Download CHAPTER 14: TRIGONOMETRIC FUNCTIONS 1. Cosine and Sine

CHAPTER 14: TRIGONOMETRIC FUNCTIONS
1. Cosine and Sine in School Mathematics
In school, the cosine and sine of an angle are introduced as follows:
Suppose that A is the angle in a triangle one of whose other angles is a
right angle. The right angle is then the largest angle in the triangle and the
side opposite it, the longest side, is called the hypotenuse. The other side
bounding the angle A is called the side adjacent to A. The third side of the
triangle is then the side opposite A. The cosine of the angle A is then given
by the formula
length of adjacent
cos A =
length of hypotenuse
and the sine of A by the formula
sin A =
length of opposite
.
length of hypotenuse
When it comes to higher mathematics these definitions are not adequate for
at least two reasons.
(1) The input into these functions is an angle, not a number. The
operations of calculus apply to functions whose inputs (and outputs)
are numbers.
(2) The definitions only make sense for a limited range of angles (those
greater than 0 and smaller than a right angle). We would like to
give a definition that makes sense for all possible angles.
In school mathematics these functions are used primarily to solve problems
about triangles, but in fact their role in mathematics and its applications is
much more important than this. They arise wherever one observes periodic
phenomena in nature or mathematics. As we will see, these functions are
more properly associated to the circle rather than to triangles.
2. Radian Measure of Angles
Angles are not numbers. However, we can use numbers to measure angles.
This can be done in several ways. In the first instance, one can use degree
measurement. The angle in one complete revolution is divided into 360
degrees. This is the traditional way of measuring angles and it is adequate
for most elementary purposes. But it is not adequate for higher mathematics
(i.e when one is doing calculus). There is nothing natural about degree
measure. The number 360 is a fairly arbitrary choice, having its origins in
ancient Babylonian astronomy. If we persisted in using degrees, however,
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the resulting formulae for the derivatives of the sine and cosine functions
would involve awkward irrational constants.
The mathematically natural measure of angle is the radian and using this
measure we get the ‘correct’ (i.e. most simple) formulae in calculus.
Definition 2.1. An angle of one radian is the angle subtended at the centre
of a circle by an arc of the circle whose length is the radius.
The angle φ in the diagram is one radian.
Recall that the length of the circumference of a circle is π × diameter =
2π × radius. Thus there are 2π radians in one complete revolution.
3. Radians versus Degrees
Since there are 360 degrees in one complete revolution we can easily compare
radians and degrees:
2π radians = 360◦
360
degrees ≈ 57.296◦
1 radian =
2π
2π
1◦ =
radians ≈ 0.01745 radians
360
90◦ = π/2 radians (right angle)
180◦ = π radians (straight line)
45◦ = π/4 radians
60◦ = π/3 radians
4. The functions ‘sine’ and ‘cosine’
From now on all angles are measured in radians (unless otherwise stated).
Here is the ‘correct’ way of defining the mathematical functions ‘sine’ and
‘cosine’:
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The functions sine and cosine are the y and x coordinates respectively of a
moving point on the unit circle in the (x, y)-plane; to be more precise:
Definition 4.1. Suppose that θ is any real number. Let P be the point on
the unit circle obtained by starting at (1, 0) and turning (counterclockwise)
through an angle of θ radians.
Then the cosine of θ, cos(θ) is the x-coordinate of P , and the sine of θ,
sin(θ) is the y-coordinate of P .
Suppose that P is a point on the unit circle corresponding to an angle of θ
radians. Then P = (cos θ, sin θ) lies on the unit circle x2 + y 2 = 1, and thus
we have the following important formula relating the cosine and the sine of
θ:
(cos θ)2 + (sin θ)2 = 1 for any θ ∈ R.
We usually write cos2 θ, sin2 θ instead of (cos θ)2 or (sin θ)2 . Thus
cos2 θ + sin2 θ = 1 for all θ ∈ R
In practice, we can use this formula to determine sin θ given cos θ, or vice
versa.
