Download Chapter 6: Trigonometric equations

MQ Maths B Yr 11 Ch 06 Page 237 Thursday, October 25, 2001 3:59 PM
Trigonometric
equations
6
syllabus ref
efer
erence
ence
Topic:
• Periodic functions and
applications
In this cha
chapter
pter
6A Simple trigonometric
equations
6B Equations using radians
6C Further trigonometric
equations
6D Identities
6E Using the Pythagorean
identity
MQ Maths B Yr 11 Ch 06 Page 238 Thursday, October 25, 2001 3:59 PM
238
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
Introduction
Sudhira is a keen fisherman. The ideal depth for fishing in Sudhira’s favourite tidal lake
is 3 metres. The depth of water in the lake can be found using the equation
p
D = 5 − 4 sin  --- t
 6
where t is the time in hours after midnight. What is the best time of day for Sudhira to
fish?
To solve this problem we need to solve a trigonometric equation
Simple trigonometric equations
From your earlier work on trigonometry, you will be familiar
with problems of the type: ‘Find the size of the angle marked
θ in the figure at right’.
The solution to this problem is set out as:
13 cm
adj
cos θ = --------hyp
θ
9 cm
9
cos θ = -----13
θ = 46°.
9
The equation cos θ = ------ is an example of a trigonometric equation. This trigonometric
13
equation had to be solved in order to find the size of the angle in the triangle. In this
particular case we knew that the angle θ was acute from the triangle that was drawn.
In the earlier chapter on graphing periodic functions we saw that the cos function
was periodic. This means that there are values of θ, other than the one already found
9
for which cos θ = ------ . There will, in fact, be an infinite number of solutions to this
13
trigonometric equation, so for practical reasons we are usually given a domain within
which to solve the equation. This domain will often be in the form 0° ≤ θ ≤ 360°,
meaning that we want solutions within the first positive revolution.
If the trigonometric ratio is positive the calculator will
y
give a first quadrant answer. To complete the solution we
need to consider all quadrants for which the trigonometric
ratio is positive.
Sine
All
9
positive positive
In the case of cos θ = ------ the cosine ratio is positive in
13
x
Tangent Cosine
the first and fourth quadrants. We found earlier that the
positive
positive
first quadrant solution to this equation was 46°. The fourth
quadrant solution will therefore be 360° − 46° = 314°.
For a negative trigonometric ratio we solve the
corresponding positive equation to find a first quadrant angle to use, then find the
corresponding angles in the negative quadrants.
MQ Maths B Yr 11 Ch 06 Page 239 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
239
WORKED Example 1
Solve the following trigonometric equations over the domain 0∞ ≤ θ ≤ 360∞, correct to the
nearest degree.
a sin θ = 0.412
4
b tan θ = − -----11
THINK
WRITE
a
a sin θ = 0.412
First quadrant angle = 24°
1
2
Write the equation.
Use your calculator to find the first
quadrant angle.
y
180
°−
24°
24°
x
5
The sine ratio is positive in the first and
second quadrants.
Find the second quadrant angle by
subtracting 24° from 180°.
Write the answer.
1
Write the equation.
2
Use your calculator to find the first
quadrant angle.
3
4
b
180° − 24° = 156°
θ = 24° or 156°
4
b tan θ = − -----11
First quadrant angle = 20°
y
180
°−2
0°
20°
x
360
°−2
0°
3
4
5
The tangent ratio is negative in the
second and fourth quadrants.
Find the second quadrant angle by
subtracting 20° from 180° and the
fourth quadrant angle by subtracting
20° from 360°.
Write the answer.
180° − 20° = 160°
360° − 20° = 340°
θ = 160° or 340°
In the earlier chapter we also found that we were able to find exact values of special
angles using the triangles on the next page.
MQ Maths B Yr 11 Ch 06 Page 240 Thursday, October 25, 2001 3:59 PM
240
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
45°
30°
2
2
1
45°
Line of bisection
60°
1
1
These special angles should be used where possible in the solution to a trigonometric
equation. They are used when we recognise any of the values produced by the triangles.
1
sin 30° = --2
3
cos 30° = ------2
3
tan 30° = ------3
2
sin 45° = ------2
2
cos 45° = ------2
tan 45° = 1
3
sin 60° = ------2
1
cos 60° = --2
tan 60° =
3
WORKED Example 2
3
Solve the equation cos q = – ------- over the domain 0∞ ≤ θ ≤ 360∞.
2
THINK
WRITE
1
Write the equation.
2
Use the special triangles to find the first
quadrant angle.
