Download 4 Trigonometric and Inverse Trigonometric Functions

MATH19832/19542 Mathematics 0C2/1C2.
General information for the academic year 2012-2013:
Lecturer: Dr Theodore Voronov, Room 2.109, School of Mathematics, Alan Turing
Building, email: [email protected] .
Lectures: Tuesdays 11am in Renold/D7 and Wednesdays 9am in Sackville/C14.
Tutorials: Thursdays 11am : George Begg/C3 (optometry only) and Fridays 9am : rooms
announced on the webpage (foundation).
Assessment: coursework 1 (deadline in week 4): 10%, coursework 2 (deadline in week
10): 10%, 2 hour end of semester 2 exam: 80%.
Course webpage: http://www.maths.manchester.ac.uk/service/MATH19832/ .
§4
Trigonometric and Inverse Trigonometric Functions
4.1
Recollection of trigonometric functions
4.1.1
Angular measure
Consider a circle of radius r. It is known that its circumference equals 2πr (in fact, this the
definition of the number π) and area of the disc equals πr2 .
circumf.
= 2πr
r
1
θ
r
r
rad
r
Area = πr2
Angles can be measured in degrees and in radians. Notation: a◦ or b rad. The notation
for the unit rad is commonly omitted. (So if no units are indicated, that means radians.) A
full circle is defined to be 360◦ . In particular, it follows that a right angle (a quarter of a full
circle) is 90◦ . Radians are less arbitrary units of angle because they are defined in terms of arc
length. An angle of 1 radian is defined to be the angle which makes an arc of length r on a
circle of radius r. Since the total arc length of a circle is 2πr, there are 2π radians in a circle.
So 2π rad = 360◦ . Angles are normally measured anti-clockwise from the x-axis as indicated.
1
4.1.2
Definitions of trigonometric functions
Given a right angled triangle as in the diagram:
r
y
θ
x
The side labelled r is called the hypotenuse, the side labelled x the adjacent and the side
labelled y the opposite. The following functions are defined for the variable θ:
sin θ
=
y
r
cos θ
=
x
r
tan θ
=
y
x
=
sin θ
cos θ
cosec θ
=
r
y
=
1
sin θ
sec θ
=
r
x
=
1
cos θ
cot θ
=
x
y
=
1
tan θ
Note that cos θ = sin
π
2
−θ ,
sin θ = cos
π
2
− θ . The angles θ and
π
2
− θ are called
complementary. (Hence the names: cosine, i.e., ‘cosinus’ means ‘complementi sinus’.)
These functions are called trigonometric or circular.
Alternative notations: tan θ = tg θ, cot θ = ctg θ, sec θ = sc θ, and cosec θ = csc θ.
The main trigonometric functions, which are sine and cosine, are define as above for an
acute angle θ, i.e., for 0 ≤ θ ≤ π2 . However, we may notice that cos θ and sin θ are respectively
the x- and the y-coordinates of a point on the unit circle. That immediately allows to extend
2
them to other values of the angles and in particular note their periodicity:
cos(θ + 2π) = cos θ ,
sin(θ + 2π) = sin θ ,
tan(θ + 2π) = tan θ .
4.1.3
Main identities
You should be familiar with the following result.
Theorem 1 (Pythagoras Theorem). In a right-angled triangle, the sum of the square on the
hypotenuse is equal to the sum of the squares on the other two sides, i.e.,
r 2 = x2 + y 2
where r is the length of the hypotenuse, x, y, the lengths of the other two sided.
Proof. Start with one triangle:
r
and place three more identical ones around it
3
x
y
SS
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
r
S
S
S
S
SS
y
y
S
x
The area of the outer square can be expressed as the area of the inner square plus the areas
of the four triangles:
(x + y)2 = r2 + 4 ·
1
xy
2
from which we obtain
x2 + 2xy + y 2 = r2 + 2xy ,
x2 + y 2 = r 2 ,
which is the statement of the Pythagoras Theorem.
We can use the definitions of the trigonometric functions, together with the Pythagoras
Theorem to obtain the following main identities satisfied by trigonometric functions.
Theorem 2. For all values of the argument θ,
cos2 θ + sin2 θ = 1 .
Proof. From the Pythagoras theorem we have
cos 2 θ + sin 2 θ =
x 2
r
+
y 2
r
=
4
x2 y 2
x2 + y 2
r2
+
=
=
=1
r2
r2
r2
r2
Corollary 1. For all values of θ for which the functions are defined:
1 + tan2 θ = sec2 θ
cot2 θ + 1 = cosec2 θ
Proof. These formulas are obtained by dividing throughout by cos2 θ and sin2 θ respectively.
