Download 5.8 Trigonometric Equations

5.8 Trigonometric Equations
To calculate the angle at which a curved section of
highway should be banked, an engineer uses the
v2 , where x is the angle of the
equation tan x = 224 000
bank and v is the speed limit on the curve, in kilometres
per hour.
2
v
The equation tan x = is an example of a
224 000
trigonometric equation. It will be used in Example 6.
A trigonometric equation is an equation that contains
one or more trigonometric functions. Other examples
of trigonometric equations include the following.
2
2
sin x + cos x = 1 2cos x − 1 = 0
The equation sin2 x + cos2 x = 1 is an identity. Recall
that a trigonometric identity is an equation that is true
for all values of the variable for which the expressions
on both sides of the equation are defined.
The equation 2cos x − 1 = 0 is not an identity and is
only true for certain values of x. To solve a trigonometric
equation that is not an identity means to find all values
of the variable, x, that make the equation true.
I NVESTIGATE & I NQUIRE
To graph the system y = 2x + 3 and y = 9 − x, use the Y= editor to enter
Y1 = 2X + 3 and Y2 = 9 − X. Graph the equations in the standard viewing
window.
1.
Use the intersect operation to find the coordinates of the point of
intersection.
2.
Equate the right sides of the two equations from question 1 and solve the
resulting equation for x algebraically.
3.
Compare the value of x you found in question 3 to the value of x you
found in question 2. Explain your findings.
4.
402 MHR • Chapter 5
graph the equations y = 4cos x and y = 1 + 2cos x, use the Y= editor
to enter Y1 = 4cos(X) and Y2 = 1 + 2cos(X). Using the mode settings, select
the degree mode. Using the window variables, adjust the window to
0° ≤ x ≤ 360° and −5 ≤ y ≤ 5 by setting Xmin = 0, Xmax = 360,
Ymin = −5, and Ymax = 5. Display the graph.
5. To
6. Use the intersect operation to find the coordinates of the points of
intersection.
7. Equate the right sides of the two equations from question 5, and solve the
resulting equation for cos x algebraically.
8. If 0° ≤ x ≤ 360°, what values of x, in degrees, give the value of cos x
from question 7?
Compare the values of x you found in question 8 to the values of x you
found in question 6. Explain your findings.
9.
EXAMPLE 1 Solving a Trigonometric Equation
sin θ − 1 = 0 on the interval 0° ≤ θ ≤ 360°.
Solve 2
SOLUTION 1 Paper-and-Pencil Method
2sin θ − 1 = 0
Add 1 to both sides:
2sin θ = 1
1
Divide both sides by 2:
sin θ = 2
The sine of an angle is positive in the first and
second quadrants.
y
y
r
0
θ
y
x
y
x
r
θ
–x 0
x
y
sin θ = –
r
The solutions are 45° and 135°.
π
3π
In radians, the solutions are and .
4
4
Use the memory aid CAST to remember which trigonometric
ratios are positive in each quadrant.
5.8 Trigonometric Equations • MHR 403
SOLUTION 2 Graphing-Calculator Method
Graph the related trigonometric function
y = 2sin x − 1 for 0° ≤ x ≤ 360°.
Using the mode settings, select the degree mode.
Find the x-intercepts using the zero operation.
The graph intersects the x-axis at 45° and 135°.
The solutions are 45° and 135°.
π
3π
In radians, the solutions are θ = and .
4
4
The window variables include Xmin = 0, Xmax = 360,
Ymin = –3, Ymax = 1.
EXAMPLE 2 Solving a Trigonometric Equation
Find the exact solutions for 3cos θ = cos θ + 1, if 0° ≤ θ ≤ 360°.
SOLUTION 1 Paper-and-Pencil Method
3cos θ = cos θ + 1
Subtract cos θ from both sides: 2cos θ = 1
Divide both sides by 2:
cos θ = 1
2
The cosine of an angle is positive in the first
and fourth quadrants.
y
y
r
0
θ
y
x
x
θ
r
cos θ = x–r
The exact solutions are 60° and 300°.
5π .
In radians, the exact solutions are π and 3
3
404 MHR • Chapter 5
x
x
0
–y
SOLUTION 2 Graphing-Calculator Method
Graph y = 3cos x and y = cos x + 1 in the
same viewing window for 0° ≤ x ≤ 360°.
Use the intersect operation to determine
the coordinates of the points of
intersection.
The window variables include Xmin = 0, Xmax = 360,
Ymin = –4, Ymax = 4.
The graphs intersect at x = 60° and x = 300°.
The solutions are 60° and 300°.
5π .
In radians, the solutions are π and 3
3
Note that an alternative graphing-calculator
method for solving Example 2 involves
rewriting 3cos x = cos x + 1 as 2cos x − 1 = 0.
The equation 2cos x − 1 = 0 can be solved
by graphing y = 2cos x − 1 and using the
zero operation to find the x-intercepts.
The window variables include Xmin = 0, Xmax = 360,
Ymin = –4, Ymax = 2.
