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6syllabussyllabusrrefefererenceenceTopic:• Periodic functions and
applications
In thisIn this chachapterpter6A Simple trigonometric
equations6B Equations using radians6C Further trigonometric
equations6D Identities6E Using the Pythagorean
identity
Trigonometricequations
MQ Maths B Yr 11 Ch 06 Page 237 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
IntroductionSudhira is a keen fisherman. The ideal depth for fishing in Sudhira’s favourite tidal lakeis 3 metres. The depth of water in the lake can be found using the equation
D = 5 − 4 sin t
where t is the time in hours after midnight. What is the best time of day for Sudhira tofish?
To solve this problem we need to solve a trigonometric equation
Simple trigonometric equationsFrom 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:
cos θ =
cos θ =
θ = 46°.
The equation cos θ = is an example of a trigonometric equation. This trigonometric
equation had to be solved in order to find the size of the angle in the triangle. In thisparticular 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 functionwas periodic. This means that there are values of θ, other than the one already found
for which cos θ = . There will, in fact, be an infinite number of solutions to this
trigonometric equation, so for practical reasons we are usually given a domain withinwhich 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 give a first quadrant answer. To complete the solution we need to consider all quadrants for which the trigonometric ratio is positive.
In the case of cos θ = the cosine ratio is positive in
the first and fourth quadrants. We found earlier that the 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 thecorresponding angles in the negative quadrants.
p6---
13 cm
9 cmθ
adjhyp---------
913------
913------
913------
y
x
Sinepositive
Allpositive
Tangentpositive
Cosinepositive
913------
MQ Maths B Yr 11 Ch 06 Page 238 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
In the earlier chapter we also found that we were able to find exact values of specialangles using the triangles on the next page.
Solve the following trigonometric equations over the domain 0∞ ≤ θ ≤ 360∞, correct to the nearest degree.a sin θ = 0.412
b tan θ = −
THINK WRITE
a Write the equation. a sin θ = 0.412Use your calculator to find the first quadrant angle.
First quadrant angle = 24°
The sine ratio is positive in the first and second quadrants.Find the second quadrant angle by subtracting 24° from 180°.
180° − 24° = 156°
Write the answer. θ = 24° or 156°
b Write the equation. b tan θ = −
Use your calculator to find the first quadrant angle.
First quadrant angle = 20°
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°.
180° − 20° = 160°360° − 20° = 340°
Write the answer. θ = 160° or 340°
411------
12
y
x24°
180° − 24°
3
4
5
14
11------
2
y
x20°
360° − 20°
180° − 20°
3
4
5
1WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 239 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
These special angles should be used where possible in the solution to a trigonometricequation. They are used when we recognise any of the values produced by the triangles.
sin 30° = cos 30° = tan 30° =
sin 45° = cos 45° = tan 45° = 1
sin 60° = cos 60° = tan 60° =
Similarly, we must be aware of when theboundary angles should be used in the solutionof the equation. Remember from the work onthe unit circle that y = sin θ, x = cos θ and
tan θ = .
45°
45°
2
1
1
30°
60°
2
1
Line of bisection
12--- 3
2------- 3
3-------
22
------- 22
-------
32
------- 12--- 3
Solve the equation cos q = over the domain 0∞ ≤ θ ≤ 360∞.
THINK WRITE
Write the equation. cos θ =
Use the special triangles to find the first quadrant angle.
First quadrant angle = 30°
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°.
