Download §2 Trigonometric functions

Mathematics 1G1. Autumn 2013
Theodore Voronov.
§2
Trigonometric functions
Trigonometry provides methods to relate angles and lengths but the functions we define have
many other applications in mathematics.
2.1
Measuring angles
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.
This gives us a formula for converting between the two measurements.
angle in degrees =
For example 90◦ =
2.2
π
π
rads, 30◦ = rads.
2
6
180
× (angle in radians)
π
Definitions of trigonometric functions
Let ABC be a right angled triangle containing the angle θ (exercise: draw and label a triangle
to fit with the information below).
We define
opposite
AC
=
hypotenuse
AB
adjacent
BC
cos(θ) =
=
hypotenuse
AB
opposite
AC
tan(θ) =
=
.
adjacent
BC
sin(θ) =
Note that
sin(θ)
.
cos(θ)
The main trigonometric functions, which are sine and cosine, are defined as above for an
acute angle θ, i.e., for 0 6 θ 6 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
them to other values of the angles and in particular note their periodicity:
tan(θ) =
cos(θ + 2π) = cos θ ,
sin(θ + 2π) = sin θ ,
tan(θ + 2π) = tan θ .
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
Note some further 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(π − θ) = sin θ
cos(π ± θ) = − cos θ
sin(
All of them can be seen from the diagram of a unit circle at the xy plane.
Sine and cosine values lie between -1 and +1. Tangent values can take any value (and are
undefined for certain values of θ).
Note that cos θ = sin π2 − θ , sin θ = cos π2 − θ . The angles θ and π2 − θ are called
complementary. (Hence the names: cosine, i.e., ‘cosinus’ means ‘complementi sinus’.)
Useful special values:
√
3
1
◦
◦
◦
◦
sin(60 ) = cos(30 ) =
sin(30 ) = cos(60 ) = ,
2
2
1
sin(45◦ ) = cos(45◦ ) = √ ,
2
tan(45◦ ) = 1
Example. Convert the following from radians to degrees:
3π
= ... .
4
Examples. Find the exact value of the following:
2π
π
(i) sin(135o )
(ii) cos( )
(iii) tan(− )
3
3
Besides the main trigonometric functions, sine, cosine and tangent (in fact, it is sufficient to
consider only sine and cosine), there are also
sec(θ) =
1
cos(θ)
cosec(θ) =
cot(θ) =
1
sin(θ)
1
tan(θ)
Graphs of the trigonometric functions are as follows.
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
f (θ) = sin θ
1
0
2π
−1
f (θ) = cos θ
1
0
2π
−1
4
f (θ) = tan θ
1
0
2π
−1
−4
We can see that the trigonometric functions are periodic, meaning that the functions repeat
the same values over a fixed period. Sine and cosine have period 2π and tangent has period π.
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
Example 2.1.
1. Sketch the graphs of the functions sin(2x) and cos(x −
differ from sin(x) and cos(x)?
π
). How do they
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2. Sketch the graphs of cosec(x), sec(x) and cot(x).
3. Find all the solutions of the following equations for x in radians.
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(ii) cos(x) = 0
(iii) tan(x) = −1
(i) sin(x) =
2
2.3
Geometric applications
Take any triangle (not necessarily right-angled) with angles A, B, C and sides a, b, c opposite
to A, B, C respectively. Given partial information about certain angles and lengths of sides we
can use sine and cosine to determine the missing angles and lengths.
Theorem 2.1 (Sine Rule).
a
b
c
=
=
sin(A)
sin(B)
sin(C)
(To prove the sine rule, draw the altitude from B to side b. Denote it by h. Then, for two
right-angled triangles, we have h = c sin A = a sin C. Hence sina A = sinc C . Similar argument
gives the remaining equality.)
This rule is useful if we know (i) two angles and a side or (ii) two sides and a non-included
angle.
Theorem 2.2 (Cosine Rule).
a2 = b2 + c2 − 2bc cos(A)
(Later we shall explain this by using vector algebra.)
This rule is useful if we know (i) three sides or (ii) two sides and an included angle.
Examples. Calculate all the missing angles and sides in the following triangles ABC.
1. A = 45◦ , B = 30◦ , a = 10
2. a = 3, b = 7, c = 5
3. a = 4, b = 6, B = π/6
4. b = 5, c = 8, A = 100◦
2.4
Trigonometric identities
You should be familiar with the following result.
Theorem 2.3 (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.
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
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 2.4. 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
=
x2 y 2
x2 + y 2
r2
+
=
=
=1
r2
r2
r2
r2
Corollary 2.1. 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.
Further identities are based on the following “addition formulas” (or “compound angle”
formulas).