EXAMPLE 3 Solving by Factoring
2
Solve the equation 2cos x − cos x − 1 = 0 on the interval 0 ≤ x ≤ 2π.
SOLUTION
2cos2 x − cos x − 1 = 0
Factor the left side:
(2cos x + 1)(cos x − 1) = 0
Use the zero product property:
2cos x + 1 = 0 or cos x − 1 = 0
2cos x = −1
cos x = 1
cos x = – 1
x = 0 or 2π
2
The solution can be modelled
2π or 4π
x=
graphically. The window variables
3
3
include Xmin = 0, Xmax = 360,
2π , 4π , and 2π.
The solutions are 0, 3 3
Ymin = –3, Ymax = 3.
5.8 Trigonometric Equations • MHR 405
EXAMPLE 4 Solving by Factoring
2
Solve the equation 2sin x − 7sin x + 3 = 0 for 0 ≤ x ≤ 2π. Express answers as
exact solutions and as approximate solutions, to the nearest hundredth of a radian.
SOLUTION
2sin2 x − 7sin x + 3 = 0
Factor the left side:
(sin x − 3)(2sin x − 1) = 0
Use the zero product property: sin x − 3 = 0 or 2sin x − 1 = 0
sin x = 3
2sin x = 1
sin x = 1
2
5π
x = π or 6
6
There is no solution to sin x = 3, since
all values of sin x are ≥ −1 or ≤ 1.
5π .
The exact solutions are π and 6
6
The approximate solutions are 0.52 and 2.62.
EXAMPLE 5 Using a Trigonometric Identity
Solve 6cos2 x − sin x − 5 = 0 for 0° ≤ x ≤ 360°. Round approximate solutions
to the nearest tenth of a degree.
SOLUTION
Use the Pythagorean identity cos2 x = 1 − sin2 x to write an equivalent equation
that involves only the sine function.
6cos2 x − sin x − 5 = 0
2
2
Substitute 1 − sin x for cos x: 6(1 − sin2 x) − sin x − 5 = 0
Expand:
6 − 6sin2 x − sin x − 5 = 0
Simplify:
−6sin2 x − sin x + 1 = 0
Multiply both sides by –1:
6sin2 x + sin x − 1 = 0
Factor the left side:
(2sin x + 1)(3sin x − 1) = 0
Use the zero product property: 2sin x + 1 = 0 or 3sin x − 1 = 0
2sin x = −1 or 3sin x = 1
sin x = – 1
or sin x = 1
2
3
x = 210° or 330°
x =⋅ 19.5° or 160.5°
The solutions are 19.5°, 160.5°, 210°, and 330°.
406 MHR • Chapter 5
EXAMPLE 6 Bank Angles
v2
Engineers use the equation tan x = to calculate the angle at which
224 000
a curved section of highway should be banked. In the equation, x is the angle
of the bank and v is the speed limit on the curve, in kilometres per hour.
a) Calculate the angle of the bank, to the nearest tenth of a degree, if the
speed limit is 100 km/h.
b) The four turns at the Indianapolis Motor Speedway are banked at an angle
of 9.2°. What is the maximum speed through these turns, to the nearest
kilometre per hour?
SOLUTION
2
v
tan x = 224 000
2
100
=
224 000
⋅
x = 2.6°
The angle of the bank is 2.6°, to the nearest tenth of a degree.
a)
2
b)
v
= tan x
224 000
v2 = 224 000tan x
v = 224
0tan
00x
v = 224
0tan
002°
9.
⋅
v = 190
The maximum speed is 190 km/h, to the nearest kilometre per hour.
Key
Concepts
• To solve a trigonometric equation that is not an identity, find all values of the
variable that make the equation true.
• Trigonometric equations can be solved
a) with paper and pencil using the methods used to solve algebraic equations
b) graphically using a graphing calculator
• Answers can be expressed in degrees or radians.
5.8 Trigonometric Equations • MHR 407
Communicate
Yo u r
Understanding
Describe how you would solve 2sin x – 3 = 0, 0° ≤ x ≤ 360°.
Describe how you would solve 2cos2 x − 3cos x + 1 = 0, 0 ≤ x ≤ 2π.
Justify your method.
3. Explain why the equation cos x − 2 = 0 has no solutions.
1.
2.
Practise
A
1.
a)
b)
c)
d)
e)
f)
Solve each equation for 0 ≤ x ≤ 2π.
sin x = 0
2cos x + 1 = 0
tan x = 1
2
sin x + 1 = 0
2cos x − 3 = 0
2sin x + 3 = 0
2.
a)
b)
c)
d)
e)
f)
Solve each equation for 0 ≤ x ≤ 360°.
sin x + 1 = 0
2
cos x − 1 = 0
2sin x − 3 = 0
2
cos x + 1 = 0
2sin x + 1 = 0
tan x = –1
3.
a)
b)
c)
d)
e)
f)
g)
Solve each equation for 0° ≤ x ≤ 360°.
