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TEM Journal. Volume 8, Issue 4, Pages 1339-1344, ISSN 2217-8309, DOI: 10.18421/TEM84-34, November 2019.
One Approach for Solving Trigonometric
Equations Using Complex in the
Mathematical Education
1 2 1
Ivo Andreev , Ivan Georgiev , Margarita Varbanova
1Department of Algebra and Geometry, University of Veliko Tarnovo, 3A, Arh. G. Kozarev Blvd. 5000 Veliko
Tarnovo, Bulgaria
2Department of Applied Mathematics and Statistics, University of Ruse, 8 Studentska str. 7000 Ruse, Bulgaria
Abstract – The goal of this development is These things request to enforce necessity of
introducing a reader the solution of one class creation, education and making of talented pupils
comprising trigonometric equations in the Teaching with outstanding mathematical accomplishment and
Course of Mathematics by using trigonometric form of interest for creative development. In is maybe the
the complex numbers. An exemplary approach to cases in which there is a concern that mathematics
solving these equations, suitable for students from 11th and motivation for studying are provoked from the
to 12th grade, as well as for pupils participating in practical application of the mathematical knowledge
mathematical camps, olympiads, mathematical
competitions, computer mathematics olympiads is and their relations with other spheres of the science.
considered. Apart from the mathematics is the trigonometry,
which is in close cooperation with the ancient
Keywords – trigonometric equations, talented pupils, development of astronomy, surveying ships,
complex number, complex analysis, function. building, cartography and other fields of the social
life.
1. Introduction
On the contemporary stage in the class and
Intensive development of the technique, outclass work of mathematics in the Bulgarian
information and communication technologies secondary schools the trigonometric functions
enforces an impression on formation and sinx,cosx,tgx,cotgx, trigonometric equations and
development the interest for learning among the in equations are very important. In the teaching and
grownup. On the other hand, in contemporary in methodological literature the key competences are
conditions in which mathematicians involved in shown, which are mandatory for the pupils in
science and technique arose the necessity of secondary schools:
continuous looking for and implementation in
different areas of well qualified specialists of know the basic trigonometric equations and to
mathematics with interest and capacity for creative solve the equations switch – it can be reduced to
activity. them;
can find values of a parameter given in advance
for certain conditions comprising trigonometric
DOI: 10.18421/TEM84-34 equations and inequalities, inequalities that can
https://dx.doi.org/10.18421/TEM84-34
be reduced to rational ones;
Corresponding author: Ivo Andreev, can take the properties of trigonometric functions
Department of Algebra and Geometry, University of Veliko [8], [11];
Tarnovo, Bulgaria acquire trigonometric formulas [3], [4], [11];
Email: i.andreev2010@abv.bg have a background for complex numbers and the
correspondence with the points of the plane;
Received: 04 April 2019.
Revised: 08 August 2019. know the algebraic and trigonometric forms of
Accepted: 15 August 2019. the complex numbers and notion related to them,
Published: 30 November 2019. as well as, know how to use them;
know the operations with the complex numbers
© 2019 Ivo Andreev, Ivan Georgiev, and apply them [7], [9], [14];
Margarita Varbanova; published by UIKTEN. This work is know the Moivre formulas and find zeroes of
licensed under the Creative Commons Attribution- polynomials with real coefficients [7], [9], [14].
NonCommercial-NoDerivs 3.0 License.
The expected results are as follows: They are able
The article is published with Open Access to solve trigonometric equations and inequalities by
at www.temjournal.com
TEM Journal – Volume 8 / Number 4 / 2019. 1339
TEM Journal. Volume 8, Issue 4, Pages 1339-1344, ISSN 2217-8309, DOI: 10.18421/TEM84-34, November 2019.
suitable for pupils of 11th th
using of different methods; to perform operation with - 12 grades form,
complex number. moreover for pupils who participate in math campus,
In the trigonometry exists a great variety of Olympiads, mathematical completions, Olympiads
trigonometric equations, that require different within computer mathematics field.
methods for solving them. However, in the school The implementation of this approach for solution
course only a few are considered – trigonometric of trigonometric equations aims to give pupils
equations and methods for their solving. Only additional mathematical knowledge, to prepare one
equations, that could be reduced to the basic directional preparation for studying elements of the
trigonometric equations: sinx =a, cosx =a, mathematical analysis in the next stages of education.
tgx = a , cotgx = a are considerd. Also, equations On the base of this good training the pupils will take
which contain only one function with the same interest in mathematics and its applications, they will
argument are considered as well. Then the solving is get a fundament for successful continuation of their
connected with substitution, which leads to education at the universities.