The cosine ratio is negative in the
second and third quadrants.
Find the second quadrant angle by
subtracting 30° from 180° and find the
third quadrant angle by adding 30° to
180°.
Write the answer.
3
4
5
3
cos θ = – ------2
First quadrant angle = 30°
180° − 30° = 150°
180° + 30° = 210°
θ = 150° or 210°
Similarly, we must be aware of when the
boundary angles should be used in the solution
of the equation. Remember from the work on
the unit circle that y = sin θ, x = cos θ and
y
tan θ = -- .
x
y
1 90°
180°
–1
0
–1 270°
0° or 360°
x
1
MQ Maths B Yr 11 Ch 06 Page 241 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
241
WORKED Example 3
Solve the equation sin θ = -1 in the domain 0∞ ≤ θ ≤ 360∞.
THINK
1
2
WRITE
sin θ = −1
θ = 270°
Write the equation.
y = sin θ so find the angle with a
y-value of −1.
remember
remember
1. Trigonometric equations are equations that use the trigonometric ratios.
2. The trigonometric functions are periodic and so they have an infinite number of
solutions. The equation is usually written with a restricted domain to limit the
number of answers.
3. There are two solutions to most trigonometric equations with a domain
0° ≤ θ ≤ 360°.
4. Remember the special triangles as they are used in many solutions.
5. Boundary angles may also provide the solution to an equation.
Simple trigonometric equations
Example
2 Find exact solutions to each of the following trigonometric equations over the domain
0° ≤ θ ≤ 360°.
2
1
3
2
d sin θ = – --a sin θ = ------b cos θ = ------c tan θ = 3
2
2
2
WORKED
Example
1
e cos θ = – --2
f
3
tan θ = − ------3
2
g sin θ = − ------2
3
h cos θ = ------2
3 multiple choice
2
If sin x = cos x = − ------- and 0° ≤ x ≤ 360°, then x is:
2
A 150° or 210°
B 135° or 225°
D 135° or 315°
E 120°
C 225°
4 It is known that sin θ < 0 and that tan θ > 0. Which quadrant does the angle θ lie in?
Explain your answer.
HEET
6.1
SkillS
1 Solve each of the following trigonometric equations over the domain 0° ≤ θ ≤ 360°,
correct to the nearest degree.
1
b cos θ = −0.25
c tan θ = 5.72
d sin θ = −0.85
a sin θ = 0.6
e cos θ = 0.195
f tan θ = −0.837 g sin θ = −0.333 h cos θ = 0.757
WORKED
HEET
SkillS
6A
6.2
MQ Maths B Yr 11 Ch 06 Page 242 Thursday, October 25, 2001 3:59 PM
242
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
5 Yvonne is doing a trigonometric problem that has reduced to the equation sin θ = 1.5.
a When Yvonne tries to solve this equation her calculator returns an error message.
Why?
b When checking her working Yvonne realises that she should have used the tangent
ratio. Why is it now possible to achieve a solution to the equation tan θ = 1.5?
6 Solve each of the following equations over the domain 0° ≤ θ ≤ 360°.
b cos θ = 0
c tan θ = 0
d sin θ = 0
a sin θ = 1
3
e cos θ = −1
f sin θ = −1
WORKED
Example
7 Solve the following trigonometric equations over the domain −360° ≤ θ ≤ 360°.
a sin θ = 0.5
b cos θ = 0.35
c tan θ = −1
d sin θ = −0.87
e cos θ = −0.87
f tan θ = 1.4
Equations using radians
We have seen that a radian is an alternative method of measuring an angle. A trigonometric equation can be solved using radians as well as degrees. Usually the domain
given will indicate whether it is expected that you will solve the equation in degrees or
in radians.
For example, if you are asked to solve an equation over the domain 0° ≤ θ ≤ 360°
then degrees are expected for the answer. However, if the given domain is 0 ≤ θ ≤ 2π
then it is expected that the answer will be given in radians.
The method of solving the equations is the same, but be sure that your calculator is
in radian mode before attempting to solve the problem to give an answer in radians.
WORKED Example 4
Solve the equation tan θ = 0.8 over the domain 0 ≤ θ ≤ 2π. Give the answer correct to
2 decimal places.
THINK
1
2
3
Write the equation.
Use your calculator to find the first
quadrant angle.
The tangent ratio is positive in the first
and third quadrants.
WRITE
tan θ = 0.8
First quadrant angle = 0.67
y
0.67
π + 0.67
4
5
Find the third quadrant angle by adding
0.67 to π.
Write the answer.