2cos2 x − 7cos x + 3 = 0
3sin x = 2cos2 x
2sin2 x − 3sin x − 2 = 0
sin2 x − 1 = cos2 x
tan2 x − 1 = 0
2sin2 x + 3sin x + 1 = 0
2cos2 x + 3sin x − 3 = 0
408 MHR • Chapter 5
Solve each equation for 0 ≤ x ≤ 2π.
Express answers as exact solutions and as
approximate solutions, to the nearest
hundredth of a radian.
2
a) sin x − 2sin x − 3 = 0
2
b) 2cos x = sin x + 1
c) 2sin x cos x + sin x = 0
2
d) sin x = 6sin x − 9
2
e) sin x + sin x = 0
f) cos x = 2sin x cos x
4.
Solve for x on the interval 0° ≤ x ≤ 360°.
Round approximate solutions to the nearest
tenth of a degree.
a) 4cos x − 3 = 0
b) 1 + sin x = 4sin x
2
c) 6cos x − cos x − 1 = 0
2
d) 9sin x − 6sin x + 1 = 0
2
e) 16cos x − 4cos x + 1 = 0
2
f) 6cos x + sin x − 4 = 0
5.
Solve for x on the interval 0 ≤ x ≤ 2π. Give
the exact solution, where possible. Otherwise,
round to the nearest hundredth of a radian.
a) sin x − 3sin x cos x = 0
2
b) 6cos x + cos x − 1 = 0
2
c) 8cos x + 14cos x = −3
2
d) 8sin x − 10cos x − 11 = 0
6.
Apply, Solve, Communicate
7. Refraction of light The diagram shows how a light ray bends as it
travels from air into water. The bending of the ray is a process known as
refraction. In the diagram, i is the angle of incidence and r is the angle of
refraction. For water, the index of refraction, n, is defined as follows.
sin i
n=
sin r
a) The index of refraction of water has a value of 1.33. For a ray with an
angle of incidence of 40°, find the angle of refraction, to the nearest tenth of
a degree.
b) State any restrictions on the value of r in the above equation. Explain.
i
r
B
Show that the area, A, of an isosceles triangle
with base b and equal angles x, can be found using the equation
2
b tan x .
A=
4
2
b) For an isosceles triangle with A = 40 cm and b = 8 cm, use the
equation to find x, to the nearest tenth of a degree.
8. Isosceles triangle a)
In right triangle ABC, BD ⊥ AC, AC = 4, and BD = 1.
1
a) Show that sin A cos A = .
4
b) Describe how you would use a graphing calculator to find the
measure of ∠A, in degrees.
c) Find the exact measure of ∠A, in degrees.
x
x
9. Application
A
B
D
C
Solve for x on the interval 0 ≤ x ≤ 2π. Give the exact solution, if
possible. Otherwise, round to the nearest hundredth of a radian.
2
2
a) tan x − 4tan x = 0
b) tan x − 5tan x + 6 = 0
2
2
c) tan x − 4tan x + 4 = 0
d) 6tan x − 7tan x + 2 = 0
10.
11. Solve for x on the interval 0° ≤ x ≤ 360°.
a) 2sin x tan x − tan x − 2sin x + 1 = 0
b) cos x tan x − 1 + tan x − cos x = 0
12. a) Graph y = sin x and y = 0.5 for
b) For what values of x is sin x ≥ 0.5?
c) For what values of x is sin x ≤ 0.5?
Check your solutions.
0° ≤ x ≤ 360°.
5.8 Trigonometric Equations • MHR 409
Write a short paragraph, including examples, to
distinguish the terms trigonometric identity and trigonometric equation.
2
2
b) Inquiry/Problem Solving If (cos x + sin x) + (cos x − sin x) = k, for
what value(s) of k is the equation an identity?
13. a) Communication
C
Solve each equation on the interval 0° ≤ x ≤ 360°.
a) sin 2x = 1
b) cos 2x = −1
c) 2sin 2x = 1
d) 2cos 2x = 1
e) 2
cos 2x = 1
f) 2sin 2x = 3
g) 2cos 2x = −3
h) sin 2x = 0
i) 2sin 0.5x = 1
14.
Use a graphing calculator to find the exact solutions, in
degrees, for 0° ≤ x ≤ 360°.
a) cos 2x = cos x
b) sin 2x = 2cos x
c) tan x = sin 2x
15. Technology
Describe how you would use a graphing calculator to
solve the equation sin (cos x) = 0, 0° ≤ x ≤ 360°.
b) Find the exact solutions to this equation.
16. Technology a)
A C H I E V E M E N T Check
Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
A storm drain has a cross section in the shape of an isosceles trapezoid.
x
The shorter base and each of the two equal sides measure 2 m, and x is
2m
the angle formed by the longer base and each of the equal sides.
a) Write an expression for the area, A, of the cross section in terms of
sin x and cos x.
b) Describe how you would use a graphing calculator to find the value of x,
in degrees, if the area of the cross section is 5 m2.
c) Find the value of x, to the nearest degree.
NUMBER
Power
Place the digits from 1 to 9 in the boxes to make the statements true.
Use the order of operations.
■×■−■=2
(■ + ■) ÷ ■ = 2
■+■−■=2
410 MHR • Chapter 5
x
2m
2m