Reform of Teaching a Trigonometry Course

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Harshdeep Singh
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1. Georgia Journal of Science
Volume 73 No. 2 Scholarly Contributions from the
Article 3
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Reform of Teaching a Trigonometry Course
Sudhir Goel
Valdosta State University, [email protected]
Iwan R. Elstak
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Goel, Sudhir and Elstak, Iwan R. (2015) "Reform of Teaching a Trigonometry Course," Georgia Journal of Science, Vol. 73, No. 2,
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2. 136 Goel and Elstak: Reform of Teaching a Trigonometry Course
Sudhir Goel* and Iwan R. Elstak
Valdosta State University, Valdosta, GA 31698
*Corresponding Author
E-mail: [email protected]
Why do some students do well in College Algebra, but poorly in Trig-
onometry? Reasons include the array of new mathematical symbols,
after years of working with the same symbols (+, –, ×, ÷), and the
belief that trigonometry is unrelated to College Algebra and Calculus.
Some students consider trigonometry to be a weed-out math course
that expedites failure of Calculus courses and that their real downfall in
mathematics is due to Trigonometry. Some calculus II students complain
that: “As soon as I see a unit circle, or bizarre symbols such as sin θ or
cos θ, my mind freezes” and “I failed your calculus course because I hate
trig, otherwise I had no problems with the calculus.” This paper tries
to mitigate the students’ fears about trigonometry by presenting it like
college algebra. The ideas presented will seamlessly introduce algebra
students to trigonometry.
Keywords: Trigonometry, Pythagorean Triples, Vedic mathematics.
Teaching a Trigonometry course is a daunting task, especially at the colle-
giate level, since professors have only half the time to cover the material as the
teachers in high schools. However, Trigonometry is such a beautiful and inter-
linked subject that the authors believe it is one of the easiest math courses to
teach as a core curriculum course, as we will demonstrate below. Our motivation
to write this paper is to show students that trigonometry is not a foreign object.
Many students give up on a calculus problem if the problem contains any trigo-
nometry in it. “I hate trigonometry”; “I took it two semesters ago. I hated it then
and I hate it even more now”; “I cannot prove a single trigonometric identity
even though I want to become a mechanical engineer”; and the list goes on and
on. Students complain that they failed their calculus course because of trigonom-
etry. Knowing that students have this mindset we should encourage them to get
interested in, and not be afraid of learning new symbols.
Trigonometry has many applications and has numerous interconnections
with other subjects. It is one of the most applicable mathematics courses and
it is utilized in Physics, Engineering, Chemistry, Aeronautics and much more.
Trigonometry is a subject, interconnected and application oriented, that exten-
sively uses College Algebra. We present this paper to show how closely College
Algebra and Trigonometry are interconnected and how we could teach trigo-
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3. Georgia Journal of Science, Vol. 73 [2015], Art. 3 137
nometry and a college algebra course in a similar fashion. We will do this by
elaborating on some theorems.
THEOREM 1: The Pythagorean Trigonometric Identities are the same
as the equation of a unit circle.
Let us consider the unit circle x2 +y2 = 1
Let θ be any angle as shown. Then:
x =ycosyθ → x = cos θ and
1 1 x
x y =ysin θ → y = sin θ. Thus
1 1 x
x θy= y
1 1 x
The equation of the unit circle is x2 + y2 = 1 and
using the equations above, we get:
(1) cos2θ + sin2 θ = 1
(2) 1 + tan2θ = sec2θ. Dividing equation (1) by cos2θ
(3) cot2θ + 1 = csc2θ. Dividing equation (1) by sin2θ
Thus three Pythagorean Trigonometric Identities and the equation of the unit
circle are the same.
Remark: The students should realize that the Pythagorean trigonometric
identities are one and the same identity. Students should practice to verify
this and should get to a point that they would not forget them and use them
spontaneously. Unfortunately it is true that even in a Calculus II course students
do not remember them or do not know how to use them. We believe that the
common problem among students is probably the time they spend to understand
the identities.
