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Trigonometry |
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Trigonometric
Equations |
Equations
of the Type a
cos
x +
b
sin
x = c |
Introducing
an auxiliary angle method |
Introducing
an auxiliary angle method example |
Introducing new unknown
t
= tan x/2 |
Introducing new unknown
t
= tan x/2
example |
The
Basic Strategy for Solving Trigonometric Equations |
Trigonometric
equations examples |
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Trigonometric
equations of the type
a
cos
x +
b
sin
x = c |
To
solve the trigonometric equations which are linear in sin
x and cos
x, and where, a,
b,
and c
are real numbers
we can use the two methods, |
a)
introducing an auxiliary angle, and b)
introducing new unknown. |
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a)
Introducing an auxiliary angle method to solve trigonometric equation |
Consider the constants
a
and b
as rectangular coordinates of a point expressed by polar coordinates (r,
j), |
then,
a =
r cos j
and b =
r sin j, |
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By substituting for
a
and b
in the given equation |
a
· cos
x + b · sin
x = c |
obtained is, r
cos
x · cos j
+ r sin
x · sin j
= c or |
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using
addition formula yields,
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since
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obtained is
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the basic trigonometric equation whose solution is known. |
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Note that the given equation,
a
cos x + b sin x = c
will have a solution if |
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it follows that the constants,
a,
b
and c
should satisfy relation c2
<
a2 + b2. |
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Introducing an auxiliary angle
method example |
Example: Solve the equation,
sin x +
Ö3
· cos x = 1.
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Solution: Comparing corresponding parameters of the given equation with
a
cos x + b sin x = c
it follows,
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a
= Ö3,
b =
1 and c
= 1. |
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By substituting given quantities to the basic equation |
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or
x -
30° = +
60° + k · 360° thus,
solutions are, x = 90° + k · 360° and
x′ = -
30° + k · 360°, kÎ Z. |
The same solution can be obtained using following
procedure, from sin x +
Ö3
· cos x = 1 |
and a
cos x + b sin x = c | ¸ b |
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that means that we can introduce an auxiliary angle
j
that is |
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sin x
· sin 30° + cos x · cos 30° = sin 30°,
cos (x
-
30°) = 1/2 |
and this is the same basic equation obtained above. |
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b)
Introducing new unknown t
= tan x/2 |
If in the equation a
cos x + b sin x = c
we substitute the sine and cosine functions by
tan x/2 = t
that is, |
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the equation becomes |
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and after rearranging |
(a +
c) · t2
-
2b · t + (c -
a) = 0. |
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Obtained quadratic equation will have real solutions
t1,2
if its discriminant is greater then or equal to
zero, |
that is if
(-2b)2
-
4 (a + c)(c -
a) >
0 or
c2
< a2
+ b2,
which is earlier mentioned condition. |
If this condition is satisfied, the solutions,
t1
and t2
can be substituted into tan x/2 =
t1 and
tan x/2 = t2. |
Thus,
obtained are the basic trigonometric equations. |
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Introducing new unknown
t
= tan x/2
example |
Example: Solve the equation,
5 sin x
-
4 cos x = 3.
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Solution: Given equation is of the form
a
cos x + b sin x = c
therefore parameters are,
a
= -
4, b =
5 and
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c
= 3, after introducing new unknown
tan x/2 = t
and substituting the values of the parameters into equation |
(a +
c) · t2
-
2b · t + (c -
a) = 0
gives (-
4 + 3) · t2
-
2 · 5 · t + [3 - (-
4)] = 0 |
or
t2
+ 10t -
7 = 0,
t1,2
= -
5 +
Ö25
+ 7 = -
5 +
4Ö2. |
Obtained values for variable
t
we plug into substitutions,
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tan x/2 = t1,
x/2 = tan-1
(t1)
or x =
2arctan(-
5 -
4Ö2) |
x = 2 · (-
84°38′21″ + k · 180°) =
-
169°16′42″ + k · 360°,
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tan x/2 = t2,
x/2 = tan-1
(t2)
or x′ =
2arctan(-
5 + 4Ö2) |
x′ =
2 · (33°17′56″ + k · 180°) =
66°35′53″ + k · 360°. |
The same result we obtain using the method of introducing the auxiliary angle
j. Plug the given parameters
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a
= -
4, b =
5 and c
= 3 into tan
j
= b/a,
tan j =
5/(-
4) => j =
-51°20′24″,
cos j =
0.624695, |
then
from the equation |
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cos
(x + 51°20′24″) =
(3/-
4) · 0.624695 |
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or x
+ 51°20′24″ =
arccos [(3/-
4) · 0.624695], |
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thus,
x +
51°20′24″ =
+ 117°56′18″ + k · 360° => x =
+ 117°56′18″
-
51°20′24″ + k · 360°. |
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Trigonometric equations of the form a
cos x + b sin x = c
we do not solve using the identity
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since that way given equation becomes quadratic with four solutions but only two of
them satisfy it. |
We will solve the equation from the previous example using this method anyway. |
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Example: Solve the equation,
5 sin x
-
4 cos x = 3
by substituting |
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Solution: Squaring both sides of an equation may introduce extraneous
or redundant (not needed) solutions.
