been through an angle of  225. If this clockwise rotation had been carried out
twice it would have been through an angle of  450.
If the point (1, 0), on the positive half of the x-axis, is rotated through an angle , the
resulting point is defined to be (cos, sin). This now extends the definition of sin and cos
to all . For example, as the above diagram shows, if the point (1, 0) is rotated through 225
the resulting point is (1/2, 1,2). Therefore cos225 =  1/2 and sin225 = 1/2.
If we rotated (1, 0) through 585, that is 225 plus 360, the resulting point would still
be (1/2, 1,2). An extra 360 in either direction does not make any difference to the
position of the rotated point and therefore makes no difference to the values of sin and cos.
Theorem 7: For all angles, , measured in degrees:
sin(  360) = sin and cos(  360) = cos,
sin( + 180) =  sin and cos( + 180) =  cos,
sin( + 90) = cos and cos( + 90) =  sin.
Proof: These can easily be proved by considering the x-y plane.
The absolute values of sin and cos lie between 0 and 1 but in different quadrants
they have different signs. By considering the x-y plane we can see that these signs are as
follows.
sin > 0
sin > 0
cos < 0
cos > 0
sin < 0
cos < 0
sin < 0
cos > 0
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sin
and so tan( + 180) =  tan etc. The value of tan is positive
cos
in the 1st and 3rd quadrants and negative in the others. A useful diagram, to help you
remember in which quadrants sin, cos and tan are positive is the following.
We define tan =
S
A
T
C
Here A stands for ALL, S stands for SIN, T stands for TAN and C stands for COS. Students
are often taught a mnemonic “All Stations To ...” where C is represented by some local
railway station that starts with “C”. As I learnt this at Canterbury Boys High School,
naturally our teachers taught us to remember “All Stations To Canterbury.”
Since most of our proofs of trigonometric identities relied on diagrams where angles
were in the range 0 to 90 we should provide new proofs to guarantee that they work for all
angles. We will not do that because we are not yet at the final stage of our development of
trigonometry.
§3.6. Radian Measure
Degrees are a very suitable unit of measure for angles. Other systems have been
proposed such as the one that assigns 100 “grads” to a right-angle, but it hasn’t caught on.
What about a unit that assigns 1.5707963 ... units to a right-angle. For practical purposes it
would be a nightmare. An yet that is the natural unit of measure for reasons that we shall
now see. In fact, unless you are a surveyor or a navigator you will be leaving degrees behind
and adopting radians as the unit.
One way to define angles is in terms of arc lengths. If we have a sector of a circle
with unit radius we define the angle as the length of the arc.
1


Since the circumference of a circle with radius 1 is 2, 360 is 2 radians and 90 is
/2. The special angles in radians are as follows.
degrees
radians
0
0
30
/6
45
/4
60
/3
90
/2
We choose not to evaluate these as decimals. It is easier, and more exact, to leave
them as multiples of . From now on all angles will be in radians and therefore we write:
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sin(  2) = sin, cos(  2) = cos, tan( + 2) = tan,
sin( + ) =  sin, cos( + ) =  cos, tan( + ) = tan,
sin(/2  ) = cos.
§3.7. The Calculus of the Trigonometric Functions
Theorem 2.8:
lim  sin x 

 = 1.
x  0 x 
Proof:
1
x
x
sin x
x
sin x.cos x
Using the facts that the shortest distance between two points is a straight line and that the
shortest distance from a point to a line is the perpendicular distance we see that:
sin x.cos x < sin x < x.
sin x
Hence cos x < x < 1.
Since the limit of cos x, as x approaches zero is 1, the result follows.
A quicker proof would come by applying L’Hôpital’s Rule:
lim  sin x 
lim  cos x 