application of knowledge for algebraic equations
solving. If the algebraic equation has a solution, then 2. Exposition the Method
it is reduced again to solving the basic trigonometric
equations. But if the equation contains trigonometric This section considers conditions necessary for the
functions with a common argument, then again it is existence of solutions for equation, as well as
suitable the substitution method to reduce the methods for their finding on the case of Euler's
original equation to equation with only one unknown formula.
x Theorem 1 [5]. Let a, b, c are real numbers and let
tg . the inequality a2 + b2 ≥c2, holds then the
2 trigonometric equations
For studying the trigonometric knowledge the
investigation and transformation of different acosx+bsinx=c (1)
trigonometric expressions for the solution of the
trigonometric equations are very important [12]. have real solutions.
It is useful to consider alternative approaches [2], Proof: We multiply the two sides of equations (1)
to solve equations and prove the issues. It is by 1 to consider the equality
interesting for the pupils with outstanding a2 + b2
mathematical capibilities the equations of the form
acosx+bsinx=c. The methodological directives a cosx+ b sin x = c .
for solution of such kind of equations are connected a2 + b2 a2 + b2 a2 + b2
with decomposition of mulipliers (the equations of
this kind [8] one solves by decomposition, as all The equality
addends one transfers at one hand-side and the
resulting expression is presented as a product) or by a 2 b 2
application of the formula ( ) or + =1
sin α ± β a2 + b2 a2 + b2
( ), for this one must divide the both sides
cos α ± β
implies, the existence of angle , such that
of the equation with a2 +b2 [10]. In [5] the ψ
equation is divided by a2 +b2 and a sinψ = a and cosψ = b .
complimentary angle is added. 2 2 2 2
ψ a +b a +b
Some pupils have difficulties with solving this
kind of equations, because it is a problem to find the In this case we have
expression for multiplication or division of the both
sides of the equation. In the present development we sinψ cosx +cosψ sinx = c
propose a new approach, that activity facilitates, in a2 + b2
which pupils for solving certain substitution prepare
concrete formulas in advance deduced from the Then, there exists a number x, such that
teacher.
In Bulgaria this approach did not apply up to now ( ) c ,
in the school course of mathematics. It is not known sin ψ + x = a2 + b2
even, in other countries’ publications, in the teaching
and methodology literature for application of this i.e. the equation (1) has a solution.
approach to solvе this class of equations. In fact the
problems, concerning solution of these equations are
1340 TEM Journal – Volume 8 / Number 4 / 2019
TEM Journal. Volume 8, Issue 4, Pages 1339-1344, ISSN 2217-8309, DOI: 10.18421/TEM84-34, November 2019.
Further, we use the Euler's formula [6]: It follows from Theorem 1, that in order equation
(1) to have real solutions
eix = cosx + isin x , (2)
2 ( 2 2 ) 2 ( 2 2 ) 2 2 2
to prove the Theorem 2. c − a +b ≤0⇒ c − a +b =i a +b −c .
Theorem 2. If a2 + b2 ≥ c2, then the solutions of t = a c±i a2 +b2 −c2 +
1/2 2 2
equation (1) are as follows: a +b
b 2 2 2
ac−b a2 +b2 −c2 +i 2 2 c±i a +b −c (4)
;
x1 = u1arccos 2 2 +2kπ a +b
a +b
acb a2 +b2 −c2 bc±a a2 +b2 −c2
ac+b a2 +b2 −c2 = 2 2 + 2 2 i
= ,
x2 u2 arccos 2 2 +2kπ a +b a +b
a +b
A B
in which
bc+a a2 +b2 −c2 The signs and ± must be taken in
u =sign ; correspondent sequence. One takes the positive value
1 a2 +b2 of the root.
By letting t = eix , one obtains for (2) and (4)
bc−a a2 +b2 −c2
u2 =sign . ix
a2 +b2 t = e =cosx+isinx
acb a2 +b2 −c2 bc±a a2 +b2 −c2
Proof: By the change of x with −x in (2) and = 2 2 + 2 2 i
summing by terms and subtraction of the results, we a +b a +b
get sin x and cosx: = A+Bi.
eix + e−ix eix − e−ix One equalizes the real and imagined parts to find:
cosx = 2 , sin x = 2i . (3)
acb a2 +b2 −c2
We substitute (3) in (1): cosx = a2 + b2 (5.1)
2 2 2
eix + e−ix eix −e−ix bc±a a +b −c
a +b =c /.eix sin x = (5.2)
2 2i a2 + b2
a 2 b 2
( ix ) ( ix ) ix 2 2
( ) ( )
2 e +1 + 2i e −1 =ce It is easy to be seen, that A + B =1.
