Download Notes on Trigonometric Functions for Math 126 1 The Trigonometric

Notes on Trigonometric Functions for Math 126
In this notes we shall extend our study of calculus to an important class of functions
called the trigonometric functions. These functions are widely used in natural sciences
to study periodic or rhythmic phenomena such as motions, sound waves, business cycles,
etc.
1
The Trigonometric Functions
Angles
An angle is formed when one ray in the plane is rotated into the other about their
common endpoint. The angle generated is a positive angle if the ray is rotated counterclockwise, and a negative angle if the ray is rotated clockwise.
The size of an angle can be represented in either degrees or radians. We have used
“degrees” for years in high school. They are simple. How big is a radian? A radian
is the measure of the central angle angle corresponding to an arc length of 1 in a unit
circle (a unit circle is the circle centered at the origin with radius 1). Since most of you
are probably more comfortable using degrees, we must deal with the key issue: how to
convert degrees to radians and vice versa? Here is the catch: The circumference of the
unit circle is 2πr, but r = 1, so is 2π. Since there are 360 degrees in one full revolution,
we have
360 degrees = 2π radians
so that
1 degree =
π
radians
180
1 radian =
180
degrees
π
and
For example,
30 degrees
5π
radians
6
−120 degrees
π
π
= radians
180
6
π
π
= 45 ·
= radians
180
4
π
2π
= (−120) ·
=−
radians
180
3
= 30 ·
In calculus all angles are in radian measurement. Remember to set your calculators to
radian mode.
The Trigonometric Functions
Let P (x, y) be a point on the unit circle so that the radius OP forms an angle of θ
radians (0 ≤ θ < 2π) with respect to the positive x-axis. We define sine of θ, denoted
by sin θ, to be the y-coordinate of P . Similarly the cosine of θ, written as cos θ, is
1
defined to be the x-coordinate of P . The other trigonometric functions are defined in
terms of sine and cosine functions.
Trigonometric Functions
If P is a point on the unit circle and the coordinates of P are (x, y), then
cos θ = x
and
sin θ = y
y
sin θ
=
(x 6= 0)
x
cos θ
1
1
csc θ = =
(y 6= 0)
y
sin θ
tan θ =
sec θ =
1
1
=
x
cos θ
(x 6= 0)
cot θ =
x
cos θ
=
y
sin θ
(y 6= 0)
As we work with trigonometric functions, it is useful to remember the values of the
sine, cosine, and tangent of some important angles, such as θ = 0, π/6, π/4, π/3, π/2,
and so on. These values can be found using elementary trigonometry. Following table
summarizes the values of the sine, cosine and tangent of some of these angles.
θ in radians
0
sin θ
0
cos θ
1
tan θ
0
π
6
1
√2
3
2
1
√
3
π
√4
2
√2
2
2
1
π
√3
3
2
1
2
√
3
π
2
1
0
∞
2π
√3
3
2
1
−
2
√
− 3
3π
√4
2
2
√
2
−
2
−1
5π
6
1
2
√
3
2
1
−√
3
−
π
0
−1
0
Note that the signs of trigonometric functions depend on the quadrant in which the point
P (x, y) lies. For example, if P (x, y) lies in the second quadrant, the first coordinate x
is negative, and the second coordinate y is positive, so cosine and tangent are negative,
whereas sine is positive. In the first quadrant, all trigonometric functions are positive. In
the second quadrant, only sine function is positive. In the third quadrant, only tangent
function is positive. In the fourth quadrant, only cosine function is positive. We usually
refer to this result as the ASTC rule, which can be remembered as the mnemonic “All
Students Take Calculus”.
2
Graphs of Trigonometric functions
Observe that the trigonometric functions repeat their values in a regular fashion. Let
P (x, y) be a point on the unit circle so that the radius OP forms an angle of θ radians
with the positive x-axis. Since the rotation of the radius OP by 2π radians leaves it in
its original configuration, we see that
sin(2π + θ) = sin θ,
cos(2π + θ) = cos θ
That is, the sine and cosine functions are periodic with period 2π. Note that sine and
cosine functions are defined for every real number in R so the domain of sine (cosine)
functions is (−inf ty, +∞). To draw the graphs of sine and cosine functions, it suffices
to focus on one period and repeating it as necessary.