Example 1. Three common right angled triangles are:
√
"
"
"
2
"
"
"
"
2
2
1
1
"
"
"
" ◦
" 30
√
3
30◦ =
π
6
π
= 12
6
√
cos π6 = 23
tan π6 = √13
sin
√
3
◦
60
45◦
1
45◦ =
1
60◦ =
π
4
π
= √12
4
cos π4 = √12
tan π4 = 1
sin
π
3 √
π
= 23
3
cos π3 = 21
√
tan π3 = 3
sin
They are obtained by the Pythagoras theorem. In particular, one takes an equilateral
triangle and divide it into two to get the values for trigonometric functions of π/3 and π/6.
4.1.4
Graphs of trigonometric functions
f (θ) = sin θ
1
0
2π
−1
5
f (θ) = cos θ
1
0
2π
−1
4
f (θ) = tan θ
1
0
2π
−1
−4
6
Note some useful relations:
π
− θ) = cos θ
2
π
cos( − θ) = sin θ
2
π
sin( + θ) = cos θ
2
π
cos( + θ) = − sin θ
2
sin(π + θ) = − sin θ
sin(
sin(π − θ) = sin θ
cos(π ± θ) = − cos θ
All of them can be seen from the diagram of a unit circle at the xy plane.
7
4.2
Addition formulas and further identities
4.2.1
Addition formulas
Theorem 3 (Addition formulas). For any angles A and B,
sin (A + B) = sin A cos B + cos A sin B
sin (A − B) = sin A cos B − cos A sin B
cos (A + B) = cos A cos B − sin A sin B
cos (A − B) = cos A cos B + sin A sin B
Proof. To prove sin (A + B) = sin A cos B + cos A sin B consider the following diagrams.
#
#
#
cos B##
#
#
sin A cos B
#
#
#
#
#
#
#
cos A cos B
# A
a
aa
B
aa
aa
aa
cos A aaa
a
cos A sin B
a
a
#
#
# # B
#
#
cos B#
#
#
#
#
#
p
q
#
#
A+B
#
aa
aa
aa
aa ◦
90
a
aa
aa
a
a
In both diagrams the lower angle is 90◦ − B, which proves that the angle at the top of the
p
p
right–hand diagram really is B. Now we have
= sin (A + B) and = cos B. Together
cos B
q
this gives
p
q cos B
sin (A + B) =
=
=q
cos B
cos B
and so: sin (A + B) = sin A cos B + cos A sin B as required.
To prove cos (A + B) = cos A cos B − sin A sin B, note that sin (θ + π2 ) = cos θ and
cos (θ + π2 ) = −sin θ. (This can be seen from the graphs.) So now
π
π
π
cos (A + B) = sin (A + B + ) = sin A cos (B + ) + cos A sin (B + )
2
2
2
Thus
π
cos (A + B) = sin (A + B + ) = −sin A sin B + cos A cos B
2
giving the result.
8
4.2.2
Further trigonometric identities
Other identities may be obtained from the formulas sin (A + B) and cos (A + B).
Theorem 4 (Addition formulas for tangent).
tan(A + B) =
tanA + tanB
1 − tanA tan B
tan(A − B) =
tan A − tan B
1 + tanA tan B
Proof. Use the addition formula for sine and cosine:
tan(A + B) =
sin(A + B)
sin A cos B + sin B cos A
=
=
cos(A + B)
cos A cos B − sin A sin B
sin A cos B
sin B cos A
sin A
sin B
+ cos
+ cos
tanA + tanB
cos A cos B
A cos B
cos A
B
=
=
sin A sin B
sin A sin B
1 − tanA tan B
1 − cos
1 − cos
A cos B
A cos B
The second formula follows in the same way.
Theorem 5 (Double angle formulas).
sin 2θ = 2 sin θ cos θ
cos 2θ = cos 2 θ − sin 2 θ
The latter identity may also be written
cos 2θ = 2cos2 θ − 1 = 1 − 2 sin2 θ .
Proof. Let θ = A = B in the sum identities
sin 2θ = sin (θ + θ) = sin θ cos θ + cos θ sin θ = 2sin θcos θ
cos 2θ = cos (θ + θ) = cos θ cos θ − sin θ sin θ = cos2 θ − sin 2 θ
Using the identity cos2 θ + sin2 θ = 1 to eliminate either cos2 θ or sin2 θ from the identity for
cos 2θ completes the proof.