180° − 30° = 150°180° + 30° = 210°
Write the answer. θ = 150° or 210°
32
-------–
13
2-------–
2
3
4
5
2WORKEDExample
yx--
y
x0
0° or 360°
270°
180°
90°
1
1
–1
–1
MQ Maths B Yr 11 Ch 06 Page 240 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
Simple trigonometric equations
1 Solve each of the following trigonometric equations over the domain 0° ≤ θ ≤ 360°,correct to the nearest degree.
2 Find exact solutions to each of the following trigonometric equations over the domain0° ≤ θ ≤ 360°.
3
If sin x = cos x = − and 0° ≤ x ≤ 360°, then x is:
4 It is known that sin θ < 0 and that tan θ > 0. Which quadrant does the angle θ lie in?Explain your answer.
a sin θ = 0.6 b cos θ = −0.25 c tan θ = 5.72 d sin θ = −0.85e cos θ = 0.195 f tan θ = −0.837 g sin θ = −0.333 h cos θ = 0.757
a sin θ = b cos θ = c tan θ = d sin θ =
e cos θ = f tan θ = − g sin θ = − h cos θ =
A 150° or 210° B 135° or 225° C 225°D 135° or 315° E 120°
Solve the equation sin θ = -1 in the domain 0∞ ≤ θ ≤ 360∞.
THINK WRITE
Write the equation. sin θ = −1y = sin θ so find the angle with a y-value of −1.
θ = 270°12
3WORKEDExample
remember1. 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 domain0° ≤ 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.
remember
6A
SkillSH
EET 6.1
SkillSH
EET 6.2
WWORKEDORKEDEExample
1
WWORKEDORKEDEExample
23
2------- 2
2------- 3
12---–
12---– 3
3------- 2
2------- 3
2-------
mmultiple choiceultiple choice
22
-------
MQ Maths B Yr 11 Ch 06 Page 241 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°.
7 Solve the following trigonometric equations over the domain −360° ≤ θ ≤ 360°.
Equations using radiansWe have seen that a radian is an alternative method of measuring an angle. A trigon-ometric equation can be solved using radians as well as degrees. Usually the domaingiven will indicate whether it is expected that you will solve the equation in degrees orin 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 ≤ 2πthen 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 isin radian mode before attempting to solve the problem to give an answer in radians.
a sin θ = 1 b cos θ = 0 c tan θ = 0 d sin θ = 0e cos θ = −1 f sin θ = −1
a sin θ = 0.5 b cos θ = 0.35 c tan θ = −1 d sin θ = −0.87e cos θ = −0.87 f tan θ = 1.4
WWORKEDORKEDEExample
3
Solve the equation tan θ = 0.8 over the domain 0 ≤ θ ≤ 2π. Give the answer correct to 2 decimal places.
THINK WRITE
Write the equation. tan θ = 0.8Use your calculator to find the first quadrant angle.
First quadrant angle = 0.67
The tangent ratio is positive in the first and third quadrants.
Find the third quadrant angle by adding 0.67 to π.
π + 0.67 = 3.81
Write the answer. θ = 0.67 or 3.81
12
y
x0.67
+ 0.67π
3
4
5
4WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 242 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 243When the special angles are used, it is still important to recognise them and recog-
nise their radian equivalents in terms of π.
sin = cos = tan =
sin = cos = tan = 1
sin = cos = tan =
All of the equations that we have dealt with so far have been one-step solutions. Inmany examples we may need to rearrange the equation before we are able to use thecalculator to solve it.
When rearranging the equation, we attempt to place the trigonometric ratio alone onone side of the equation, as in the example above.
π6--- 1
2--- π
6--- 3
2------- π
6--- 3
3-------
π4--- 2
2------- π
4--- 2
2------- π
4---
π3--- 3
2------- π
3--- 1
2--- π
3--- 3
Solve the equation sin θ = over the domain 0 ≤ θ ≤ 2π.
THINK WRITE
Write the equation. sin θ =
Use the special triangles to find the first quadrant angle.
First quadrant angle =
The sine ratio is negative in the third and fourth quadrants.
Find the third quadrant angle by adding
to π and the fourth quadrant angle by
subtracting from 2π.