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By plugging the results into given equation show that only solutions
b) and c) satisfy the equation what match with previous results obtained using another two methods.
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The
basic strategy for solving trigonometric equations |
When solving trigonometric equations we usually use some of the following
procedures, |
- apply known
identities to modify given equation to an equivalent expressed in terms of one function, |
- rearrange the given
equation using different trigonometric formulae to an equivalent, until you recognize one |
of the known types. |
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Trigonometric equations
examples |
Example: Find the solution
of the equation, 2
sin
(x
+ 60°) · cos
x = 1.
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Solution: Applying products as sums formula
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2 · (1/2)
[sin
(x
+ 60°
+ x)
+ sin
(x
+ 60° - x)] = 1
or sin
(2x
+ 60°)
+ sin 60° = 1
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then
sin
(2x
+ 60°) = 1
- Ö3/2,
2x
+ 60° =
sin-1(1
- Ö3/2) + k · 360° |
and 2x′
+ 60° =
180°
- sin-1(1
- Ö3/2) + k · 360° |
so,
the solution is 2x =
-
60°
+ sin-1(1
- Ö3/2) + k · 360°,
x =
-
26°9′1″ + k · 180°, |
2x′ =
120° - sin-1(1
- Ö3/2) + k · 360°,
x′ = 56°9′1″ + k · 180°,
k Î Z. |
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Example: Find the solution
of the equation, |
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Solution: Using
identities |
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given
equation becomes 2 · (1
+ cos x) - Ö3
· cot x/2 = 0, |
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therefore,
1
+ cos x
= 0,
cos
x
= - 1,
x
= p
+ k · 2p, |
and 2sin
x - Ö3
= 0,
sin
x
= Ö3/2,
x
= p/3
+ k · 2p
and x′ = 2p/3
+ k · 2p,
k Î Z. |
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Example: Solve the equation,
cos
2x
+ cos
6x - cos
8x -
1 = 0.
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Solution: To the first two terms apply the sum to product formula and remaining two terms transform using
known identity, thus
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and
since 1
+ cos
2x = 2cos
2 x
then, 1
+ cos
8x = 2cos
2 4x plugging into the
given equation |
2cos
4x
· cos
2x - 2cos
2 4x
= 0
or 2cos
4x
· (cos
2x - cos
4x)
= 0 |
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it
follows that 4cos
4x
· sin
3x
· sin
x
= 0
meaning,
cos
4x
= 0,
sin
3x
= 0
and/or sin
x
= 0.
Since |
sin
3x
= sin (x
+ 2x)
= sin
x
· cos
2x + cos
x
· sin
2x
= sin
x
· (cos
2 x -
sin
2 x) + cos
x
· 2sin
xcos
x |
=
sin
x
· (1 -
2sin
2 x) + 2sin
x
· (1 -
sin
2 x)
=
3sin
x - 4sin
3 x,
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solutions of the equation
sin
x
= 0
are included in the solutions of sin
3x
= 0.
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If sin
x
= 0
then,
sin
3x
= 3sin
x - 4sin
3 x
= sin
x
· (3 -
4sin
2 x)
= 0.
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Therefore, the solution set of the given equation is the union of the solutions of each of the equations, that is
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cos
4x
= 0,
4x
=
+ p/2
+ k · 2p,
x
=
+
p/8
+ k · p/2
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and sin
3x
= 0,
3x
= k · p,
x
= k · p/3,
k Î Z.
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Functions
contents D |
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