It is possible to deduce general formulas for cos nx and sin nx. We shall not do that, but
consider particular examples instead.
Example 2. cos 3x = cos(2x + x) = cos 2x cos x − sin 2x sin x = (cos2 x − sin2 x) cos x −
2 sin x cos x sin x = (2 cos2 x − 1) cos x − 2 cos x(1 − cos2 x) = 4 cos3 x − 3 cos x. Therefore
cos 3x = 4 cos3 x − 3 cos x.
9
Example 3. cos 4x = cos(3x + x) = (4 cos3 x − 3 cos x) cos x − (3 sin x − 4 sin3 x) sin x =
4 cos4 x − 3 cos2 x − 3 sin2 x + 4 sin4 x = 4 cos4 x + 4 sin4 x − 3 = 4 cos4 x + 4(1 − cos2 x)2 − 3 =
4 cos4 x + 4(1 − 2 cos2 x + cos4 x) − 3 = 8 cos4 x − 8 cos2 x + 1 . Therefore
cos 4x = 8 cos4 x − 8 cos2 x + 1 .
From the formulas for multiple angles, we can deduce formulas for powers (useful for integration).
Example 4. Express cos3 x in terms of multiple angles. Solution. We have cos 3x = 4 cos3 x −
3 cos x, hence 4 cos3 x = cos 3x + 3 cos x and finally
cos3 x =
1
(cos 3x + 3 cos x) .
4
Theorem 6.
sin 2θ =
2 tan θ
,
1 + tan2 θ
cos 2θ =
1 − tan2 θ
.
1 + tan2 θ
Proof. Consider the standard formulas for double angle and divide by 1 = cos2 θ + sin2 θ:
sin 2θ = 2 sin θ cos θ =
θ cos θ
sin θ
2 sincos
2 cos
2 tan θ
2 sin θ cos θ
2θ
θ
=
=
.
2θ
2θ =
2
2
sin
sin
1 + tan2 θ
cos θ + sin θ
1 + cos2 θ
1 + cos2 θ
Similarly for cos 2θ:
cos2 θ − sin2 θ
1 − tan2 θ
.
=
1 + tan2 θ
cos2 θ + sin2 θ
cos 2θ = cos2 θ − sin2 θ =
Remark 1. The above group of formulas is often re-written in terms of x = 2θ and referred
to as “tangent half-angle formulas” :
sin x =
2 tan x2
1 + tan2
Theorem 7.
x
2
,
cos x =
sin X + sin Y
sin X − sin Y
cos X + cos Y
cos X − cos Y
1 − tan2
1 + tan2
x
2
x
2
.
X +Y
X −Y
= 2 sin
cos
2
2
X −Y
X +Y
sin
= 2 cos
2
2
X +Y
X −Y
= 2 cos
cos
2
2
X +Y
X −Y
= −2 sin
sin
2
2
10
Proof.
sin (A + B) = sin A cos B + cos A sin B
sin (A − B) = sin A cos B − cos A sin B
Adding,
sin (A + B) + sin (A − B) = 2 sin A cos B
Let X = A + B, Y = A − B then A = 21 (X + Y ), B = 12 (X − Y ) and
sin X + sin Y = 2 sin
X +Y
2
cos
X −Y
2
Subtracting,
sin (A + B) − sin (A − B) = 2 cos A sin B
i.e.
X +Y
2
sin X − sin Y = 2 cos
sin
X −Y
2
cos (A + B) = cos A cos B − sin A sinB
cos (A − B) = cos A cos B + sin A sinB
Adding,
cos (A + B) + cos (A − B) = 2 cos A cos B
i.e.
cos X + cos Y = 2 cos
X +Y
2
cos
X −Y
2
Subtracting,
cos (A + B) − cos (A − B) = −2 sin A sin B
i.e.
cos X − cos Y = −2 sin
X +Y
2
sin
X −Y
2
One can use the above formulas to do the converse: to express the product of sines and
cosines via the sum.
Example 5. Consider the formula
cos X + cos Y = 2 cos
X +Y
2
11
cos
X −Y
2
.
Introduce α =
X+Y
2
and β =
X−Y
2
. We have α + β = X and α − β = Y . Therefore in the new
notation the above formula becomes
cos α cos β =
1
cos(α + β) + cos(α − β) .
2
This formula is useful for integration.
Let us consider some more examples.