Example 4.1. Suppose that we know cos θ = 3/5 and that sin θ is negative.
What is the value of sin θ?
Solution: We have 1 = cos2 θ + sin2 θ = (3/5)2 + sin2 θ = 9/25 + sin2 θ.
2
Thus
p sin θ = 1 − (9/25) = 16/25. Hence, taking square-roots, sin θ =
± 16/25 = ±4/5. Since we are given sin θ < 0, we must have sin θ = −4/5.
5. Some values of these functions
We now have two new functions whose domain is the whole real line and
whose target is the interval [−1, 1]:
cos : R → [−1, 1],
sin : R → [−1, 1]
The definition of sine and cosine above tell us how to go about calculating
the values of these functions for any input θ:
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Example 5.1. θ = π/2:
π/2 radians is one quarter of a circle. Thus if we start at the point (1, 0)
on the x-axis and rotate through π/2 radians we arrive at the point P on
the y-axis with coordinates (0, 1).
By definition, cos π/2 is the x-coordinate of the point P . So cos(π/2) = 0.
Similarly, sin π/2 is by definition the y-coordinate of the point P . So
sin(π/2) = 1
Example 5.2. θ = π:
So cos π = −1, sin π = 0
Example 5.3. θ = 3π/2:
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So cos(3π/2) = 0, sin(3π/2) = −1
Example 5.4. θ = 2π:
So cos 2π = cos 0 = 1, sin 2π = sin 0 = 0
Example 5.5. θ = 5π/2:
So cos(5π/2) = cos(π/2) = 0,
sin(5π/2) = sin(π/2) = 1
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In general
cos(θ + 2π) = cos(θ)
sin(θ + 2π) = sin(θ)
for all θ ∈ R
We say that these two functions are periodic with period 2π. This means
that the functions (and hence their graphs) repeat themselves on adjacent
intervals of length 2π.
Example 5.6.
θ = −π/2 :
Exercise: Find sin and cos of the following numbers: 7π/2, 9π, 21π, 50π,
−7π/2.
Example 5.7. θ = π/4 :
The diagram shows that the triangle is isosceles so that the x- and ycoordinates of the point P are equal. Thus cos θ = sin θ (and both arep
posi2
2
2
2
tive). So 1 =pcos θ + sin
1/2
√ θ = 2 cos θ =⇒ cos θ = 1/2 =⇒ cos θ = ± √
=⇒ cos θ = 1/2 = 1/ 2 since cos θ > 0. Thus cos π/4 = sin π/4 = 1/ 2
Example 5.8. θ = 3π/4
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√
From the diagram,
√ cos θ < 0, cos θ = − sin θ. Thus cos(3π/4) = −1/ 2,
sin(3π/4) = 1/ 2.
Exercise: Calculate cos and sin of the following numbers: θ = 5π/4, 7π/4, −π/4.
By definition, cos θ is an x-coordinate and sin θ is a y-coordinate. So it is
easy to tell when cos and sin are negative or positive:
Thus, cos is positive to the right of the y-axis, and sin is positive above the
x-axis.
Example 5.9. θ = π/3:
This is a trickier computation. We want to find the x- and y-coordinates of
the point P in the diagram:
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To do this we join P to the point (1, 0) and observe (by counting angles)
that the resulting large triangle is equilateral with side 1 = the radius of
our circle. Furthermore, the perpendicular from P to the x-axis bisects the
equilateral triangle:
It follows that the x-coordinate of P is 1/2 and thus cos π/3 = 1/2.
Knowing the value of the cosine, we can solve to find the sine:
sin π/3 > 0 and sin2 π/3 + cos2 π/3 = 1
=⇒ sin2 π/3 =p
3/4
√
=⇒ sin π/3 = 3/4 = 3/2.
√
Exercise: Show sin π/6 = 1/2, cos π/6 = 3/2. Find the sin and cos of
2π/3, −π/6, −π/3.