The solutions of (1) are given by the systems:
We set eix = t
ac−b a2 +b2 −c2
a b cosx = 2 2
( 2 ) ( 2 ) a +b
2 t +1 + 2i t −1 =ct
A
a −ib a +ib 1 ∪
t2 −ct + =0 bc+a a2 +b2 −c2
2 2 sinx = a2 +b2
2 1 1 2 2 2
( ) ( ) ( ) B
D=c −4.2 a−ib 2 a+ib =c − a +b 1 (6)
2 ( 2 2 ) ac+b a2 +b2 −c2
t = c± c − a +b .a+ib cosx = a2 +b2
1/2 a−ib a+ib
a A2
( 2 ( 2 2 )) 2 2 2
= a2 + b2 c± c − a +b + bc−a a +b −c
b sinx = a2 +b2
( 2 ( 2 2 ))
+i 2 2 c± c − a +b . B
a +b 2
TEM Journal – Volume 8 / Number 4 / 2019. 1341
TEM Journal. Volume 8, Issue 4, Pages 1339-1344, ISSN 2217-8309, DOI: 10.18421/TEM84-34, November 2019.
Therefore, the solutions of (1) are: II approach: According Theorem 1 [5] in order the
equation 3cosx+3sinx =3, to have real roots,
( ) ( )
x =sign B arccos A +2kπ;
1 1 1 (7) a b c
( ) ( ) 2 2 2
x2 = sign B2 arccos A1 + 2kπ. the inequality a +b −c ≥0. Therefore
3+9−9≥0 is fulfilled.
This deducing is a training of the algebra on Taking into account (6), we write down, we note:
complex numbers. It can be done with pupils of the
last classes and as corollary the formulas for direct 3 3−3 3 3 3+3 3
solving of equation (1). If they are not taught cosx = 12 cosx = 12
profoundly, the teacher gives prepared formulas (6) ∪
and (7). sin x = 9+3 sin x = 9−3
Using elements of the higher mathematics is not 12 12
always a good idea. For this, a compromise must be 3
reached between the comprehensibility of a given cosx = 0 cosx =
situation presented at a lower level and their ability ∪ 2
to participate in a broader sphere at a higher level [1]. sin x =1 1
sin x = 2
3. Application
In this section the applications of the theorems Therefore, the solution of the equations are:
proved are illustrated. In the first problem the ( ) ( ) π ;
equation has two solutions. At second problem one x1 = sign 1 arccos 0 + 2kπ = 2 + 2kπ,k∈Ζ
obtains one, root and the third one the solutions are
connected by a parameter. In some parametric 1 3 π
equations, apart from the traditional approach to x2 = sign arccos +2kπ = +2kπ,k∈Ζ.
2 2 6
solution, it is appropriate to give a graphical
approach to the solution [13].
Problem 1. Solve the equation 3cosx +3sinx = 3. Problem 2. Solve the equation cosx +sin x = 2.
Solution: Solution:
I-st approach: Write the equation in the form I approach: One rewrite the equation in the form
3 3 3 1 3 3
cosx+ sin x = ⇒ cosx+ sin x = .
2 3 2 3 2 3 2 2 2 2 cosx+ 2 sinx =1. Since sinπ = 2 and
π 1 π 3 2 2 4 2
Since sin 6 = 2 and cos 6 = 2 , the expression π 2 2 2
cos 4 = 2 , the expression 2 cosx+ 2 sin x
1cosx+ 3sinx can be written as can be written as
2 2 π π π
sin π cosx +cosπ sinx = sinπ + x. sin cosx+cos sinx =sin +x.
6 6 6 4 4 4
Therefore the given equation is equivalent to the
Therefore, the equation is equivalent to the equation
π 3 equation sinπ + x =1. The solutions of the last
sin +x= . The solutions of the last one are 4
6 2
as follows: equation are as follows:
π +x = π +2kπ ⇒ x = π +2kπ, k =0, ±1, ±2, x =x =π +2kπ, k =0, ±1, ±2,
6 3 6 1 2 4
π +x=π −π +2kπ ⇒x=π +2kπ, II approach: It follows from Theorem 1
6 3 2 a2 + b2 −c2 ≥ 0. Let a =1, b =1 and c = 2 . In
k =0, ±1, ±2, view of (6), we have:
1342 TEM Journal – Volume 8 / Number 4 / 2019
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