The basic premise behind the identities is the Unit Circle. To our surprise,
even some students in a set theory course could not tell us the equation of a unit
circle. Students should be serious about learning and make learning their first
priority. We believe that making students’ primary and secondary curriculum
much stronger would be a step in the right direction to fixing this problem.
We will now present some examples of trigonometry identity problems and
show how to solve them as college algebra problems.
Example 1: Verify cotθ + tanθ = secθ • cscθ
We begin with the left hand side (LHS) of the equation and recall that cotθ =
. We know that for any angleθ, cosθ = x in the unit circle. Likewise sinθ = y.
. Similarly we have tanθ = .
We thus get for the LHS: = secθ • cscθ
If we consider the right hand side (RHS) of the identity, we see that it is equal to
what we found for the LHS, namely: secθ • cscθ. 2
4. 138 Goel and Elstak: Reform of Teaching a Trigonometry Course
The basic idea is to change the ‘new’ symbols (secθ, cscθ, tanθ, cotθ, …)
back to the algebraic symbols and coordinates x and y in the unit circle. Now we
have familiar symbols that our students are used to, and then solve the identity
as an algebraic problem.
Example 2: Verify: .
We start with the LHS of the identity replacing cotθ, cscθ and sinθ by coordi-
nates in the unit circle: LHS = . We then multiply the numerator and the
denominator by y2 and get: . Since x2 + y2 = 1 it follows that
x2 = 1 – y2. Therefore, the RHS of the equation becomes: after factoring
the denominator.
The numerator 1 – y2 can also be factored and becomes (1 + y)(1 – y).
After reducing the rational form to simplest terms we find that:
Once again we solved the problem in example 2 as an exercise in algebra so
that we did not have to deal with unknown symbols.
Example 3: Verify:
Beginning with the LHS of the identity we need to prove, we remind the students
that cos A = x1; cos B = x2 and similarly for sin A we write y1 and for sin B: y2
The LHS of the identity is then .
We also would like to point out that if we can get the left hand corner of the
denominator equal to 1, we make a small step forward. We do that by dividing
numerator and denominator by x1x2.
We then obtain: = RHS
The students need to be reminded that x12 + y12 = 1 and x22 + y22 = 1 since
(x1,y1) and (x2,y2) are points on the unit circle.
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Example 4:
= (y2 + x2) - yx
= 1 - sinβ cosβ = RHS
Remark: These examples should depict to our students that trigonometry
is not a foreign object. They might become aware that it is in fact algebra, up to
Pythagorean identities, dealing with sines and cosines that translate into x and
y coordinates. It may alleviate the fear of students who hate trigonometry. The
unit circle may seem friendlier. The authors believe that for a majority of our
students this work should be refreshing. Some students may still like the older
way better as change is harder to adapt to.
We will now consider an example that is more involved.
Example 5: Verify
Based on our experience with the course this example seems to ask for a lot
more from the students than the previous examples. The first question we have
to ask is: how do we find sin 2t, cos 2t and tan 2t? In order to do so we first work
with Pythagorean triples to obtain their sum and difference.
THEOREM 2: Given two Pythagorean triples, we can obtain two new
Pythagorean triples by “adding” in the Vedic style or “subtracting” two given
Pythagorean triples to obtain new Pythagorean triples.
Proof: Students had valid questions when the form with cos 2t and sin 2t
appeared and they wondered what to do next. For this we would use the Pythag-
orean triples. We propose an elegant solution by adding two Pythagorean triples
to obtain a new Pythagorean triple. For example, how do we add the triples
12, 5, 13 and 4, 3, 5, to get a new Pythagorean triple? First of all we explain
the formula known as “Vertical and Crosswise.” This formula originated in
Vedic mathematics (mathematics derived from Hindus’ sacred Scriptures called
“Vedas”). We begin with two sets of any three numbers (not necessarily Py-
thagorean triples) and produce a third row.