6. The graphs of sin and cos
Since we have calculated many values, we can easily sketch the graph y =
cos x.
For instance, since cos(0) = 1, the point (0, 1) lies on the graph. Similarly,
according to our computations above the following points lie on the graph:
(π/2, 0),√(π, −1), (3π/2,√0), (2π, 1), (5π/2, 0), (−π/2, 0), (−π, −1), (−3π/2, 0),
(π/4, 1/ 2), (−π/4, 1/ 2) . . ..
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Plotting these points and joining them together gives the graph of y = cos x
The graph is a regular wave pattern with the ‘crest’ aligned with y-axis.
The wave-length is 2π. The ‘amplitude’ of the wave is 2, since the highest
point is 1 and the lowest point is −1.
Note the ‘periodicity’ (i.e. that the pattern repeats itself over intervals of
length 2π), since cos(x + 2π) = cos x.
The graph of f (x) = sin x:
Note that this is an identically shaped graph but shifted to the right by π/2
units. (Can you explain this phenomenon?).
7. Other Trigonometric Functions
sin x
if x ∈ R and cos x 6= 0.
Definition 7.1. tan(x) =
cos x
What is the domain of tan x?
cos x = 0 if x = . . . , −3π/2, −π/2, π/2, 3π/2, . . .
So the domain is all x except x = ±π/2, ±3π/2, . . . ;i.e. the domain of tan x
is the real line with the odd multiple of π/2 removed.
It is easy to calculate values of tan x from the corresponding values of sine
and cosine:
sin 0
0
• x = 0: tan 0 =
= =0
cos 0
1
√
sin π/4
1/ 2
• x = π/4: tan π/4 =
= √ =1
cos π/4
1/ 2
√
sin(−π/4)
−1/ 2
√ = −1
• x = −π/4: tan(−π/4) =
=
cos(−π/4)
1/ 2
√
√
tan π/3 = 3, tan π/6 = 1/ 3.
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Note that tan π/2 is not defined.
What happens as x gets closer to π/2?
If x is close to π/2, then
sin x
tan x =
cos x
very close to 1
=
very close to 0
= a very large number
The closer x gets to π/2, the larger tan x. The vertical line x = π/2 is a
vertical asymptote.
The graph of f (x) = tan x:
Definition 7.2. sec x =
domain as tan x).
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for x 6= . . . − π/2, π/2, 3π/2, 5π/2, . . . (same
cos x
As for tan, the vertical lines x = π/2, x = 3π/2, x = −π/2 etc are all
vertical asymptotes of the function.
√
√
Some values of sec: sec 0 = 1/1 = 1. sec π/4 = 1/(1/ 2) = 2. sec π =
1/ − 1 = −1 . . . etc.
The graph of y = sec x
cos x
(also denoted cotan x).
sin x
The domain is every x except the points where sin x = 0;i.e. except . . . x =
−π, x = 0, ±π, ±2π, . . ..
Definition 7.3. cot x =
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√
1/ 2
cot π/2 = 0/1 = 0, cot π/4 = √ = 1 . . . etc.
1/ 2
1
(also denoted cosecx ).
sin x
It has the same domain as cot x. √
√
csc π/2 = 1/1 = 1, csc π/4 = 1/(1/ 2) = 2 . . . etc.
Exercise: Sketch the graphs of csc x and cot x.
Definition 7.4. csc x =
8. The derivatives of the trigonometric functions
We have the following two fundamental formulae:
d
sin x = cos x
dx
d
cos x = − sin x
dx
These formulae can be ‘guessed’ by examining the graphs of sin and cos:
this will be explained in the lecture. See the appendix to this chapter for a
sketch of the proof of these formulae.