θ = π + or θ = 2π +
θ = or θ =
Write the answer. θ = or
32
-------–
13
2-------–
2π3---
y
x3—π
2 –π3—π
+ 3—ππ
3
4
π3---
π3---
π3--- π
3---
4π3
------ 5π3
------
54π3
------ 5π3
------
5WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 243 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
Equations using radians
1 Solve each of the following equations over the domain 0 ≤ x ≤ 2π. Give your answerscorrect to 2 decimal places.
2 Solve each of the following over the domain 0 ≤ x ≤ 2π.
a sin x = 0.8 b cos x = −0.5 c tan x = 1.5 d sin x = −0.327e cos x = 0.707 f tan x = −0.39
a sin x = b cos x = − c tan x = 1 d cos x =
e tan x = − f sin x = −
Find x if 2 sin x = 0.984 over the domain 0 ≤ x ≤ 2p.
THINK WRITE
Write the equation. 2 sin x = 0.984Divide both sides by 2 to get sin x by itself.
sin x = 0.492
Use your calculator to find the first quadrant angle.
First quadrant angle = 0.514
The sine ratio is positive in the first and second quadrants.
Find the second quadrant angle by subtracting 0.514 from π.
θ = π − 0.514θ = 2.628
Write the answer. θ = 0.514 or 2.628
12
3
4 y
x0.514
– 0.514
π
5
6
6WORKEDExample
remember1. 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.
remember
6BWWORKEDORKEDEExample
4
WWORKEDORKEDEExample
5 32
-------12--- 2
2-------
312---
MQ Maths B Yr 11 Ch 06 Page 244 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 2453 Solve each of the following over the domain 0 ≤ x ≤ 2π.
4 Find exact solutions to each of the following equations over the domain 0 ≤ x ≤ 2π.
5
The solution to the equation 2 cos x + 1 = 0 over the domain 0 ≤ x ≤ 2π is:
6 Solve each of the following, to the nearest degree, over the domain 0 ≤ x ≤ 360°.
Further trigonometric equationsIn many cases the equation that we have to solve may not be in the domain0 ≤ x ≤ 2π. We may be asked to solve the equation in the domain 0 ≤ x ≤ 4π (2 revol-utions) 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 wesimply add or subtract 2π to the first revolution solutions.
a sin x = 0 b tan x = 0 c cos x = 0 d sin x = 1e cos x = 1 f cos x = −1 g sin x = −1
a 2 sin x = 1 b 2 cos x = c 2 tan x = 2d 2 sin x + = 0 e 2 cos x + = 0 f tan x + 3 = 0
A B C
D E
a 4 sin x = 1 b 3 cos x = −2 c 2 tan x − 7 = 0d 4 + sin x = 3 e 1 + 2 cos x = f 3 tan x + 9 = 0
WWORKEDORKEDEExample
6 33 2 3
mmultiple choiceultiple choice
π3--- 2π
3------, 5π
6------ 7π
6------, π
6--- 5π
6------,
2π3
------ 4π3
------, 4π3
------ 5π3
------,
2
Find α if sin α = 0.7 in the domain 0 ≤ α ≤ 4p.
Continued over page
THINK WRITE
Write the equation. sin α = 0.7Use your calculator to find the first quadrant angle.
First quadrant angle = 0.7754
The sine ratio is positive in the first and second quadrants.
12
y
x
– 0.7754
π
0.7754
3
7WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 245 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
In many equations you will first need to make the trigonometric ratio the subject ofthe equation.
THINK WRITE
Find the second quadrant angle by subtracting 0.7754 from π.
α = π − 0.7754α = 2.3662
Find the solutions between 2π and 4π adding 2π to each of the first revolution solutions.
α = 0.7754 + 2π α = 2.3662 + 2πα = 7.0586 α = 8.6494
Write the answer. α = 0.7754, 2.3662, 7.0586, 8.6494
4
5
6
Find x if cos x + 1 = 0 over the domain -2π ≤ x ≤ 2π.
THINK WRITE
Write the equation. cos x + 1 = 0
Make cos x the subject of the equation. cos x = −1
cos x =
Use the special triangles to find the first quadrant angle.
