Download Trigonometric Functions

STUDENT’S COMPANIONS IN BASIC MATH: THE FOURTH
Trigonometric Functions
Let me quote a few sentences at the beginning of the preface to a book by David
Kammler entitled ‘A First Course in Fourier Analysis’ to emphasize the importance of
trigonometric functions (or, in engineering literature, sinusoidal functions):
“You can synthesize a variety of complicated functions from pure sinusoids in much
the same way that you produce a major chord by striking nearby C, E, G keys on a
piano. . . . It was Joseph Fourier, however, who developed modern methods of using
trigonometric series and integrals as he studied the flow of heat in solids. Today, Fourier
analysis is a highly evolved branch of mathematics with an incomparable range
of applications and with an impact that is second to none.”
Due to this clear message, in selecting materials in elementary aspects of trigonometric
functions here, your FOURTH COMPANION keeps Fourier analysis in mind.
Trigonometric functions are sometimes called circular functions, because they have
something to do with a circle (just as hyperbolic functions have something to do with a
hyperbola). As you know, the ratio between the circumference C and the diameter D
of a circle is π: C/D = π. If R is the radius of the circle, then D = 2R and hence
C/D = π gives C = πD = π (2R), or
C = 2πR.
From now on we assume that R = 1, that is, the circle we are looking at is a unit
circle. Through the center O of this unit circle we draw a horizontal line and a vertical
line, giving us the x–axis and the y–axis of a cartesian coordinate system. It follows from
Pythagoras theorem that the unit circle is described by the equation
x2 + y 2 = 1
in this cartesian plane.
QUESTION 1. How do we get this equation? Explain.
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Take an arbitrary point P (x, y) on the circle. The radial line OP and the x–axis
determines an angle θ, measured by radians, that is, the size of this angle is the arc
length of the portion from the circle facing the angle. The cosine function and the sine
function at θ is just the x–coordinate and the y–coordinate respectively of point P :
cos θ = x,
sin θ = y.
Notice the basic relation: cos2 θ + sin2 θ = 1. (It is a bit unfortunate that cosine should
lead the way instead of sine, but people tend to let sine lead.)
QUESTION 2. Why do we have cos2 θ + sin2 θ = 1?
EXAMPLE 3. (In each of the following cases, draw a figure to visualize the situation.)
When P = (1, 0), we have θ = 0, x = 1 and y = 0. Hence cos 0 = 1 and sin 0 = 0.
When P = (0, 1), we have x = 0 and y = 1. In this case θ (in degrees, is 90o )
corresponds to a quarter of a circle and hence is 2π/4 = π/2 in radians. Thus we have
cos π/2 = 0 and sin π/2 = 1. Similarly, when P = (−1, 0) we have θ = π from which
we get cos π = −1 and sin π = 0. When P = (0, −1) we can take either θ = 3π/2 or
θ = −π/2 and hence cos 3π/2 = cos(−π/2) = 0 and sin 3π/2 = sin(−π/2) = −1.
EXERCISE 4. (a) Check that if θ corresponds to one-eighth of the circle, we have θ = π/4
√
√
and x = y. Deduce that x = y = 2/2 and consequently cos π/4 = sin π/4 = 1/ 2.
(b) Check that if θ is one sixth of the circle, we have θ = π/3 and 2x = 1. (Hint: draw
√
an appropriate equilateral triangle.) Deduce that cos π/3 = 1/2 and sin π/3 = 3/2.
√
(c) Verify that cos π/6 = 3/2 and sin π/6 = 1/2.
The cosine function is an even function and the sine function is an odd function. In
other words,
cos(−x) = cos x
for all x. Recall that, a function f
sin(−x) = − sin x
defined on the real line is an even function if
f (−x) = f (x) for all x and f is an odd function if f (−x) = −f (x) for all x.
QUESTION 5. Which of the following functions is an even function, an odd function, or,
neither even nor odd? (a) cos x + sin x, (b) cos x sin x, (c) sin2 x, (d) cos3 x sin3 x.
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Both sine function and cosine function are periodic functions of period 2π:
cos(x + 2πn) = cos x,
sin(x + 2πn) = sin x;
n = 0, ±1, ±2, . . .
for all x. Also, their averages over one period are zero:
1
2π
Z
2π
1
2π
cos x dx = 0,
0
Z
2π
sin x dx = 0.
0
(These identities can be understood easily from the graphs of sine and cosine functions.)
QUESTION 6. In each of the following cases, what is the period of the given function?
(a) cos 4x, (b) sin 6x, (c) cos 4x + sin 6x, (d) cos x4 , (e) sin x6 , (f) cos x4 + sin x6 .