The four other trigonometric functions are derived from the two basic ones
by division. So their derivatives can be found by using the quotient rule for
differentiation:
d
d sin x
(tan x) =
dx
dx cos x
cos x cos x − sin x(− sin x)
=
cos2 x
2
cos x + sin2 x
1
=
=
cos2 x
cos2 x
2
= sec x
Thus
d
(tan x) = sec2 x
dx
Similarly,
d
1
d
(sec x) =
dx
dx cos x
cos x · 0 − 1 · (− sin x)
=
cos2 x
sin x
1
sin x
=
·
=
2
cos x
cos x cos x
= sec x tan x
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d
(sec x) = sec x tan x
dx
Exercise: Show
d
(cot x) = − csc2 x
dx
d
(csc x) = − csc x cot x
dx
d
Example 8.1. Find dx
tan(x3 ).
Solution: We must use the chain rule:
d
tan(x3 ) = sec2 (x3 ) · 3x2
dx
= 3x2 sec2 (x3 )
Example 8.2. Find
Solution:
√
d
(csc( x)).
dx
√
√
√
d
1
(csc( x)) = − csc( x) cot( x) · √
dx
2 x
√
√
csc( x) cot( x)
√
= −
.
2 x
9. Appendix: The derivative of cos and sin
.
The purpose of this section is to give some idea of the proof of the following
two fundamental formulae:
d
sin x = cos x
dx
d
cos x = − sin x
dx
(You must know these formulae, but will not be examined on the rest of
the material in this section.)
We begin with a very important limit.
Consider the function f (x) = sin(x)/x. This is not defined at x = 0. What
happens as x gets closer to 0?:
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x
(sin x)/x
0.5
0.958851 . . .
0.4
0.973545 . . .
0.2
0.993346 . . .
0.1
0.998334 . . .
0.01 0.9999833 . . .
0.001 0.99999983 . . .
↓
↓
0
1
This table suggests that
sin x
= 1.
x→0 x
This can indeed be shown to be true.
[Note that if we measured our angles in degrees rather than radians, this
limit would have a different value (try this on your calculator).
To be precise: suppose we let deg sin(x) be the function deg sin x = sin(x◦ ),
then
deg sin x
π
lim
=
≈ 0.01745329 . . . .
x→0
x
180
Why is this?: x◦ = (π/180)x = πx/180 radians. Thus deg sin(x) =
sin(x◦ ) = sin(πx/180) and hence
lim
deg sin x
sin(πx/180)
= lim
x→0
x→0
x
x
π sin(πx/180)
·
= lim
x→0 180
πx/180
π
sin(πx/180)
=
lim
180 x→0 πx/180
π
π
=
·1=
.
180
180
lim
This is the reason why it is essential to use radians and not degrees in
calculus.]
Now we will compute the derivative of f (x) = sin x from first principles.
In the process, we will need to use the following formulae:
(1) There is a formula which expresses the sin of a sum of two numbers
in terms of the sin and cos of the numbers:
sin(A + B) = sin A cos B + cos A sin B.
sin(h)
= 1 (This is just the limit formula above.)
h→0
h
cos(h) − 1
(3) lim
=0
h→0
h
(2) lim
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[ Proof of (3): Multiply above and below by cos(h) + 1:
cos2 (h) − 1
h→0 h · (cos(h) + 1)
− sin2 (h)
= lim
h→0 h · (cos(h) + 1)
sin(h)
sin(h)
= − lim
·
h→0
h
cos(h) + 1
0
= −1 · = 0 ]
2
lim
Thus
d
sin(x + h) − sin x
(sin x) = lim
h→0
dx
h
sin x · cos h + cos x · sin h − sin x
= lim
h→0
h
sin x · (cos h − 1) + cos x · sin h
= lim
h→0
h
cos h − 1
sin h
= lim sin x ·
+ cos x ·
h→0
h
h
sin h
cos h − 1
+ cos x · lim
= sin x · lim
h→0
h→0
h
h
( sum and constant rules for limits )
= sin x · 0 + cos x · 1 (using 2 and 3)
= cos x
Conclusion:
d
sin x = cos x
dx
Similarly it is possible to show that:
d
cos x = − sin x
dx