First quadrant angle =
The cosine ratio is negative in the second and third quadrants.
Find the second quadrant angle by
subtracting from π. Find the third
quadrant angle by adding to π.
x = π − x = π +
x = x =
To find the solutions between −2π and 0, subtract 2π from each of the first revolution solutions.
x = − 2π x = − 2π
x = x =
Write the answer. x =
2
1 2
2 21–
2-------
3π4---
4
5
π4---
π4---
π4--- π
4---
3π4
------ 5π4
------
63π4
------ 5π4
------
5π–4
--------- 3π–4
---------
75π–4
--------- 3π–4
--------- 3π4
------ 5π4
------, , ,
8WORKEDExample
remember1. 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.
remember
MQ Maths B Yr 11 Ch 06 Page 246 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 247
Further trigonometric equations
1 Solve each of the following trigonometric equations over the domain 0 ≤ x ≤ 4π.
2 Solve each of the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
3 Find the solutions to the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
4 Find all the solutions to the following equations over the domain −2π ≤ x ≤ 2π. Giveeach answer correct to 2 decimal places.
5 A particle moves in a straight line so that its distance, x metres, from a point O is givenby the equation x = 3 + 4 sin t, where t is the time in seconds after the particle beginsto 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 2decimal places.
IdentitiesAn identity is a relationship that holds true for all legitimate values of a pronumeral orpronumerals. For example, a simple identity is x + x = 2x. The identities described inthis section are far more interesting and useful than this, as you will see.
a cos x = −0.6591 b sin x = 0.9104 c cos x = 0.48 d sin x = −0.371
e tan x = 0.58 f tan x = −2.1
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
a 2 sin x − 1 = 0 b 3 cos x = 0 c 2 sin x + = 0 d tan x + = 0e cos x = 1 f tan x − 1 = 0
a 4 sin x + 2 = 6 b 3 cos x − 3 = 0 c cos x + 4 = 4.21
d 2 sin x − 5 = −4 e cos x + 2 = 3 f 2 cos x + = 0
FishingYou 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
D = 5 − 4 sin t
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.
6CWWORKEDORKEDEExample
7
WWORKEDORKEDEExample
8 3 3
2 3
12---
2 3
WorkS
HEET 6.1
p6---
MQ Maths B Yr 11 Ch 06 Page 247 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 identityConsider the right-angled triangle in the unit circle shown.
Applying Pythagoras’ theorem to this triangle gives the identity:
sin2 θ + cos2 θ = 1
The tangentConsider the unit circle on the right.
A tangent is drawn at A and extended to the pointC, so that OC is an extension of OP. This tangent iscalled tangent θ, which is abbreviated to tan θ.
Triangles ODP and OAC are similar, because theyhave their three corresponding angles equal.
It follows that: (corresponding sides)
or (as mentioned in an earlier section).
Another relationship between sine and cosine — complementary functionsConsider the unit circle shown on the right:
The triangles OAB and ODC are congruent becausethey have all corresponding angles equal and thehypotenuse equal (radius = 1).
Therefore all corresponding sides are equal and itfollows that:
sin (90 − θ )° = cos θ = x
and cos (90 − θ )° = sin θ = y
OR
sin ( − θ ) = cos θ
and cos ( − θ ) = sin θ
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 )° = ?
sin
cos
y
P( )1
DO
θ
θθ
θ
sin tan
cos
y
x
P( )
1
D A1
1
O
B
Cθ
θθ
θθ
tan θ1
------------ sin θcos θ------------=
θtan sin θcos θ------------=
y
x
y
y
x
x 11
A 1
–1
–1
1
O
D C(90 – )
B( )
θ
θθ
θ
π2---
π2---
MQ Maths B Yr 11 Ch 06 Page 248 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 249
If sin θ = 0.4 and 0° < θ < 90°, find, correct to 3 decimal places: a cos θ b tan θ. THINK WRITEa Use the identity sin2 θ + cos2 θ = 1. a sin2 θ + cos2 θ = 1
Substitute 0.4 for sin θ. (0.4)2 + cos2 θ = 1Solve the equation for cos θ correct to 3 decimal places.
cos2 θ = 1 − 0.16= 0.84
cos θ = ±= 0.917 or −0.917
Retain the positive answer only as cosine is positive in the first quadrant.
