Transcript Chapter 8 Applications of Trigonometric Functions

Chapter 8
Applications of Trigonometric
Functions
By Hannah Chung and Evan Jaques
Section 8.1
Applications Involving Right Triangles
• A right triangle is a triangle with a 90⁰ angle.
The side opposite the right angle is called the
hypotenuse and the other sides are legs.
• The angle Θ is an acute angle, or less than 90⁰.
Hypotenuse
Leg
Leg
Section 8.1
Applications Involving Right Triangles
The 6 Trigonometric Functions
Hypotenuse
Opposite
Adjacent
Sin Θ=Opposite
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Hypotenuse
Cos Θ=Adjacent
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Hypotenuse
Tan Θ=Opposite
Adjacent
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Θ
Csc=Hypotenuse
Opposite
-------------------------------------------------------------------
Sec Θ=Hypotenuse
Adjacent
-------------------------------------------------------------------
Cot Θ=Adjacent
Opposite
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Section 8.1
Applications Involving Right Triangles
• The sum of two complementary angles is a
right angle.
• Cofunctions are the functions sine and cosine,
tangent and cotangent, and secant and
cosecant.
• Cofunctions of complementary angles are
equal.
Section 8.2
The Law of Sines
• Oblique triangles are triangles without a right
angle.
• In order to solve an oblique triangle, you need
to know the length of 1 side and: 2 angles, 1
angle and 1 more side, or 2 more sides.
• The Law of Sines is: sin A = sin B = sin C
•
•
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a
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b
-----------------------------------------------
c
Section 8.2
The Law of Sines
Angle C found by 180⁰=40⁰+60⁰+C
Angle C=80⁰
60⁰
4
c
Side b found by sin 40⁰=sin 60⁰
4
b
b=4 sin 60°
sin 40°
--------------------------------------
--------------------------------------
--------------------------------------------------------
C
40⁰
b
Side c found by sin 40⁰=sin 80⁰
4
c
c=4 sin 80°
sin 40°
-------------------------------------
---------------------------------------------------------
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Section 8.2
The Law of Sines
Angle C found by 35°+15°+C=180°
C=130°
15⁰
5
a
Side a found by sin 35° = sin 130°
a
5
a=5 sin 35°
sin 130°
---------------------------------------------
----------------------------------------------
-------------------------------------------------------
35⁰
C
b
Side b found by sin 15° = sin 130°
b
5
b=5 sin 15°
sin 130°
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---------------------------------------------
Section 8.2
The Law of Sines
b
a
SSA triangles can be solved by using
the relationship between side a,
side b, and height
b
A
A
a<h
h=b sin A
Results in no triangle
b
A
a a
b sin A < a and a<b
Results in 2 different triangles
a
a=h=b sin A
Results in right triangle
b
a
A
a>b
Results in one triangle
---------
Section 8.2
The Law of Sines
Angle C found by 25.4°+40°+C=180°
C=114.6°
Angle B found by sin 40° = sin B
3
2
Sin B = 2 sin 40° =.43
3
B=25.4°
------------------------------------------
------------------------------
Side c found by sin 40° = sin 114.6°
3
c
c=3 sin 114.6° = 4.24
sin 40°
--------------------------------------------------------------
-----------------------------------------------
-------------------------------------------------------------------------
3
C
2
40°
B
c
--------------------------------------------------------
Section 8.3
The Law of Cosines
Derive the law of cosines to solve oblique triangles with
two sides and the included angle (SAS) and where three
sides are known (SSS)
Theorem: Law of Cosines
For a triangle with sides a, b, c, and opposite angles A, B, C, respectively
c²= a²+ b² - 2ab cos C
b²= a²+ c² - 2ac cos B
a²= b²+ c² - 2bc cos A
Theorem: Law of Cosines
The square of one side of a triangle equals the sum of the squares of the
other two sides minus twice their product times the cosine of their
included angle.
Section 8.3
The Law of Cosines
Proof:
Use distance formula to compute c².
y
(a cosC, a sinC)
C² = (b- a cos C)² + (0 – a sinC)²
= b² - 2ab cos C + a²cos²C + a²sin²C
= b² - 2ab cosC + a²(cos²C + sin²C)
= a² + b² - 2ab cos C
a
c
C
(0,0)
b
(b, 0)
x
Section 8.3
The Law of Cosines
Use Law of Cosines to solve a SAS Triangle
Solve triangle with a=2, b=3, C=60⁰
Law of cosines to find third side, c.