For the next four exercises (EXERCISES 7, 8, 9, 10), you need to know a little bit of
R
R
integration, especially
cos x dx = sin x + C and sin x dx = − cos x + C.
EXERCISE 7. Verify that, if n is any nonzero integer,
Z
Z
2π
2π
sin nx dx = 0,
cos nx dx = 0.
0
0
Why do we require n to be nonzero?
The most basic identity in trigonometry is the following addition formulas:
cos(α + β) = cos α cos β − sin α sin β
sin(α + β) = sin α cos β + cos α sin β.
Other trigonometric identities are more or less consequences of these two identities.
Let us consider a special case of (1): β = α = θ. They become
cos 2θ = cos2 θ − sin2 θ
and
sin θ = 2 sin θ cos θ.
The first identity should be put together with cos2 θ + sin2 θ = 1:
cos2 θ + sin2 θ = 1
cos2 θ − sin2 θ = cos 2θ
3
(1)
If we take the sum and the difference of these two identities and then divide by 2, we get
cos2 θ =
1 + cos 2θ
2
sin2 θ =
1 − cos 2θ
.
2
(2)
What is the big deal of the above identities? Well, they tell you how to find the indefinite
R
R
R
integrals
cos2 x dx and
sin2 x dx from the known identities
cos x dx = sin x + C
R
and
sin x dx = − cos x + C.
EXERCISE 8. Find
EXERCISE 9. Find
R
R
cos2 x dx and
cos4 x dx and
R
R
sin2 x dx.
sin4 x dx.
(Hint: cos4 x = (cos2 x)2 = · · · )
Replacing β by −β in (1), and noticing cos(−β) = cos β, sin(−β) = − sin β, we obtain
cos(α − β) = cos α cos β + sin α sin β
sin(α − β) = sin α cos β − cos α sin β
(3)
Adding or subtracting various identities from (1) and (3) we can convert “products in
cosines and sines” to “sums or differences in cosines and sines”. The detail is left to you
as the following exercise.
EXERCISE 10. Deduce from (1) and (3) the following
1
cos α cos β = (cos(α − β) + cos(α + β))
2
1
sin α sin β = (cos(α − β) − cos(α + β))
2
1
sin α cos β = (sin(α + β) + sin(α − β))
2
From these identities and EXERCISE 7 deduce that, for all integers m, n
Z
2π
sin mx cos nx dx = 0
(4)
0
and for all integers m, n with m 6= n,
Z
Z
2π
2π
cos mx cos nx dx = 0,
sin mx sin nx dx = 0.
0
0
Why do we need to specify m 6= n in the last two identities?
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(5)
Identities (4) and (5) constitute the orthogonality relations for trigonometric functions, which are fundamental in Fourier analysis.
The most basic fact about trigonometric functions for differential calculus is, the limit
of sin h/h is 1 as h tends to zero:
lim
h→0
sin h
= 1.
h
(6)
Here, h is “dummy” in the sense that nothing is changed if we replace h by another
letter in all of its occurences. For example, we can rewrite (6) as limx→0 sin x/x = 1.
EXERCISE 11. Use (6) to derive
d
sin x = cos x
dx
and
d
cos x = − sin x.
dx
(Note that, according to the definition of derivative,
d
sin(x + h) − sin x
sin x = lim
.
h→0
dx
h
Now you realize that (1) is needed.
PROBLEM 12. What makes you think that (6) is correct?
(Hint: The total length of all
sides of a regular n–gon inscribed in a unit circle tends to 2π as n → ∞.)
EXERCISE 13. Find the limit of
1 − cos h
as h → 0.
h2
(Hint: use the following trick:
1 − cos h
(1 − cos h)(1 + cos h)
sin2 h
1
=
=
.
2
2
2
h
h (1 + cos h)
h 1 + cos h
If you know L’Hospital’s rule, you can avoid this trick.)
EXERCISE 14. According to Newton’s second law and Hooke’s law, the motion of a
mass–spring system is described by the differential equation
m
d2 x
+ kx = 0
dt2
(7)
where m is the mass and k is Hooke’s constant. Both m and k are positive numbers.
Check that
x = A sin(at) + B cos(at)
5
r
k
is the frequency and
m
A, B are arbitrary constants. Verify that this solution can be rewritten as M sin(at + b),
√
where M = A2 + B 2 is the amplitude and b is an appropriate constant.
(representing oscillatory motion) is a solution to (7), where a =
In Fourier analysis, one crucial step is to find a closed form for the sum
Sn = 1 + 2(cos x + cos 2x + cos 3x + · · · + cos nx)
(8)
where n is an arbitrary positive integer. A VERY SLICK way to deal with it is to see
what happens if both sides are multiplied by sin(x/2):
³
x´
x
x
x
x
x
sin
Sn = sin + 2 sin cos x + 2 sin cos 2x + sin cos 3x + · · · + 2 sin cos nx.