The graph of sine function
y
6
1 r
r
− π2
r
y = sin θ
π
2
r
r
π
r
r
3π
2
2π
r -
5π
2
θ
−1 r
The graph of cosine function
y
6
1 r
y = cos θ
r
− π2
r
π
2
r
r
π
r
r
3π
2
2π
r
5π
2
-
θ
−1 r
To sketch the graph of tangent function, recall that tan θ = sin θ/ cos θ, so tangent is
π
not defined when cos θ = 0, that is, when θ = + nπ (n = ±1, ±2, · · · ). We can show
2
π
that vertical lines with equation θ = + nπ (n = ±1, ±2, · · · ) are vertical asymptotes
2
of the tangent function y = tan θ. Also observe that the tangent function is periodic
with period π. Now we can sketch the graph of y = tan θ.
3
y
6
y = tan θ
r
− 3π
2
r
r
−π
r
− π2
π
2
r
r
π
r
-
3π
2
θ
Transformations of the Graphs of Trigonometric Functions
We now consider the graphs of functions that are transformations of the sine and cosine
functions. It’s traditional to use the letter x to denote the independent variable in the
domain of a function. So from here on we shall use x and write y = sin x, y = cos x, etc.
Suppose we are given the sine and cosine functions
y = A sin B(x − C)
and y = A cos B(x − C)
The number |A| is called the amplitude and is the largest value these functions attain. If
A < 0, then the graphs of basic sine and cosine functions reflect in the x-axis. Horizontal
1
, B 6= 0. The graphs of the basic sine and cosine functions are
stretching is given by
|B|
translated horizontally right by C units if C > 0, and translated horizontally left by |C|
units if C < 0. For example,
sin(x +
π
) = cos x
2
sin(x + π) = − sin x
y
6
1 r
y = sin(x)
y = sin(x + π2 )
r
− π2
r
π
2
r
r
π
r
r
3π
2
2π
r -
5π
2
x
y = sin(x + π)
−1 r
4
Similarly we have
cos(t +
π
) = − sin t
2
cos(t + π) = − cos t
Inverse Trigonometric Functions
By their nature of being periodic functions, trigonometric functions are not one-to-one.
However, we can restrict the domain so that the trigonometric function is one-to-one
and then we can find its inverse function.
Trigonometric Inverse Functions
Range
h π πi
− ,
2 2
Domain
y = arcsin x
[−1, 1]
y = arccos x
[−1, 1]
y = arctan x
[−∞, ∞]
[0, π]
h π πi
− ,
2 2
Note that the inverse sine functions is sometimes denoted by y = sin−1 x. But when
this notation is used, do not confuse with y = (sin x)−1 . The inverse sine function is
y = sin−1 x, while the reciprocal of sine function is
y = (sin x)−1 =
1
= csc x
sin x
The following relations follows from the inverse function properties.
sin(arcsin x) = x
cos(arccos x) = x
for − 1 ≤ x ≤ 1,
arcsin(sin x) = x
for − 1 ≤ x ≤ 1,
arccos(cos x) = x
for −
π
π
≤x≤ ,
2
2
for 0 ≤ x ≤ π,
1
Example: Evaluating the inverse trigonometric functions (a) arcsin , (b) arccos 0
2
Solution:
(a) The number in the interval [−π/2, π/2] whose sine is 1/2 is π/6. Thus arcsin(1/2) =
π/6.
(b) The number in the interval [0, π] whose cosine is 0 is π/2. Thus arccos 0 = π/2. 5
Trigonometric Identities
Equations expressing the relationship between trigonometric functions are called trigonometric identities. Some important trigonometric identities are listed in the following
table.