π + 0.67 = 3.81
θ = 0.67 or 3.81
x
MQ Maths B Yr 11 Ch 06 Page 243 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
243
When the special angles are used, it is still important to recognise them and recognise their radian equivalents in terms of π.
π 1
sin --- = --6 2
3
π
cos --- = ------6
2
3
π
tan --- = ------6
3
2
π
sin --- = ------4
2
2
π
cos --- = ------4
2
π
tan --- = 1
4
3
π
sin --- = ------3
2
π 1
cos --- = --3 2
π
tan --- =
3
3
WORKED Example 5
3
Solve the equation sin θ = – ------- over the domain 0 ≤ θ ≤ 2π.
2
THINK
WRITE
1
Write the equation.
2
Use the special triangles to find the first
quadrant angle.
The sine ratio is negative in the third
and fourth quadrants.
3
3
sin θ = – ------2
π
First quadrant angle = --3
y
π
—
3
x
π
π+—
3
4
Find the third quadrant angle by adding
π
--- to π and the fourth quadrant angle by
3
π
subtracting --- from 2π.
3
5
Write the answer.
π
2π – —
3
π
π
θ = π + --- or θ = 2π + --3
3
5π
4π
θ = -----or θ = -----3
3
5π
4π
θ = ------ or -----3
3
All of the equations that we have dealt with so far have been one-step solutions. In
many examples we may need to rearrange the equation before we are able to use the
calculator to solve it.
When rearranging the equation, we attempt to place the trigonometric ratio alone on
one side of the equation, as in the example above.
MQ Maths B Yr 11 Ch 06 Page 244 Thursday, October 25, 2001 3:59 PM
244
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
WORKED Example 6
Find x if 2 sin x = 0.984 over the domain 0 ≤ x ≤ 2p.
THINK
1
2
3
4
WRITE
Write the equation.
Divide both sides by 2 to get sin x by
itself.
Use your calculator to find the first
quadrant angle.
The sine ratio is positive in the first and
second quadrants.
2 sin x = 0.984
sin x = 0.492
First quadrant angle = 0.514
y
π
–0
.51
4
0.514
x
5
6
Find the second quadrant angle by
subtracting 0.514 from π.
Write the answer.
θ = π − 0.514
θ = 2.628
θ = 0.514 or 2.628
remember
remember
1. Many trigonometric equations will need to be solved using radians.
2. The domain within which you are asked to solve the equation will tell you
whether to use degrees or radians.
3. You will need to know the special angle results as they apply to radians.
4. You must isolate the trigonometric ratio before you can solve any equation
using either your calculator or the special angles.
6B
Equations using radians
1 Solve each of the following equations over the domain 0 ≤ x ≤ 2π. Give your answers
correct to 2 decimal places.
4
a sin x = 0.8
b cos x = −0.5
c tan x = 1.5
d sin x = −0.327
e cos x = 0.707
f tan x = −0.39
WORKED
Example
2 Solve each of the following over the domain 0 ≤ x ≤ 2π.
1
3
5
c tan x = 1
b cos x = − --a sin x = ------2
2
WORKED
Example
e tan x = − 3
f
1
sin x = − --2
2
d cos x = ------2
MQ Maths B Yr 11 Ch 06 Page 245 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
3 Solve each of the following over the domain 0 ≤ x ≤ 2π.
a sin x = 0
b tan x = 0
c cos x = 0
e cos x = 1
f cos x = −1
g sin x = −1
245
d sin x = 1
4 Find exact solutions to each of the following equations over the domain 0 ≤ x ≤ 2π.
a 2 sin x = 1
b 2 cos x = 3
c 2 tan x = 2
6
d 2 sin x + 3 = 0
e 2 cos x + 2 = 0
f
3 tan x + 3 = 0
WORKED
Example
5 multiple choice
The solution to the equation 2 cos x + 1 = 0 over the domain 0 ≤ x ≤ 2π is:
5π 7π
π 2π
π 5π
A ---, -----B ------, -----C ---, -----3 3
6 6
6 6
4 π 5π
2π 4π
D ------, -----E ------, -----3 3
3 3
6 Solve each of the following, to the nearest degree, over the domain 0 ≤ x ≤ 360°.
a 4 sin x = 1
b 3 cos x = −2
c 2 tan x − 7 = 0
d 4 + sin x = 3
e 1 + 2 cos x = 2
f 3 tan x + 9 = 0
Further trigonometric equations
In many cases the equation that we have to solve may not be in the domain
0 ≤ x ≤ 2π. We may be asked to solve the equation in the domain 0 ≤ x ≤ 4π (2 revolutions) or −2π ≤ x ≤ 2π (also 2 revolutions, but one in the negative sense).