2
2
2
2
2
2
Starting with the second term, apply the recipe 2 sin α cos β = sin(α + β) + sin(α − β) (see
EXERCISE 10) to each term. Then we realize that many terms can be canceled. After
some simplification and rearranging, we arrive at
sin
Sn ≡ 1 + 2(cos x + cos 2x + cos 3x + · · · + cos nx) =
(2n + 1)x
2
.
x
sin
2
PROBLEM 15. Follow the suggestion here to derive the last identity. (A more natural
way to derive this identity is to regard cos kx as the real part of eikx so that Sn given by
(8) becomes a geometric progression. See EXAMPLE 21 from the SIXTH COMPANION.)
PROBLEM 16. By applying sin 2θ = 2 sin θ cos θ repeatedly, we have
1 = sin
π
π
π
π
π
π
= 2 cos sin = 22 cos cos sin = · · ·
2
4
4
4
8
8
By noticing that limn→∞ 2n sin π/2n = π, derive
2
π
π
π
= cos cos
· · · cos n · · ·
π
4
8
2
(an infinite product)
√
1 + cos 2θ
2 + 2 cos 2θ
as cos θ =
. Starting
Rewrite the double angle formula cos θ =
2
2
√
with cos π/4 = 2/2, use the last identity to churn out:
q
p
p
√
√
√
2+ 2+ 2
π
π
π
2
2+ 2
, cos =
, cos
=
, ...
cos =
4
2
8
2
16
2
2
6
Hence derive the following
π
2
2
2
=√ p
√ q
p
√ ······
2
2 2+ 2
2+ 2+ 2
which is called Vieta’s formula for π/2.
The tangent function is the quotient ‘sine over cosine’ (in some books tg x is used
instead of tan x) and the secant function is the reciprocal of cosine
tan x =
sin x
cos x
sec x =
1
.
cos x
Notice that the cosine function vanishes at half-integral multiples of π; in other words,
cos(n + 21 )π = 0 for all integers n. So both of tan x and sec x are undefined for
x = (n + 12 )π.
QUESTION 17. Why do we have 1 + tan2 x = sec2 x?
EXERCISE 18. Use (1) to verify the addition formula for the tangent function:
tan(α + β) =
Use this identity and the fact that lim
h→0
tan α + tan β
.
1 − tan α tan β
tan h
sin h
= lim
cos h = 1 to deduce that
h→0
h
h
d
tan x = 1 + tan2 x ≡ sec2 x.
dx
sin h
1 − cos h
1
d
= 1, lim
=
to deduce
sec x = sec x tan x. (Cer2
h→0
h→0
h
h
2
dx
tainly, these two identities can also deduced from (d/dx) sin x = cos x and (d/dx) cos x =
Also, use lim
− sin x by means of the quotient rule.)
A rather amazing fact about the tangent function is that all trigonometric functions
sin t, cos t, tan t, sec t etc. can be expressed as a rational function of tan(t/2). Indeed,
sin t = 2 sin
and
sin 2t
tan 2t
2 tan 2t
t
t
2 t
cos = 2
cos
=
2
=
2
2
2
cos 2t
sec2 2t
1 + tan2
1 − tan2
t
2 t
2 t
2 t
− sin
= cos (1 − tan ) =
cos t = cos
2
2
2
2
sec2 2t
2
7
t
2
t
2
1 − tan2 2t
=
.
1 + tan2 2t
EXERCISE 19. Verify that, setting u = tan 2t , we have
sin t =
2u
1 − u2
2u
1 + u2
2
,
cos
t
=
,
tan
t
=
,
sec
t
=
, dt =
du.
2
2
2
2
1+u
1+u
1−u
1−u
1 + u2
These identities can be used to integrate all rational functions of trigonometric functions.
But this topic, interesting though, is normally not covered in a calculus course today.
The identities in the above exercise seem to be rather mysterious. Actually, they can be
visualized by means of so–called stereographic projection. Again, consider the unit
circle x2 + y 2 = 1 in the cartesian plane. The topmost point on the circle is N = (0, 1)
and let us call it the North Pole. Take any point P (x, y) on the circle and let Q(u, 0)
be the intersection of the line N P and the x–axis. Then we have
u=
x
,
1−y
x=
2u
,
2
u +1
y=
u2 − 1
.
u2 + 1
(9)
(Draw a picture!)
EXERCISE 20. Verify the above identities. (Hint: To verify u = x/(1 − y), use the fact
that N , P and Q are colinear. Then compute 1 + u2 = ((1 − y)2 + x2 )/(1 − y)2 = · · ·
and use the fact that x2 + y 2 = 1.)