Trigonometric Identities
Pythagorean Identities
sin2 θ + cos2 θ = 1,
tan2 θ + 1 = sec2 θ
Even-Odd Identities
sin(−θ) = − sin θ,
cos(−θ) = cos θ,
tan(−θ) = − tan θ
Double-Angle Formulas
sin 2θ = 2 sin θ cos θ,
Reduction Formulas
π
− θ = cos θ,
sin
2
π
sin
+ θ = cos θ,
2
cos 2θ = cos2 θ − sin2 θ
cos
cos
π
2
π
2
− θ = sin θ
+ θ = − sin θ
6
2
Derivatives of Trigonometric Functions
In this section we develop rules for differentiating trigonometric functions. To derive the
rule for differentiating the sine function, we need the following two important limits.
sin h
=1
h→0 h
lim
cos(h) − 1
=0
h→0
h
and
lim
Since the sine and cosine functions are periodic, their derivatives must be periodic also.
Let’s look at the graph of f (x) = sin x and estimate the derivative graphically.
y
6
1 r
r
− π2
r
f (x) = sin x
π
2
r
r
π
r
r
3π
2
2π
r -
5π
2
θ
−1 r
π
Observe that, from the graph, the derivative of f (x) = sin x is zero at x =
and
2
3π
x=
. Using the definition of the derivative, we have the derivative of f (x) = sin x at
2
x=0
sin h
sin(0 + h) − sin 0
= lim
=1
f 0 (0) = lim
h→0 h
h→0
h
the derivative of f (x) = sin x at x = π
sin(π + h) − sin π
− sin h
= lim
= −1
h→0
h
h
and the derivative of f (x) = sin x at x = 2π
f 0 (0) = lim
h→0
sin(2π + h) − sin 2π
sin h
= lim
=1
h→0
h→0 h
h
Now we have the graph of the derivative of f (x) = sin x
f 0 (0) = lim
y
6
1 r
f 0 (x)
r
− π2
r
π
2
r
r
π
r
r
3π
2
2π
−1 r
7
r
5π
2
-
θ
This graph looks like the graph of cosine function. In fact, it can be proved that the
derivative of the sine is the cosine by using sine rule for sums (sin(α + β) = sin α cos β +
cos α sin β) and the definition of the derivative.
Similarly we can draw the graph of the derivative of f (x) = cos x.
y
6
1 r
f (x) = cos x
r
− π2
r
π
2
r
r
π
r
r
3π
2
2π
r
-
5π
2
θ
−1 r
y
6
1 r
r
− π2
r
f 0 (x)
π
2
r
r
π
r
r
3π
2
2π
r -
5π
2
θ
−1 r
The graph of the derivative of the cosine looks exactly like the graph of sine, except
reflected about the x-axis.
d
Example: Prove that
cos x = − sin x
dx
π
(that is,
Proof: Since the cosine function is the sine function shifted to the left by
2
cos x = sin(x + π/2)), we differentiate sin(x + π/2) using Chain Rule:
d
d
cos x =
sin(x + π/2) = cos(x + π/2)
dx
dx
But cos(x+π/2) is the cosine shifted to the left by π/2, which gives a sine curve reflected
about the x-axis. So we have
d
cos x = cos(x + π/2) = − sin x
dx
8
Example: Differentiate the following functions
1. 2 sin(3θ)
2. cos2 x
3. cos(x2 )
4. e− sin x
Solution:
1.
d
d
d
2 sin(3θ) = 2 (sin(3θ)) = 2 cos(3θ) (3θ) = 2(cos(3θ))3 = 6 cos 3θ
dθ
dθ
dθ
2.
d
d
d
(cos2 x) =
(cos x)2 = 2(cos x) cos x = 2 cos x(− sin x) = −2 sin x cos x
dx
dx
dx
3.
d
d
(cos(x2 )) = − sin(x2 ) (x2 ) = −2x sin(x2 )
dx
dx
4.