To find the solutions to a trigonometric equation beyond the first revolution we
simply add or subtract 2π to the first revolution solutions.
WORKED Example 7
Find α if sin α = 0.7 in the domain 0 ≤ α ≤ 4p.
THINK
1
2
3
Write the equation.
Use your calculator to find the first
quadrant angle.
The sine ratio is positive in the first and
second quadrants.
WRITE
sin α = 0.7
First quadrant angle = 0.7754
y
π–
0.7
75
4
0.7754
x
Continued over page
MQ Maths B Yr 11 Ch 06 Page 246 Thursday, October 25, 2001 3:59 PM
246
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
THINK
4
5
6
WRITE
Find the second quadrant angle by
subtracting 0.7754 from π.
Find the solutions between 2π and 4π
adding 2π to each of the first revolution
solutions.
Write the answer.
α = π − 0.7754
α = 2.3662
α = 0.7754 + 2π
α = 7.0586
α = 2.3662 + 2π
α = 8.6494
α = 0.7754, 2.3662, 7.0586, 8.6494
In many equations you will first need to make the trigonometric ratio the subject of
the equation.
WORKED Example 8
Find x if
2 cos x + 1 = 0 over the domain -2π ≤ x ≤ 2π.
THINK
1
2
3
4
5
6
7
WRITE
Write the equation.
Make cos x the subject of the equation.
Use the special triangles to find the first
quadrant angle.
The cosine ratio is negative in the
second and third quadrants.
Find the second quadrant angle by
π
subtracting --- from π. Find the third
4
π
quadrant angle by adding --- to π.
4
To find the solutions between −2π and
0, subtract 2π from each of the first
revolution solutions.
Write the answer.
remember
remember
2 cos x + 1 = 0
2 cos x = −1
–1
cos x = ------2
π
First quadrant angle = --4
π
x = π − --4
3π
x = -----4
π
x = π + --4
5π
x = -----4
5π
3π
x = ------ − 2π
x = ------ − 2π
4
4
–5 π
–3 π
x = --------x = --------4
4
–5 π –3 π 3 π 5 π
x = ---------, ---------, ------, -----4
4 4 4
1. To find solutions to trigonometric equations between 2π and 4π we add 2π to
any solutions in the first revolution.
2. To find solutions to trigonometric equations between −2π and 0 we subtract 2π
from any solutions in the first revolution.
3. In many cases it may be necessary to rearrange an equation to make the
trigonometric ratio the subject.
MQ Maths B Yr 11 Ch 06 Page 247 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
6C
247
Further trigonometric
equations
1 Solve each of the following trigonometric equations over the domain 0 ≤ x ≤ 4π.
a cos x = −0.6591 b sin x = 0.9104
c cos x = 0.48
d sin x = −0.371
7
e tan x = 0.58
f tan x = −2.1
WORKED
Example
2 Solve each of the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
a sin x = 0.2686
b cos x = −0.7421 c tan x = −0.4776 d sin x = −0.5432
e cos x = 0.1937 f tan x = 3
3 Find the solutions to the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
a 2 sin x − 1 = 0 b 3 cos x = 0
c 2 sin x + 3 = 0 d tan x + 3 = 0
8
e
f
2 cos x = 1
3 tan x − 1 = 0
WORKED
Example
4 Find all the solutions to the following equations over the domain −2π ≤ x ≤ 2π. Give
each answer correct to 2 decimal places.
b 3 cos x − 3 = 0
d 2 sin x − 5 = −4
e
2 cos x + 2 = 3
cos x + 4 = 4.21
c
1
--2
f
2 cos x +
3 =0
5 A particle moves in a straight line so that its distance, x metres, from a point O is given
by the equation x = 3 + 4 sin t, where t is the time in seconds after the particle begins
to move.
a Find the distance from O when the particle begins to move.
b Find the time when the particle first reaches O. Give your answer correct to 2
decimal places.
Fishing
You should now be able to solve the fishing problem given at the start of this
chapter. The depth of water in the lake was given by
p
D = 5 − 4 sin  --- t
 6
Substitute D = 3 and solve to find the best time for Sudhira to fish. The solutions
should be found in the domain 0 < t < 24.
Identities
An identity is a relationship that holds true for all legitimate values of a pronumeral or
pronumerals. For example, a simple identity is x + x = 2x. The identities described in
this section are far more interesting and useful than this, as you will see.