For 0° < θ < 90°, cos is positive so cos θ = 0.917.
b Use the identity tan θ = . b tan θ =
Substitute 0.4 for sin θ and 0.917 for cos θ.
=
Calculate the solution correct to 3 decimal places.
= 0.436
123
0.84
4
1sin θcos θ------------ sin θ
cos θ------------
20.4
0.917-------------
3
9WORKEDExample
Find all possible values of sin θ if cos θ = 0.75.
THINK WRITEUse the identity sin2 θ + cos2 θ = 1. sin2 θ + cos2 θ = 1Substitute 0.75 for cos θ. sin2 θ + (0.75)2 = 1Solve the equation for sin θ correct to 3 decimal places.
sin2 θ = 1 − 0.5625= 0.4375
sin θ = ±Retain both the positive and negative solutions, since the angle could be in either the first or fourth quadrants.
= 0.661 or −0.661
123
0.43754
10WORKEDExample
Find a if 0° < a < 90° and a sin a = cos 42° b cos a = sin 73°.
THINK WRITEa Write the equation. a sin a = cos 42°
Replace cos 42° with sin (90 − 42)° (complementary functions).
sin a = sin (90 − 42)°sin a = sin 48°
a = 48°b Write the equation. b cos a = sin 73°
Replace sin 73° with cos (90 − 73)°. cos a = cos (90 − 73)°cos a = cos 17°
a = 17°
12
12
11WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 249 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
If 0° < a < 90° and cos a = , find the exact values of:
a sin a b tan a c cos (90 − a) d sin (180 + a).
(Note: The above results could have been obtained using the identities directly.)
THINK WRITE
Draw a right-angled triangle. Mark in angle a, its adjacent side (adj) 2 and the hypotenuse (hyp) 3.
Use Pythagoras to calculate the opposite side (opp) to a.
O2 = 32 − 22
= 5O =
a Use the right-angled triangle to find . a sin a =
Substitute opp = and hyp = 3. =
b Use the right-angled triangle to find . b tan a =
Substitute opp = and adj = 2. =
c Use the identity cos (90 − a) = sin a. c cos (90 − a)° = sin a
Substitute sin a = . cos (90 − a) =
d Use the symmetry property sin (180 + a)° = −sin a.
d sin (180 + a)° = −sin a
Substitute sin a = . sin (180 + a) = −
23---
1
opp = 5
adj = 2hyp = 3
a2
3
5
1opphyp--------- opp
hyp---------
2 55
3-------
1oppadj--------- opp
adj---------
2 55
2-------
1
25
3------- 5
3-------
1
25
3------- 5
3-------
12WORKEDExample
remember1. sin2 θ + cos2 θ = 1 3. sin (90 − θ )° = cos θ 5. sin = cos θ
2. tan θ = 4. cos (90 − θ )° = sin θ 6. sin = cos θ
π2--- θ–
sin θcos θ------------ π
2--- θ–
remember
MQ Maths B Yr 11 Ch 06 Page 250 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 251
Identities
1 Copy and complete the table below, correct to 3 decimal places:
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.