d − sin x
d
e
= e− sin x (− sin x) = e− sin x (− cos x) = −e− sin x cos x
dx
dx
Since tan x = sin x/(cos x), we differentiate tan x using the quotient rule. We have
d
(sin x)0 cos x − sin x(cos x)0
cos2 x + sin2 x
1
tan x =
=
=
= sec2 x
2
2
dx
cos x
cos x
cos2 x
The derivatives of the sine, the cosine and the tangent
d
sin x = cos x
dx
d
cos x = − sin x
dx
d
1
tan x =
= sec2 x
dx
cos2 x
Derivative of Inverse Trigonometric Functions
To find the derivative of arctan x, we use the identity tan(arctan x) = x. Differentiating
using Chain Rule gives
d
tan(arctan x) = 1
dx
that is,
d
sec2 (arctan x) (arctan x) = 1
dx
9
so
d
1
(arctan x) =
2
dx
sec (arctan x)
Using the identity 1 + tan2 θ = sec2 θ, and replace θ by arctan x, we have
d
1
1
(arctan x) =
=
2
dx
1 + tan (arctan x)
1 + x2
By a similar argument, we obtain the derivative of arcsin x and arccos x.
The derivative of the inverse trigonometric functions
d
1
, |x| < 1
arcsin x = √
dx
1 − x2
d
1
arccos x = − √
, |x| < 1
dx
1 − x2
1
d
arctan x =
dx
1 + x2
Practice Questions:
Find the derivatives of the following functions
1. f (θ) = sin θ + cos θ
2. g(θ) = cos θ sin θ
3. f (x) = sin(2 − 7x + 3x2 )
4. f (y) = esin y
5. R(θ) = θ ln(cos θ)
√
6. f (x) = 1 − tan x
7. g(x) = cos(sin x)
1
x
8. w(θ) = e sin
x
9. f (w) = arcsin(w2 )
10. g(t) = ln(cos(t − 1))
11. r(x) = arctan(
x
)
x+1
10
Use implicit differentiation to find dy/dx.
12 sin(x + y) = 2xy
13 ecos y = x2 sin x
14 sin(2y) + cos(3x) = y
In the following questions you are given f 0 (x). In each case, find all critical points (i.e.,
f 0 (x) = 0) and show which are relative maxima and which are relative minima. You
may use first or second derivative test.
15 f 0 (x) = cos(2x),
0 ≤ x ≤ 2π
16 f 0 (x) = ln(sin x),
0≤x≤π
17 f 0 (x) = cos(ex ),
0 ≤ x ≤ ln(2π)
Solutions:
1. f 0 (θ) = cos θ + (− sin θ) = cos θ − sin θ
2. g 0 (θ) = (− sin θ) sin θ + cos θ(cos θ) = cos2 θ − sin2 θ = cos(2θ)
3. Using Chain Rule, let u = 2 − 7x + 3x2 , so f = sin u. Then we have f 0 (x) =
df du
·
= cos u · (−7 + 6x) = (6x − 7) cos(2 − 7x + 3x2 ).
du dx
4. Using Chain Rule, let u = sin y, so f = eu . It follows f 0 (y) =
df du
·
= eu ·(cos y) =
du dy
esin y cos y.
d
5. Use Product Rule and Chain Rule. R0 (θ) = 1 · ln(cos θ) + θ (ln(cos θ)) =
dθ
1
θ sin θ
ln(cos θ) + θ
· (− sin θ) = ln(cos θ) −
.
cos θ
cos θ
d
Note that to find
(ln(cos θ)), we need to let u = cos θ, so ln(cos θ) = ln u and
dθ
then apply Chain Rule.