ET
SHE
Work
a 4 sin x + 2 = 6
6.1
MQ Maths B Yr 11 Ch 06 Page 248 Thursday, October 25, 2001 3:59 PM
248
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
The Pythagorean identity
y
Consider the right-angled triangle in the unit
circle shown.
Applying Pythagoras’ theorem to this triangle gives
the identity:
P(θ )
1
θ
O cos θ
sin θ + cos θ = 1
2
2
y
1
The tangent
Consider the unit circle on the right.
A tangent is drawn at A and extended to the point
C, so that OC is an extension of OP. This tangent is
called tangent θ, which is abbreviated to tan θ.
Triangles ODP and OAC are similar, because they
have their three corresponding angles equal.
It follows that:
or
sin θ
D
B
P(θ ) C
tan θ
1 sinθ
θ
O cos θ D 1A
x
sin θ
tan θ
------------ = ------------ (corresponding sides)
1
cos θ
sin θ
tan θ = -----------cos θ
(as mentioned in an earlier section).
Another relationship between sine and cosine —
complementary functions
Consider the unit circle shown on the right:
The triangles OAB and ODC are congruent because
they have all corresponding angles equal and the
hypotenuse equal (radius = 1).
Therefore all corresponding sides are equal and it
follows that:
and
and
sin (90 − θ )° = cos θ = x
cos (90 − θ )° = sin θ = y
OR
π
sin ( --- − θ ) = cos θ
2
π
cos ( --- − θ ) = sin θ
2
y
1
D
x
–1
O
y
θ
C(90 – θ )
1
θ
1
x
B(θ )
y
A1
x
–1
We say that sine and cosine are complementary functions.
Although the complementary function for tangent is not required for this course, you
may like to try to find it; that is,
tan (90 − θ )° = ?
MQ Maths B Yr 11 Ch 06 Page 249 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
WORKED Example 9
249
If sin θ = 0.4 and 0° < θ < 90°, find, correct to 3 decimal places: a cos θ
b tan θ.
THINK
WRITE
2
2
a sin2 θ + cos2 θ = 1
a 1 Use the identity sin θ + cos θ = 1.
(0.4)2 + cos2 θ = 1
2 Substitute 0.4 for sin θ.
cos2 θ = 1 − 0.16
3 Solve the equation for cos θ correct
to 3 decimal places.
= 0.84
cos θ = ± 0.84
= 0.917 or −0.917
Retain
the
positive
answer
only
as
For
0°
<
θ
<
90°,
cos is positive
4
cosine is positive in the first quadrant.
so cos θ = 0.917.
sin θ
sin θ
b tan θ = -----------b 1 Use the identity tan θ = ------------ .
cos θ
cos θ
0.4
2 Substitute 0.4 for sin θ and 0.917 for
= ------------0.917
cos θ.
3
= 0.436
Calculate the solution correct to 3
decimal places.
WORKED Example 10
Find all possible values of sin θ if cos θ = 0.75.
THINK
2
2
1 Use the identity sin θ + cos θ = 1.
2 Substitute 0.75 for cos θ.
3 Solve the equation for sin θ correct to 3
decimal places.
4
Retain both the positive and negative
solutions, since the angle could be in
either the first or fourth quadrants.
WORKED Example 11
Find a if 0° < a < 90° and
a sin a = cos 42°
THINK
a 1 Write the equation.
2 Replace cos 42° with sin (90 − 42)°
(complementary functions).
b
1
2
WRITE
sin2 θ + cos2 θ = 1
sin2 θ + (0.75)2 = 1
sin2 θ = 1 − 0.5625
= 0.4375
sin θ = ± 0.4375
= 0.661 or −0.661
Write the equation.
Replace sin 73° with cos (90 − 73)°.
b cos a = sin 73°.
WRITE
a sin a = cos 42°
sin a = sin (90 − 42)°
sin a = sin 48°
a = 48°
b cos a = sin 73°
cos a = cos (90 − 73)°
cos a = cos 17°
a = 17°
MQ Maths B Yr 11 Ch 06 Page 250 Thursday, October 25, 2001 3:59 PM
250
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
WORKED Example 12
2
If 0° < a < 90° and cos a = --3- , find the exact values of:
a sin a
b tan a
c cos (90 − a)
d sin (180 + a).
THINK
1
2
WRITE
Draw a right-angled triangle.
Mark in angle a, its adjacent side (adj) 2 and
the hypotenuse (hyp) 3.
a
hyp = 3
adj = 2
opp = 5
3
a
b
c
d
Use Pythagoras to calculate the opposite side
(opp) to a.