c²= a²+ b² - 2ab cosC
= 4+ 9 – 2 x 2 x 3 x cos60⁰
= 13-(12 x ½) = 7
c= √7
B
c
2
60⁰
For A:
a²= b²+ c² - 2bc cos A
2bc cos A= b²+ c²- a²
cos A= (b²+ c²- a²)/(2bc)= (9+ 7- 4)/(2x 3√7)= 12/(6√7)= (2√7)/7
A= cos¯¹(2√7/7)≈40.9⁰
For B:
b²= a²+ c² - 2ac cos B
cosB= (a²+ c² - b²)/2ac= (4+7- 9)/4√7= 1/(2√7)= √7/14
B= cos¯¹(√7/14)≈ 79.1⁰
Check: A+ B+ C= 40.9⁰ + 79.1⁰ + 60⁰= 180⁰
A
3
Section 8.3
The Law of Cosines
Use Law of Cosines to solve a SSS Triangle
For A:
cosA= (b²+ c²- a²)/2bc= (9+36-16)/(2x3x6)= 29/36
A= cos¯¹29/36≈ 36.3⁰
For B:
cosB= a²+c²-b²/2ac= 16+36-9/2x4x6= 43/48
B= cos¯¹43/48≈ 26.4⁰
B
6
4
C
Since A+B+C= 180⁰,
C=180⁰ - A – B= 180⁰- 36.3⁰-26.4⁰= 117.3⁰
A
3
Section 8.4
Area of a Triangle
The area K of a triangle is:
K= 1/2bh
where b is the base and h is an altitude drawn to that base
h
b
Proof: the area of the triangle with base b
and altitude h is exactly half the area of the
rectangle, which is bh.
Section 8.4
Area of a Triangle
Find the area of SAS triangles
Suppose that we know two sides a and b and the angle C.
Then the altitude can be found through:
h/a= sinC
a
h
So that
h= a sinC
C
b
Using this fact in formula (1) produces K=1/2bh= 1/2b(a sinC)= 1/2ab sinC
So:
K=1/2ab sinC (2)
By dropping altitudes from the other two vertices of the triangle, we obtain:
K=1/2bc sinA
K=1/2ac sinB
Theorem: The area K of a triangle equals one-half the product of two of its
sides times the sine of their included angle.
Section 8.4
Area of a Triangle
Find the area of SSS triangles
If three sides of a triangle are known, we use Heron’s Formula:
The area K of a triangle with sides a, b, and c is
K= √s(s-a)(s-b)(s-c)
Where s=1/2(a+b+c)
Example:
a
b
a=4, b=5, c=7
c
s=1/2(4+5+7)= 8
K= √s(s-a)(s-b)(s-c)=
K=√8x4x3x1= √96= 4√6 units
Section 8.5
Simple Harmonic Motion; Damped Motion; Combining Waves
• Simple harmonic motion is a special kind of vibrational motion
in which the acceleration a of the object is directly
proportionate to the negative of its displacement d from its rest
position. a=-kd
• An object that moves on a coordinate axis so that the distance d
from its rest position at time t is given by either d=a cos (ωt) or
d = a sin (ωt), where a and ω>0 are constants.
• The motion has amplitude |a| and period 2π/ω.
• The frequency f of an object in simple harmonic motion is the
number of oscillations per unit time. Since the period is the
time required for one oscillation, it follows that the frequency is
the reciprocal of the period.
• f=ω/2π
Section 8.5
Simple Harmonic Motion; Damped Motion; Combining Waves
• Suppose that an object attached to a coiled spring is pulled
down a distance of 5 inches from its rest position and then
released. If the time for one oscillation is 3 seconds, write
an equation that relates the displacement d of the object
from its rest position after time t. Assume no friction.
• The motion is simple harmonic. When the object is
released (t=0), the displacement of the object from the rest
position is -5 units (since the object was pulled down).
Because d=-5 when t=0, it is easier to use cosine d = a cos
(ωt) to describe the motion. The amplitude is |-5|=5 and
the period is 3 so a=-5 and 2π/ω=period=3, ω=2π/3.
• The equation of the motion of the object is d=-5 cos [2π t]
•
3
•
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Section 8.5
Simple Harmonic Motion; Damped Motion; Combining Waves
• Damped motion is simple harmonic motion with
friction.
• The displacement d of an oscillating object from
its at-rest position at time t is given by
ω^2 – b^2
4m^2
t
-------------------------
• b is the damping factor or damping coefficient
and m is the mass of the oscillating object. Here
|a| is the displacement at t=0 and 2π/ω is the
period under simple harmonic motion (no
damping).
Section 8.5
Simple Harmonic Motion; Damped Motion; Combining Waves
• Analyze the damped vibration curve d(t)=e^(-t/π) cos t, t ≥ 0
• The displacement d is the product of y=e^(-t/π) and y=cos t.
Using absolute value properties and the fact that |cos t| ≤ 1,
we find that
• |d(t)|=|e^(-t/π) cos t|=|e^(-t/π||cos t|≤|e^(-t/π|=e^(-t/π)
• As a result, -e^(-t/π)≤d(t)≤e^(-t/π)
• This means that the graph of d will lie between the graphs of
y=e^(-t/π) and y=-e^(-t/π), which are the bounding curves of
d.