5 Use the diagram at left to find the exact values of:a cb sin xc cos x.
6 Use the diagram at right to find the exact values of:a b b cos x c tan x.
7 Find the exact values of:
a cos x if sin x = and 90° < x < 180°
b sin x if cos x = − and x is in the third quadrant
c cos x if sin x = − and x is in the fourth quadrant
d sin x if cos x = and < x < 2π
8Examine the diagram at right and answer the following questions.a sin 54° is equal to:
b cos 54° is equal to:
θ° 30° 81° 129° 193° 260° 350° −47°
sin2 θ
cos2 θ
sin2 θ + cos2 θ
a cos x if sin x = 0.4 b cos x if sin x = −0.7c sin x if cos x = 0.24 d sin x if cos x = −0.9
A cos 54° B cos 36° C tan 36°D sin 36° E tan 54°
A tan 36° B cos 36° C tan 54°D sin 36° E sin 54°
6D
WWORKEDORKEDEExample
9a
WWORKEDORKEDEExample
9b
WWORKEDORKEDEExample
10
x
53
c
x78
b
2
1213------
35---
725------
32
-------3π2
------
mmultiple choiceultiple choice36°
54°
a
b
c
MQ Maths B Yr 11 Ch 06 Page 251 Thursday, October 25, 2001 3:59 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:
d tan 54° is equal to:
9 Find a° if 0° ≤ a° ≤ 90° and:
10 Copy and complete the following table.
11 If 0° < a°, b°, c° < 90° and sin a° = , cos b° = , tan c° = , find:
A B sin 36° cos 36° C
D sin 54° cos 54° E sin 36° + cos 36°
A B sin 54° − cos 54° C
D sin 54° cos 54° E sin 36° cos 36°
a sin a° = cos 20° b sin a° = cos 58° c cos a° = sin 39°d cos a° = sin 82° e sin 8° = cos a° f cos 44° = sin a°g sin 89° = cos a° h cos 17° = sin a°.
sin θ 0.8 0.28 0.77 0.573
cos θ 0.6 0.96 0.3 0.447
tan θ 3.18 1.2 2 0.7
a sin b° b tan b° c cos a°d tan a° e sin c° f cos c°g sin (90 − a)° h cos (90 − b)° i sin (90 − c)°j sin (180 − a)° k cos (180 + b)° l tan (180 + c)°.
Further trigonometric identitiesThe equations tan θ = and sin2 θ + cos2 θ = 1 are not the only non-trivial
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
3 1 + tan2 x =
4 sin (x + y) = sin x cos y + cos x sin y
cos 36°sin 36°------------------ sin 36°
cos 36°------------------
cos 36°sin 36°------------------ cos 54°
sin 54°------------------
WWORKEDORKEDEExample
11
WWORKEDORKEDEExample
12
2
5------- 3
5--- 11
5----------
sin θcos θ------------
1
cos2 x---------------
MQ Maths B Yr 11 Ch 06 Page 252 Monday, October 29, 2001 12:22 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 identityConsider the quadratic equation x2 − x = 0. This equation is solved by first factorisingthe expression then solving each factor equal to zero. Hence, there are two solutions tothe equation, as shown.x2 − x = 0x(x − 1) = 0so 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 equation2 sin2 θ = sin θ.
This equation is solved in the same way as a normal quadratic equation; however, thetwo answers are in terms of the trigonometric ratio and then have to be solved.
Some equations of this type will involve both the sin and cos ratios, but to solve theequation there must be only one ratio. We use the identity sin2θ + cos2 θ = 1.
Solve the equation 2 sin2 θ = sin θ over the domain 0 ≤ θ ≤ 2π.
THINK WRITE
Write the equation. 2 sin2 θ = sin θMove sin θ to the left of the equation. 2 sin2 θ − sin θ = 0Factorise the expression. sin θ (2 sin θ − 1) = 0Set each factor equal to zero and solve. sin θ = 0 or 2 sin θ − 1 = 0
sin θ =
Solve sin θ = 0 and 2 sin θ − 1 = 0. θ = 0, π, 2π θ =
Combine all five solutions to the equation.
θ = 0,
1234
12---
5π6--- 5π
6------,
6π6--- π 5π
6------ 2π, , ,
13WORKEDExample
Solve the equation 2 sin2 x = cos x + 1 over the domain 0 ≤ θ ≤ 2π.