Example 6. Calculate the product:
cos 20◦ · cos 40◦ · cos 80◦ .
Solution: denote the product by x; multiply through by sin 20◦ ; we get
1
sin 40◦ · cos 40◦ · cos 80◦ =
2
111
1
1
11
◦
◦
sin 80 · cos 80 =
sin 160◦ = sin(180◦ − 20◦ ) = sin 20◦ ,
22
222
8
8
x · sin 20◦ = sin 20◦ · cos 20◦ · cos 40◦ · cos 80◦ =
by a repeated application of the double angle formula. Therefore
cos 20◦ · cos 40◦ · cos 80◦ =
12
1
.
8
4.3
Inverse trigonometric functions
Firstly, recall the idea of an inverse function.
Example 7. The functions y = exp x and y = ln x are mutually inverse functions. That means
the following. Suppose x > 0. If we take ln x and then apply the exponential, the we return to
the original x:
exp(ln x) = x .
Similarly, if we take first exp x and then apply logarithm, we return to the original x:
ln(exp x) = x .
The effect of one function is undone by its inverse.
The situation is not always that simple.
Example 8. Consider the functions y = x2 and y =
mutually inverse functions. Indeed,
√
x. It is natural to think that they are
√
( x)2 = x .
Note that this makes sense for x > 0 only. Consider also
√
x2 .
Here x can be both positive and negative. Is it true that
√
The symbol
x2 = x ?
√
√
x has two interpretations. If we agree that x ≥ 0 always (the “arithmetic
square root”), then

x for x ≥ 0
√
x2 =
−x for x ≥ 0
Alternatively, we may agree that
Then
√
= | x| .
x takes two values (positive and negative), e.g.,
√
4 = ±2.
√
x2 = ±x .
√
In the sequel we stick to the first interpretation; so a for any a ≥ 0 will always mean for us
√
√
the arithmetic square root, a ≥ 0, and we write ± a for the ‘general’ square root.
13
To summarize, when we solve an equation
f (x) = y
(1)
for x (so y is given), two things may happen:
1. Solution exists not for all values of y ;
2. For a given y, there may be many solutions x.
In the latter case it is often said that the “inverse function” f −1 , which sends y to x if
f (x) = y (the inverse function for a function f ) is “multi-valued”. Typically, one particular
value (particular solution) is chosen as the principal value (or principal branch). Other solutions
of the equation (1) are expressed in terms of the principal solution given by the principal value.
Example 9. The arithmetic square root
√
y (defined for y ≥ 0) is the principal value of the
“inverse function” for the function f (x) = x2 . All solutions of the equation
x2 = y
are expressed as
√
x=± y
(where
√
y is the arithmetic square root).
Now we proceed to the inverse functions of the trigonometric functions.
Definition 1. Let sin x = y. Then write x = Arcsin y (pronounced “arc sine”; capital A) for
any angle x such that sin x = y.
Similarly for cos and tan.
Note the capital letters used here for Arcsin, Arccos and Arctan. We are not obliged to use
y for the argument of the inverse functions. In the sequel we usually write Arcsin x, Arccos x
and Arctan x or use any other letter for the argument which is suitable.
Clearly, the so defined functions (called inverse trigonometric functions or circular functions) are multi-valued because, in particular, the functions sin, cos and tan are periodic. The
graphs of y = Arcsin x and y = Arccos x are shown on the next page.
14
6
6
2π
2π
π
1
π
2
−1
Principal
y = arcsin
x
Values
1
0
y = arccos x
−1
0
1
-
− 12 π
y = Arccos x
y = Arcsin x
To get a unique value, we specify the range.
Definition 2. y = arcsin x is the value of y = Arcsin x such that − 21 π ≤ y ≤ 12 π.
y = arccos x is the value of y = Arccos x such that 0 ≤ y ≤ π
y = arctan x is the value of y = Arctan x such that − 12 π ≤ y ≤ 12 π.
Remark 2. Inverse trigonometric functions “with lower case” such as arcsin, arccos and arctan
are single-valued functions regarded as the principal values of the multi-valued Arcsin, Arccos
√ √
and Arctan. (Their position is similar to that of the arithmetic square root a, a ≥ 0, in
√
relation to the double-valued square root ± a.) There is an alternative notation for them,
namely, sin−1 , cos−1 and tan−1 . Because of the danger of confusion with powers such as
sin2 x = (sin x)2 , it is preferable to use the arc notation. (Its origin is in Latin arcus, arc, so
that, e.g., ‘arc sine’ means the ‘arc length of a unit circle corresponding to a given sine’, since
arc length for a unit circle coincides with angle measured in radians.)