1
opp
Use the right-angled triangle to find --------- .
hyp
2
Substitute opp =
1
opp
Use the right-angled triangle to find --------- .
adj
2
Substitute opp =
1
Use the identity cos (90 − a) = sin a.
2
Substitute sin a =
1
2
opp
a sin a = --------hyp
=
5 and hyp = 3.
5
------3
opp
b tan a = --------adj
=
5 and adj = 2.
5
------2
c cos (90 − a)° = sin a
cos (90 − a) =
5
------- .
3
5
------3
d sin (180 + a)° = −sin a
Use the symmetry property
sin (180 + a)° = −sin a.
Substitute sin a =
O2 = 32 − 22
=5
O = 5
5
sin (180 + a) = − -----3
5
------- .
3
(Note: The above results could have been obtained using the identities directly.)
remember
remember
1. sin2 θ + cos2 θ = 1
3. sin (90 − θ )° = cos θ
π
5. sin  --- – θ = cos θ
2

sin θ
2. tan θ = -----------cos θ
4. cos (90 − θ )° = sin θ
π
6. sin  --- – θ = cos θ
2

MQ Maths B Yr 11 Ch 06 Page 251 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
6D
251
Identities
1 Copy and complete the table below, correct to 3 decimal places:
θ°
30°
81°
129°
193°
260°
−47°
350°
sin2 θ
cos2 θ
sin2 θ + cos2 θ
WORKED
Example
9a
WORKED
Example
9b
WORKED
Example
2 If sin θ = 0.8 and 0° < θ < 90°, find, correct to 3 decimal places:
a cos θ
b tan θ.
3 If cos θ = 0.3 and 0° < θ < 90°, find, correct to 3 decimal places:
a sin θ
b tan θ.
4 Find all possible values of the following.
a cos x if sin x = 0.4
c sin x if cos x = 0.24
10
5
3
5
x
b cos x if sin x = −0.7
d sin x if cos x = −0.9
Use the diagram at left to find the exact values of:
a c
b sin x
c cos x.
b
c
6 Use the diagram at right to find the exact values of:
a b
b cos x
c tan x.
2 7
8
x
7 Find the exact values of:
a cos x if sin x =
b sin x if cos x =
c
cos x if sin x =
d sin x if cos x =
12
------ and 90° < x < 180°
13
− 3--5- and x is in the third quadrant
7
- and x is in the fourth quadrant
− ----25
3π
3
------- and ------ < x < 2π
2
2
8 multiple choice
Examine the diagram at right and answer the following questions.
a sin 54° is equal to:
A cos 54°
D sin 36°
b cos 54° is equal to:
A tan 36°
D sin 36°
B cos 36°
E tan 54°
C tan 36°
B cos 36°
E sin 54°
C tan 54°
a
36°
c
54°
b
MQ Maths B Yr 11 Ch 06 Page 252 Monday, October 29, 2001 12:22 PM
252
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
c
tan 36° is equal to:
cos 36°
A -----------------sin 36°
D sin 54° cos 54°
d tan 54° is equal to:
cos 36°
A -----------------sin 36°
D sin 54° cos 54°
WORKED
Example
11
sin 36°
C -----------------cos 36°
B sin 36° cos 36°
E sin 36° + cos 36°
cos 54°
C -----------------sin 54°
B sin 54° − cos 54°
E sin 36° cos 36°
9 Find a° if 0° ≤ a° ≤ 90° and:
a sin a° = cos 20°
b sin a° = cos 58°
d cos a° = sin 82°
e sin 8° = cos a°
g sin 89° = cos a°
h cos 17° = sin a°.
c
f
cos a° = sin 39°
cos 44° = sin a°
10 Copy and complete the following table.
sin θ
0.8
0.28
cos θ
0.6
0.96
0.77
0.3
tan θ
WORKED
Example
12
3.18
11 If 0° < a°, b°, c° < 90° and sin a° =
a
d
g
j
b
e
h
k
sin b°
tan a°
sin (90 − a)°
sin (180 − a)°
0.573
2
------- ,
5
0.447
1.2
2
cos b° = 3--5- , tan c° =
tan b°
sin c°
cos (90 − b)°
cos (180 + b)°
c
f
i
l
0.7
11
---------- ,
5
find:
cos a°
cos c°
sin (90 − c)°
tan (180 + c)°.
Further trigonometric identities
sin θ
The equations tan θ = ------------ and sin2 θ + cos2 θ = 1 are not the only non-trivial
cos θ
trigonometric identities.