For example:
7 5 4
3 6 2
-9 57 8
We use the vertical and crosswise formula to obtain the three numbers in
the third row: the first of the three new numbers, - 9, is obtained by multiplying
vertically the first two numbers 7 x 3 and 5 x 6, and then by taking their differ-
ence (21 – 30 = -9.) To obtain the second number, 57, we multiply crosswise 4
6. 140 Goel and Elstak: Reform of Teaching a Trigonometry Course
the numbers in the first two columns and add them (7 x 6 + 5 x 3 = 42 + 15
= 57). To obtain the third number, 8, we multiply the last two numbers verti-
cally (4 x 2 = 8). The question is if this simple process helps us to generate new
Pythagorean triples. At first glance it appears to be a hoax. The students had no
idea where we were going.
Vedic Mathematics. We use “Vedic” mathematics to “add” two Pythag-
orean triples to find a new Pythagorean triple. Vedic Mathematics was used
in India thousands of years ago and it was discovered from the Vedas (The
main Hindu Religious Scriptures). One of the authors learned some of it in high
school; his math teacher was very fond of Vedic mathematics. Vedic mathemat-
ics was rediscovered from the Vedas between 1911 and 1918 by Sri Bharati
Krsna Tirthaji (1884 – 1960) (1). According to his research, all of mathematics is
based on only sixteen Sutras or word-formulae. It does not seem plausible to the
authors in today’s environment that all the progress in mathematics derives from
sixteen formulas. One of the authors read a few pages from the text “TRIPLES”
by Kenneth Williams (2), and it refreshed some of his childhood memories. He
also consulted the website and read chapters
from the text “Vertically and Crosswise” (3). Now we will show how the ancient
methods from India can help us understand more trigonometry and geometry.
Adding two Pythagorean triples: Let A: (x, y, r) and B: (X, Y, R) be
two Pythagorean triples (right triangle sides that are integers) shown below in
the two triangles.
The new Pythagorean triple, the “sum”, is then obtained as follows (exactly
similar to the example shown previously):
Note that the difference in the cos column is taken to be positive.
angle cos sin hypotenuse
A x y r
A+B xX-yY xY+yX rR
Proof: (xX – yY)2 + (xY + yX)2 = x2X2 + y2Y2 + x2Y2 + y2X2 + (2 x y XY
– 2 x y XY) = x2(X2 + Y2) + y2(X2 + Y2) = (x2 + y2) • R2 = r2R2.
It shows that adding two Pythagorean triples gives us a new Pythagorean
triple. One needs to be careful that it is not vertical adding of numbers. Nor is it
a determinant from matrix theory!
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7. Georgia Journal of Science, Vol. 73 [2015], Art. 3 141
As a byproduct, we obtained the following two trig identities:
cos (A + B) = x X – y Y = cos (A) cos (B) – sin(A) sin (B) and
sin (A + B) = x Y + y X = sin(A) cos (B) + cos (A) sin (B).
Note: Before we get the cosine or sine, the Pythagorean numbers x, y, r and X,
Y, and R need to be reduced to for the cosine and for the sine, becoming
rational numbers.
Example 6: Creating New Pythagorean Triples
? cos ? sin ?
A 12 5 13
B 4 3 5
A+B (48-15) (36+20) 65
33 56 65
Observe that 332 + 562 = 652 or 1089 + 3136 = 4225. Thus “adding” the
Pythagorean triples (12, 5, 13) and (4, 3, 5), using vertically and crosswise
generated numbers produces another Pythagorean triple (33, 56, 65).
In the diagram below we show the two triangles in the “Sum” position. No-
tice that the larger triangle is not yet the triangle with Pythagorean integers. The
sides of this triangle however are rational numbers 12, 204/11 and 237/11. To find
the Pythagorean triple one needs to multiply these rational numbers by a factor
of 23/4 to get 33, 56 and 65.