Theorem 2.5 (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
Theorem 2.6 (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 A cos B
1 − cos A cos B
The second formula follows in the same way.
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
Theorem 2.7 (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.
Corollary 2.2. tan(2x) =
2 tan(x)
1 − tan2 (x)
It is possible to deduce general formulas for cos nx and sin nx. We shall not do that, but
consider particular examples instead.
Example 2.2. 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.
Example 2.3. 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 .
Example 2.4. By√using the addition formula show that
1+ 3
sin(75◦ ) = √ .
2 2
2.5
Solving trigonometric equations
From the graphs of sin(x), cos(x) and tan(x) we can see that there are infinitely many values
of x that give the same value of the function. This causes problems when we want to go back
from the value of the function to the value of the angle that it came from - because there are
infinitely many angles that could have given that value of the function. For example how do we
solve
sin(x) = 0?
There are infinitely many solutions to this equation, x = 0, ±π, ±2π, ...
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
We can define an inverse operation to taking sine if we choose a standard interval of width
π and restrict the values that this inverse can take to that interval. We’ll choose our standard
interval for the inverse sine function1 arcsin to run from − π2 to π2 (excluding − π2 itself to avoid a
π
π
repetition). That is, define arcsin(x) to be the unique value y with − < y 6 and sin(y) = x.
2
2
π
So arcsin(0.5) = rads, arcsin(−0.8) = −0.93 rads.
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Similarly we can define inverse functions for cosine and tangent.
The inverse cosine, arccos(x), of x is the unique value y with 0 6 y 6 π and cos(y) = x.
(Note we chose a different standard interval for this but, if you think about the shape of the
graph of cos, it does make sense to do so.)
π
π
The inverse tangent, arctan(x), of x is the unique value y with − < y < and tan(y) = x.
2
2
When we solve equations involving trigonometric functions we should consider all possible
solutions initially, not just those one our chosen interval. Some of these may be later discarded
due to physical restrictions on the range of possible values (e.g. we may need only positive
solutions).
For example consider the equation
cos(5y) = −0.5
We have arccos(−0.5) =
cos(x) = −0.5 for
x=
(∗)
2π
. From the graph of cos(x) we see that we get the same value of
3
4π
2π
± 2nπ and x =
± 2nπ for any n = 0, 1, 2, . . .
3
3
So the solution to (∗) is
4π 2nπ
2π 2nπ
±
and y =
±
for any n = 0, 1, 2, . . .
15
5
15
5
4π 2π 2π 4π 8π
This gives values y = . . . , − , − , , , , . . .
15
15 15 15 15
Example 2.5. Find all solutions to the following equations.
1
(i) sin(4x) = √
(ii) tan(3x) = −1
2
Example 2.6. Some
frequently√met values of√the arc functions: arcsin 0 = 0, arccos 0 =
√
2
arctan 0 = 0, arcsin 2 = arccos 22 = π4 , arcsin 23 = π3 , arcsin 12 = π6 , arctan 1 = π4 .
y=
Example 2.7. Let cos α = 12 . 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 = π3 + 2kπ or α = 5π
+ 2kπ .
3
5π
,
3
π
,
2
so
Example 2.8. Find θ in the range 0 6 θ < π such that tan θ = 5 .
Solution: The unique solution in the range between − π2 and π2 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 6 θ < π. Hence the answer is: θ = arctan 5 ≈ 1.37
The notation sin−1 (x) is also used in place of arcsin(x) but we avoid this, and corresponding notations for
the other inverse trig functions, here because that notation might (wrongly!) suggest the meaning 1/ sin(x).
1
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Mathematics 1G1. Autumn 2013
Theodore Voronov.
Example 2.9. Solve for all values: cosx = 17 .
Solution. We have cos x = √17 or cos x = − √17 . Therefore x = ± arccos √17 + 2kπ or x =
π ± arccos √17 + 2kπ. This combines into x = ± arccos √17 + kπ.
Example 2.10. 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 = 21 or sin x = −3. The second
equation has no solutions, so we have sin x = 12 ⇐⇒ x = (−1)k arcsin 21 + kπ = (−1)k π6 + kπ .
Example 2.11. 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
√
√
3
2
2
3
cos x + √
sin x = 13 √ cos x + √ sin x .
2 cos x + 3 sin x = 4 + 9 √
4+9
4+9
13
13
Now we look for x0 such that cos x0 =
arccos √313 . So
2 cos x + 3 sin x =
√3
13
and sin x0 =
√2 .
13
We can take x0 = arcsin √213 =
√
2
13 sin(x + x0 ) where x0 = arcsin √ .
13
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