Prove (or at least verify) that the equations below are also identities using one of
the following methods:
i Use the identities above, and algebraic manipulation.
ii Complete a table of values for several values of x and show that the left side
of the equation equals the right side.
iii Plot the left-hand side as Y1 and the right-hand side as Y2 using a graphics
calculator (or using graphing software) to show both sides’ graphs are
identical.
1 sin 2x = 2 sin x cos x
2 sin 3x + sin x = 2 sin 2x cos x
1
3 1 + tan2 x = -------------2
cos x
4 sin (x + y) = sin x cos y + cos x sin y
MQ Maths B Yr 11 Ch 06 Page 253 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
253
Using the Pythagorean identity
Consider the quadratic equation x2 − x = 0. This equation is solved by first factorising
the expression then solving each factor equal to zero. Hence, there are two solutions to
the equation, as shown.
x2 − x = 0
x(x − 1) = 0
so x = 0 or x − 1 = 0.
That is, x = 0 or x = 1.
A similar equation involves the use of the trigonometric ratios. Consider the equation
2 sin2 θ = sin θ.
This equation is solved in the same way as a normal quadratic equation; however, the
two answers are in terms of the trigonometric ratio and then have to be solved.
WORKED Example 13
Solve the equation 2 sin2 θ = sin θ over the domain 0 ≤ θ ≤ 2π.
THINK
WRITE
4
Write the equation.
Move sin θ to the left of the equation.
Factorise the expression.
Set each factor equal to zero and solve.
5
Solve sin θ = 0 and 2 sin θ − 1 = 0.
1
2
3
6
Combine all five solutions to the
equation.
2 sin2 θ = sin θ
2 sin2 θ − sin θ = 0
sin θ (2 sin θ − 1) = 0
sin θ = 0 or 2 sin θ − 1 = 0
1
sin θ = --2
π 5π
θ = 0, π, 2π θ = ---, -----6 6
π
5π
θ = 0, ---, π, ------, 2 π
6
6
Some equations of this type will involve both the sin and cos ratios, but to solve the
equation there must be only one ratio. We use the identity sin2θ + cos2 θ = 1.
WORKED Example 14
Solve the equation 2 sin2 x = cos x + 1 over the domain 0 ≤ θ ≤ 2π.
THINK
1
2
3
4
Write the equation.
Make the substitution sin2 x = 1 − cos2 x.
Form a quadratic equation by bringing all
terms to one side of the equation.
Factorise the quadratic.
WRITE
2 sin2 x = cos x + 1
2(1 − cos2 x) = cos x + 1
2 − 2 cos2 x = cos x + 1
1 − 2 cos2 x − cos x = 0
2 cos2 x + cos x − 1 = 0
(2 cos x − 1)(cos x + 1) = 0
Continued over page
MQ Maths B Yr 11 Ch 06 Page 254 Thursday, October 25, 2001 3:59 PM
254
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
THINK
WRITE
5
Solve each factor equal to 0.
6
1
Solve cos x = --- and cos x = −1.
2
7
Combine all solutions.
2 cos x − 1 = 0 or cos x − 1 = 0
1
cos x = --cos x = −1
2
π 5π
x = ---, -----x=π
3 3
π
5π
x = ---, π, -----3
3
remember
remember
1. Some trigonometric equations are solved as quadratic equations.
2. sin2 x + cos2 x = 1
3. Equations should have only one trigonometric ratio. The identities
cos2 x = 1 − sin2 x and sin2 x = 1 − cos2 x can be used to reduce an
equation to one trigonometric ratio.
6E
WORKED
Example
13
Using the Pythagorean identity
1 Solve the trigonometric equation 2 cos2 θ = cos θ over the domain 0 ≤ θ ≤ 2π.
2 Solve each of the following equations over the domain 0 ≤ x ≤ 2π.
b cos2 x + cos x = 0
c 2 sin2 x + 3 sin x = 0
a sin2 x − sin x = 0
d 2 cos2 x + cos x − 1 = 0 e sin2 x + 3 sin x − 4 = 0 f 2 sin2 x − sin x − 1 = 0
3 Solve the trigonometric equation 2 sin2 x = 1 − sin x over the domain 0 ≤ x ≤ 2π.
4 Solve each of the following equations over the domain 0 ≤ θ ≤ 2π.
a 2 cos2 θ = 1 + sin θ
b 2 sin2 θ + sin θ − 1 = 0 c 2 sin2 θ − 1 = 0
14
e sin2 θ = 1 + cos θ
f 2 cos2 θ = 5 + 5 sin θ
d 1 + cos θ = 2 sin2 θ
WORKED
Example
Work
ET
SHE
6.2
MQ Maths B Yr 11 Ch 06 Page 255 Thursday, October 25, 2001 3:59 PM
C h a p t e r 6 Tr i g o n o m e t r i c e q u a t i o n s
255
summary
Simple trigonometric equations
• Trigonometric equations are equations that use trigonometric ratios.