The general values of the inverse trigonometric functions are expressed via their principal
15
values as follows.
or π − arcsin x + 2kπ
Arcsin x = arcsin x + 2kπ
= (−1)n arcsin x + nπ
Arccos x = ± arccos x + 2kπ
(k, n = 0, ±1, ±2, ±3, . . .) ,
(k = 0, ±1, ±2, ±3, . . .) ,
(k = 0, ±1, ±2, ±3, . . .) .
Arctan x = arctan x + kπ
The argument x for Arcsin x and Arccos x should be between −1 and 1; the argument x for
Arctan x can be any number.
Example 10. Some frequently met values of the arc functions: arcsin 0 = 0, arccos 0 =
√
arctan 0 = 0, arcsin
2
2
√
= arccos
2
2
= π4 , arcsin
√
3
2
π
,
2
= π3 , arcsin 12 = π6 , arctan 1 = π4 .
Theorem 8 (Properties of inverse trigonometric functions).
arcsin(−x) = − arcsin x ,
arccos(−x) = π − arccos x ,
arctan(−x) = − arctan x .
Proof. Note that both sides of the above identities take values in the same range. To prove
them, it is sufficient to apply sin, cos and tan respectively to to both sides.
Theorem 9 (Relations between inverse trigonometric functions).
π
− arccos x .
2
x
arcsin x = arctan √
.
1 − x2
Proof. Consider the first identity. Since arccos x takes values between 0 and π, the expression
arcsin x =
π
2
− arccos x takes values between − π2 and
π
2
as is required for arcsin x. Therefore it is sufficient
to check that sine applied to the r.h.s. gives x. Indeed,
π
sin
− arccos x = cos(arccos x) = x .
2
So the l.h.s. is indeed equal to the r.h.s. For the second identity, note that the l.h.s. and r.h.s.
take values in the same range by the definition of arcsin and arctan. Therefore it is sufficient
to check that tan applied to both sides gives the same number. Indeed, if α = arcsin x, then
p
√
sin α = x and − π2 ≤ α ≤ π2 . In this range, cos α = 1 − sin2 α = 1 − x2 . Hence
tan arcsin x = tan α =
which completes the proof.
16
sin α
x
,
=√
cos α
1 − x2
Consider examples.
Example 11. Let cos α = 21 . Find all values of α.
Solution: We have α = Arccos 12 = ± arccos 21 + 2kπ = ± π3 + 2kπ. Note that − π3 + 2π =
the solution can be written in an alternative form α =
π
3
+ 2kπ or α =
5π
3
5π
,
3
so
+ 2kπ .
Example 12. Find θ in the range 0 ≤ θ < π such that tan θ = 5 .
Solution: The unique solution in the range between − π2 and
π
2
is θ = arctan 5 ≈ 1.37 (using a
calculator or tables). General solution of the equation tan θ = 5 is obtained by adding integral
multiples of π. However, adding any multiple of π to to θ = arctan 5 takes it out of the range
0 ≤ θ < π. Hence the answer is: θ = arctan 5 ≈ 1.37
Example 13. Solve for all values: cosx = 17 .
Solution. We have cos x =
√1
7
or cos x = − √17 . Therefore x = ± arccos √17 + 2kπ or x =
π ± arccos √17 + 2kπ. This combines into x = ± arccos √17 + kπ.
Example 14. Solve the equation: 2 sin2 x + 5 sin x − 3 = 0.
Solution. We factorize: (2 sin x − 1)(sin x + 3) = 0, so sin x =
equation has no solutions, so we have sin x =
1
2
1
2
or sin x = −3. The second
⇐⇒ x = (−1)k arcsin 12 + kπ = (−1)k π6 + kπ .
Example 15. Express 2 cos x + 3 sin x in the form A sin(x + x0 ) where A and x0 are to be
determined.
Solution. We can write
√
2 cos x + 3 sin x = 4 + 9
2
3
√
cos x + √
sin x
4+9
4+9
Now we look for x0 such that cos x0 =
√3
13
and sin x0 =
√
= 13
√2 .
13
2
3
√ cos x + √ sin x
13
13
√
2
13 sin(x + x0 ) where x0 = arcsin √ .
13
17
.
We can take x0 = arcsin √213 =
arccos √313 . So
2 cos x + 3 sin x =