√
6. Use Chain Rule. Let u = 1 − tan x, f (x) = u = u1/2 .
1
− sec2 x
f 0 (x) = u−1/2 · (0 − sec2 x) = √
2
2 1 − tan x
7. Use Chain Rule. Let u = sin x. Then g = cos u.
g 0 (x) = − sin(sin x) · cos x
11
8. Use Product Rule and Chain Rule.
1
1
1
1
ex
1
0
x
x
x
w = e sin
+ e · cos
· − 2 = e sin
− 2 cos
x
x
x
x
x
x
9. Let u = w2 . f = arcsin u.
f0 = √
1
2w
· 2w = √
1 − u2
1 − w4
10. Let u = t − 1 and v = cos u, so g = ln v. Use Chain Rule.
dg
dg dv du
1
sin(t − 1)
=
·
·
= · (− sin u) · 1 = −
= − tan(t − 1)
dt
dv du dt
v
cos(t − 1)
11. Let u =
x
, r = arctan u. Use Chain Rule
x+1
r0 =
1
(x + 1) − x
1
·
=
2
2
1+u
(x + 1)
(x + 1)2 + x2
12. differentiating both sides with respect to x, we have
cos(x + y) · (1 + y 0 ) = 2y + 2xy 0
solving y 0
y0 =
2y − cos(x + y)
cos(x + y) − 2x
13. differentiating both sides with respect to x, we have
ecos y · (− sin y) · y 0 = 2x sin x + x2 cos x
solving y 0
y0 = −
2x sin x + x2 cos x
(sin y)ecos y
14. differentiating both sides with respect to x, we have
cos(2y) · 2y 0 − sin(3x) · 3 = y 0
solving y 0
y0 =
3 sin(3x)
2 cos(2y) − 1
15. Let f 0 = 0, we have cos(2x) = 0. It follows that 2x = π/2, 3π/2, 5π/2, 7π/2, etc.
that is, x = π/4, 3π/4, 5π/4, 7π/4, etc. Over the interval x ∈ [0, 2π], there are four
critical values, x = π/4, 3π/4, 5π/4 and 7π/4.
Now we use the second derivative test.
f 00 = − sin(2x) · 2 = −2 sin(2x)
12
At x = π/4, f 00 (π/4) = −2 sin(π/2) = −2 < 0, so the function has a relative
maximum at x = π/4.
At x = 3π/4, f 00 (3π/4) = −2 sin(3π/2) = 2 > 0, so the function has a relative
minimum at x = 3π/4
At x = 5π/4, f 00 (5π/4) = −2 sin(5π/2) = −2 < 0, so the function has a relative
maximum at x = 5π/4
At x = 7π/4, f 00 (7π/4) = −2 sin(7π/2) = 2 > 0, so the function has a relative
minimum at x = 7π/4
16. Let f 0 = 0, we have ln(sin x) = 0. It follows that sin x = 1, so x = π/2, 3π/2, 5π/2
etc. Over the interval x ∈ (0, π), there is only one critical value, x = π/2. We
check if we can use the second derivative test.
f 00 =
cos x
1
· cos x =
sin x
sin x
at x = π/2, f 00 (π/2) = 0, so the second derivative test is inconclusive. Now we have
to use the first derivative test. Since sin x < 1 if x 6= π/2, so f 0 (x) = ln(sin x) < 0
for x < π/2 and x > π/2. Hence x = π/2 is neither a relative maximum nor a
relative minimum.
17. Let f 0 = 0, we have cos(ex ) = 0. It follows that ex = π/2, 3π/2, 5π/2, 7π/2 etc,
that is, x = ln(π/2), ln(3π/2), ln(5π/2). ln(7π/2), etc. Therefore over the interval
x ∈ [0, ln(2π)], there are two critical values x = ln(π/2) and ln(3π/2). Use the
second derivative test.
f 00 = − sin(ex ) · ex = −ex sin(ex )
At x = ln(π/2), (also note that eln(π/2) = π/2,)
π
π
π
f 00 (ln(π/2)) = −eln(π/2) sin(eln(π/2) ) = − sin( ) = − < 0
2
2
2
so x = π/2 gives a relative maximum.
At x = ln(3π/2), (also note that eln(3π/2) = 3π/2,)
f 00 (ln(3π/2)) = −eln(3π/2) sin(eln(3π/2) ) = −
so x = 3π/2 gives a relative minimum.
13
3π
3π
3π
sin( ) =
>0
2
2
2