• Trigonometric equations are periodic and so may have an infinite number of
solutions unless the domain is restricted.
• In a domain of one revolution most trigonometric equations will have two
solutions.
• Be aware of the special triangles as they may provide the solution to many
equations.
Radians
• A trigonometric equation may need to be solved using radians.
• The domain within which you are asked to solve the equation will tell you whether
to use degrees or radians.
Further trigonometric equations
• To find solutions to a trigonometric equation between 2π and 4π, add 2π to any
solutions in the first revolution.
• To find solutions to trigonometric equations between −2π and 0, subtract 2π from
any solutions in the first revolution.
Identities
• sin2 θ + cos2 θ = 1
sin θ
• tan θ = -----------cos θ
π
• sin (90 − θ )° = cos θ or sin  --- – θ = cos θ
2

π
• cos (90 − θ )° = sin θ or sin  --- – θ = cos θ
2

Using the Pythagorean identity
• The Pythagorean identity can be used to simplify a quadratic equation using two
trigonometric ratios when one of them is squared.
• cos2 x = 1 − sin2 x and sin2 x = 1 − cos2 x
MQ Maths B Yr 11 Ch 06 Page 256 Thursday, October 25, 2001 3:59 PM
256
M a t h s Q u e s t M a t h s B Ye a r 1 1 f o r Q u e e n s l a n d
CHAPTER
review
6A
1 Solve the following trigonometric equations over the domain 0° ≤ θ ≤ 360°, correct to the
nearest degree.
b cos θ = −0.4
c tan θ = 1.6
a sin θ = 0.9
6A
2 Find exact solutions to the following trigonometric equations over the domain 0° ≤ θ ≤ 360°.
a
3
sin θ = – ------2
1
b cos θ = --2
c
2
tan θ = ------2
6A
3 Solve each of the following equations over the domain 0° ≤ θ ≤ 360°.
b cos θ = −1
c tan θ = −1
a sin θ = −1
6A
4 Solve the following trigonometric equations over the domain −360° ≤ θ ≤ 360°.
b tan θ = −2.25
c sin θ = 0.95
a cos θ = −0.5
6B
5 Solve each of the following equations over the domain 0 ≤ x ≤ 2π. Give your answers
correct to 2 decimal places.
a sin x = 0.7
b cos x = −0.85
c tan x = 0.2
6B
6 Solve each of the following over the domain 0 ≤ x ≤ 2π.
a
3
cos x = – ------2
b tan x = – 3
c
3
sin x = – ------3
6B
7 Find exact solutions to each of the following equations over the domain 0 ≤ x ≤ 2π.
a 2 sin x = 3
b 2 cos x = −1
c 4 tan x = −4
6B
8 Solve each of the following, to the nearest degree, over the domain 0° ≤ x ≤ 360°.
a 4 sin x = 3
b 4 cos x = −3
c 2 tan x − 6 = 0
6C
9 Solve each of the trigonometric equations below over the domain 0 ≤ x ≤ 4π.
a cos x = −0.458
b sin x = −0.504
c tan x = −0.84
6C
10 Solve the trigonometric equations below over the domain −2π ≤ x ≤ 2π.
a sin x = −0.816
b cos x = 0.427
c tan x = −1.6774
6C
11 Find solutions to the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
1
a
3 cos x + 1 = 0
b 4 sin x = 0
c ------- tan x – 1 = 0
3
12 Find:
a cos θ if sin θ = 0.5 and θ lies in the second quadrant
5
b sin x if cos x = – ------ and x is in the third quadrant.
12
6D
6E
6E
13 Given that a lies in the first quadrant find a if:
a sin a = cos 30°
b cos a = cos 28°.
14 Solve the trigonometric equation 2 sin2 θ = sin θ for the domain 0 ≤ θ ≤ 2π.
15 Solve each of the following equations over the domain 0 ≤ x ≤ 2π.
b sin2 x + 3 sin x − 4 = 0
a 3 cos2 x − 2 cos x = 0
2
d 2 − sin x − 3 cos2 x = 0
c 2 cos x = 1 + sin x
test
yourself
CHAPTER
6D
6