 = 1, but the normal proof that the derivative of sin x is cos x

 =
x  0 x  x  0  1 
relies on this limit.
lim  cos x  1 

 = 0.
x  0
x

2
cos x  1 cos (x/2)  sin2(x/2)  1
2sin2(x/2) sin(x/2)
Proof:
=
=
= (x/2) . sin(x/2).
x
x
x
sin(x/2)
As x  0, (x/2)  1 and sin(x/2)  0 so the product approaches zero.
Corollary:
d
Theorem 2.9: dx sin x = cos x.
Proof: Let y = sin x and let y + y = sin(x + x).
y sin(x + x)  sin x sin x.(cosx  1) + cos x.sinx
Then
=
=
x
x
x
cosx  1
sinx
=sin x 
 + cos x 
.
 x 
 x 
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dy
cosx  1
sinx
As x  0, 
  0 and 
  1, so dx = cos x.
 x 
 x 
d
Theorem 2.10: dx cos x =  sin x and
d
2
dx tan x = sec x.
d
d
Proof: dx cos x = dx sin(/2  x)
= cos(/2  x).(1) by the Chain Rule
=  sin x.
d
d  sin x 
dx tan x = dx cos x
cosx.cos x  sin x( sin x)
=
by the Quotient Rule
cos2x
sin2x + cos2x
=
cos2x
1
= cos2x
= sec2x.
§3.8. Trigonometric Functions Without Geometry
We have discussed trigonometry at two levels so far. The first involves angles in
triangles and the second involved rotations. But now we leave any geometrical interpretation
behind and consider sin x etc as functions of a real number x.
One reason for doing this is that the majority of uses of the trigonometric functions in
advanced mathematics do not involve anything remotely geometric. Another reason is that
many of our proofs of the properties of the trig functions rely on an inspection of certain
diagrams and we should remember that diagrams can make hidden assumptions. The most
rigorous treatment avoids any geometric considerations and is purely algebraic.
The x-y plane becomes the complex plane if we consider the y-axis as the imaginary
axis. Points in the plane, instead of being represented by pairs of real numbers (x, y) are now
represented by a single complex number x + iy. The advantage of using complex numbers is
that we are able to multiply complex numbers.
The polar form for complex numbers express them as r(cos + isin) where r is the
modulus and  is the argument. Because of De Moivre’s Theorem we can write this as rei.
So cos is the real part of ei and sin is the imaginary part.
With the mention of the “argument” of a complex number we are still tied to angles
and geometric interpretation. Imagine explaining the trigonometric functions to a
“disembodied angel”, that is a being with immense intelligence but no concept of space.
Anything geometric would have to be expressed entirely algebraically. This is how we could
do it.
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It involves concepts, such as power series in a complex variable which we will not
have seen yet. But if you are willing not to worry too much about justifying the steps you
should be able to follow.
Some functions can be expressed as infinite sums. For example the exponential
function can be written as
x2 x2
x2
ex = 1 + x + 2! + 3! + ... + n! + ...
This is an infinite sum that converges in the sense that if we take more and more terms the
finite sums converge to a limit. Here, this limit is the exact value of ex. This is the best way
to define the exponential function.
Not every series converges for all x. For example it would be foolish to define
(x) = 1 + x + x2 + ... unless |x| < 1. Even for these values it would be rather silly to make
such a definition because this is a function we already know, the sum to infinity of a
1
Geometric Progression. In fact (x) =
for |x| < 1.
1 x
Now all this can be moved into the complex plane. Indeed we define the function ez
for complex z by
z2 x3
zn
z
e = 1 + z + 2! + 3! + ... + n! + ...
It is possible to show that this converges for all x  C.
We can then consider the special case where z is purely imaginary, that is z = ix.
Then:
x2
x3 x4
x5
zn
eix = 1 + ix  2!  i 3! + 4! + i 5!  ... + inn! + ...
x2 x4
x2n
x3 x5
x2n+1




= 1  2! + 4!  ... + (1)n(2n)! + ... + i x  3! + 5!  ... + (1)n(2n + 1)! + ...




We then define cos x to be the real part of eix and define sin x to be the imaginary part.
We immediately get the fact that eix = cos x + i sin x. Also, we get the power series
expansions for sin x and cos x:
x3 x5
sin x = x  3! + 5!  ... and
x2 x4
cos x = 1  2! + 4!  ...
An immediate advantage of these definitions is that, providing we have established
the validity of differentiating a power series term by term, it is easy to show that the
derivative of sin x is cos x and the derivative of cos x is  sin x.
But, of course, there are disadvantages. We would have to prove all the basic
properties of the trigonometric functions all over again, this time algebraically. For example
we would have to prove that sin(x + y) = sin x.cos y + cos x.sin y by manipulating the power
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series rather than drawing diagrams. Even a simple result such as sin 2 = 1 is not obvious.
83 325
Does 2  3! + 5!  ... indeed converge to 1?
These proofs can be quite hard and we do not present them here. But be assured that
they can be done and that the theory of the trigonometric functions does not rest on
potentially weak arguments arising from the inspection of diagrams.
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