Continued over page
THINK WRITE
Write the equation. 2 sin2 x = cos x + 1Make the substitution sin2 x = 1 − cos2 x. 2(1 − cos2 x) = cos x + 1Form a quadratic equation by bringing all terms to one side of the equation.
2 − 2 cos2 x = cos x + 11 − 2 cos2 x − cos x = 02 cos2 x + cos x − 1 = 0
Factorise the quadratic. (2 cos x − 1)(cos x + 1) = 0
1
2
3
4
14WORKEDExample
MQ Maths B Yr 11 Ch 06 Page 253 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
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.
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 sin2 x − sin x = 0 b cos2 x + cos x = 0 c 2 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
a 2 cos2 θ = 1 + sin θ b 2 sin2 θ + sin θ − 1 = 0 c 2 sin2 θ − 1 = 0
d 1 + cos θ = 2 sin2 θ e sin2 θ = 1 + cos θ f 2 cos2 θ = 5 + 5 sin θ
THINK WRITE
Solve each factor equal to 0. 2 cos x − 1 = 0 or cos x − 1 = 0
cos x = cos x = −1
Solve cos x = and cos x = −1. x = x = π
Combine all solutions. x =
5
12---
612--- π
3--- 5π
3------,
7π3--- π 5π
3------, ,
remember1. Some trigonometric equations are solved as quadratic equations.2. sin2 x + cos2 x = 13. 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 anequation to one trigonometric ratio.
remember
6EWWORKEDORKEDEExample
13
3
WorkS
HEET 6.2
WWORKEDORKEDEExample
14
MQ Maths B Yr 11 Ch 06 Page 254 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
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
• tan θ =
• sin (90 − θ )° = cos θ or sin = cos θ
• cos (90 − θ )° = sin θ or sin = cos θ
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
summary
sin θcos θ------------
π2--- θ–
π2--- θ–
MQ Maths B Yr 11 Ch 06 Page 255 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
1 Solve the following trigonometric equations over the domain 0° ≤ θ ≤ 360°, correct to the nearest degree.a sin θ = 0.9 b cos θ = −0.4 c tan θ = 1.6
2 Find exact solutions to the following trigonometric equations over the domain 0° ≤ θ ≤ 360°.
a sin θ = b cos θ = c tan θ =
3 Solve each of the following equations over the domain 0° ≤ θ ≤ 360°.a sin θ = −1 b cos θ = −1 c tan θ = −1
4 Solve the following trigonometric equations over the domain −360° ≤ θ ≤ 360°.a cos θ = −0.5 b tan θ = −2.25 c sin θ = 0.95
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
6 Solve each of the following over the domain 0 ≤ x ≤ 2π.
a cos x = b tan x = c sin x =
7 Find exact solutions to each of the following equations over the domain 0 ≤ x ≤ 2π.a 2 sin x = b 2 cos x = −1 c 4 tan x = −4
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
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
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
11 Find solutions to the following trigonometric equations over the domain −2π ≤ x ≤ 2π.
a b 4 sin x = 0 c
12 Find:a cos θ if sin θ = 0.5 and θ lies in the second quadrant
b sin x if cos x = and x is in the third quadrant.
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.a 3 cos2 x − 2 cos x = 0 b sin2 x + 3 sin x − 4 = 0c 2 cos2 x = 1 + sin x d 2 − sin x − 3 cos2 x = 0
CHAPTERreview
6A
6A3
2-------–
12--- 2
2-------
6A
6A
6B
6B3
2-------– 3– 3
3-------–
6B 3
6B
6C
6C
6C3 cos x 1+ 0= 1
3------- tan x 1– 0=
6D5
12------–
6D6E
testtest
CHAPTERyyourselfourself
testyyourselfourself
66E
MQ Maths B Yr 11 Ch 06 Page 256 Thursday, October 25, 2001 3:59 PM
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