Let t be the angle between OP and OS, where S is the is the South Pole (0, −1).
Then the angle between N P and N S is t/2.
EXERCISE 21. Check that u = tan t/2, x = sin t and y = − cos t. Then use (9) to verify
the identities in EXERCISE 19.
The identities
u2 − 1
y= 2
u +1
2u
x= 2
,
u +1
gives us a rational parameterization of the unit circle. A direct computation shows
µ
2
2
x +y =
2u
1 + u2
¶2
µ
+
u2 − 1
1 + u2
¶2
= 1.
Notice that, if u is a rational number, so are x and y, and henceforth a rational
point (x, y) on the unit circle is produced. All rational points on the unit circle can be
obtained in this way. Indeed, if x and y are rational numbers, then so is u = x/(1 − y).
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A rational point gives us a Pythagoras triple and vice versa. (A triple (a, b, c) of integers
is called a Pythagoras triple if a2 + b2 = c2 , e.g. (3, 4, 5).) For example, letting u = 4,
we have x = 2u/(u2 + 1) = 8/17, y = (u2 − 1)/(u2 + 1) = 15/17 and hence we obtain the
Pythagoras triple (8, 15, 17). You can easily check that 82 + 152 = 172 (= 289). You may
secretly take a large u to produce a huge triple to perplex your friends.
One amazing identity in calculus concerning trigonometric functions is the identity
d
ln | sec x + tan x| = sec x.
dx
(10)
It is amazing because normally we expect that the derivative of a function looks more
complicated, but in this case it is simpler! We need this in calculus because it tells us what
the integral of sec x is:
Z
sec x dx = ln | sec x − tan x| + C.
Identity (10) is quite straightforward to check by means of the chain rule and the identities
d
dx
tan x = sec2 x and
d
dx
sec x = tan x sec x. The expression ln | sec x + tan x| looks
mysterious. Actually it occurs in the parametric equations of a curve called tractrix. If
you tow an object on the y–axis by a string and walk along the x–axis, the curve traced
out by the object is a tractrix. The surface of revolution obtained by rotating this curve
about the x–axis is called a pseudo–sphere. It has a constant negative curvature. (A usual
sphere is a surface of a constant positive curvature.)
EXERCISE 22. Check (10).
R
R
R
R
QUESTION 23. What are
cos x dx, sin x dx, tan x dx and sec x dx?
Finally, we consider inverse trigonometric functions arcsine, arccosine and arctangent.
Strictly speaking, none of the trigonometric functions is invertible. But this does not
bother us, if we are willing to narrow down their domains with care. For example, the
function x = u2 is not invertible. This is because different u–values may give the same
x, e.g. both u = 2 and u = −2 give x = 4. By if we impose the restriction that u is
√
nonnegative, then u is uniquely determined by x and we can write u = x. To express
√
in a proper way, we say that u = x is the inverse function of x = u2 restricted to
[0, ∞), the set of nonnegative numbers. Now we define
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u = Arcsin x = the inverse function of x = sin u restricted to [− π2 ,
π
2 ].
u = Arccos x = the inverse function of x = cos u restricted to [0, π].
u = Arctan x = the inverse function of x = tan u restricted to (− π2 ,
π
2 ).
You should draw a picture in each case to see what is behind each restriction. To
indicate the domain and the range of each of these inverse functions, we write:
h π πi
,
Arcsin : [−1, 1] → − ,
2 2
Arccos : [−1, 1] → [0, π],
³ π π´
Arctan : R → − ,
.
2 2
The derivatives of these inverse functions are
1
d
Arcsin x = √
,
dx
1 − x2
1
d
Arccos x = − √
,
dx
1 − x2
1
d
Arctan x =
.
dx
1 + x2
EXERCISE 24. Verify the above identities.
The most fascinating inverse trig–function is arctangent. Its derivative 1/(1 + x2 )
looks like a bell curve but it is not (because of its “big tails” on both sides). The total
area of the region underneath this curve and above the x–axis is
Z
∞
−∞
dx
≡ lim
M →∞
1 + x2
Z
M
−M
¯M
dx
π ³ π´
¯
=
lim
Arctan
x
=
− −
= π.
¯
M →∞
1 + x2
2
2
−M
We normalize 1/(1 + x2 ) by dividing π. The function
f (x) =
1
1
π 1 + x2
obtained in this way is called the Cauchy distribution in probability theory.
There are some intriguing identities giving us π in terms of arctangents:
1
1
− 4Arctan
(Machin’s formula, 1706)
5
239
1
1
1
π = 12Arctan + 4Arctan
+ 4Arctan
. (Ferguson’s identity)
4
20
1985
π = 16Arctan
PROBLEM 25. Verify the above identities.
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