For the identity in Example 5 above, we need to find sin 2A, cos2A, and tan
2A. We first need to find the “sum” corresponding to the use of the same triple,
twice. Using the sum formula that we obtained above we find: 6
8. 142 Goel and Elstak: Reform of Teaching a Trigonometry Course
In particular for a unit circle
? cos ? sin ? ? cos ? sin ?
A x y r A x y 1
A x y r A x y 1
2 2 2 2 2
2A (x -y ) (2xy) r 2A (x -y ) (2xy) 1
For the sake of completeness and clarity for students we show that both
quantities from our table, (x2 – y2) and (2xy), will represent in this unit circle con-
text, actual trigonometric quantities that satisfy the requirement that they are X
and Y- coordinates taken from the unit circle. To prove that (and re-connect to
the unit circle) we show that [cos (2A)]2 + [sin(2A)]2 = 1!
Proof: x and y are on the unit circle so x2 + y2 = 1. If we square the cosine
and the sine of 2A we find:
(x2 –y2)2 + (2xy)2 = x4 + y4 – 2x2 y2 + 4x2 y2 =
x4 + 2x2 y2 + y4 = (x2 + y2)2 = 12 = 1
Note that it proves the trig identities:
We now return to Example 5 stated above: verify that tan 2t = .
After using the identities from above we find:
LHS: tan2t = which equals x2 + x2 – y2 – x2 = 2x2 – (x2 + y2) (subtract
x2 and add it at the same time) = .
Dividing both top and bottom by x2 we obtain:
The work we have discussed so far begs the question: is the difference of
two Pythagorean triples also a Pythagorean triple? Can we obtain it by using the
“Vertical and Crosswise Sutra (rule)?” The answer is yes and is shown below.
Let A: (x, y, r) and B: (X, Y, R) be two Pythagorean triples. Then the new
Pythagorean triplet (the difference) is obtained as follows:
? cos ? sin ?
A x y r
A-B (xX + yY) (yX – xY) rR
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9. Georgia Journal of Science, Vol. 73 [2015], Art. 3 143
As in the case of the sum of the Pythagorean triples, the difference also
gives the following identities:
cos(A - B) = xX + yY = cos (A) cos (B) + sin(A)sin(B) and
sin(A - B) = yX - xY = sin(A) cos(B) - cos(A)sin(B)
Example (notice the change of signs in the formulas!):
? cos ? sin ?
A 12 5 13
B 4 3 5
A-B (48+15) (20-36) 65
63 -16 65
Observe that (63)2 + (-16)2 = (65)2 and that (3969) + (256) = 4225.
In the unit circle a negative value like -16 can be convertred into the co-
ordinate -16/65 by dividing by the radius of 65 units. Thus by subtracting two
Pythagorean triples, we obtain a new Pythagorean triple. Of course this is just
an example and not a proof.
On the other hand, these ideas show that any two rows of triples that
represent sides of a right triangle (even if they are NOT Pythagorean triples),
produce a third row (a triple) that represents numbers that are still sides of
right triangles!
Some advantages of studying Pythagorean triples and their connections to
• It helps students to understand the unit circle better, especially the fact
that on a unit circle, x = cosθ and y = sinθ.
• It is obviously simpler to work with.
• It shows a connection between algebra and trigonometry.
• Students need to know just one sutra (formula), “vertical and crosswise.”
With its help they can obtain many different trigonometric identities and
thus they do not have to worry about memorizing them, which is one of
the major complaints students have about trigonometry.
• This method can be used to solve trigonometric equations.
Remark: We think that in a classroom, it may be best to present the two
methods side-by-side. The traditional method to prove trigonometric identities
and the method(s) presented in this paper may help students to appreciate the
connection between proving trigonometric identities and using algebra they have
already learned.
We redo Example 1 using the traditional method and the method in this
paper side by side. 8
10. 144 Goel and Elstak: Reform of Teaching a Trigonometry Course
Example 1 (revisited):
Verify: cot (θ) + tan (θ) = sec (θ) csc (θ). Begin with the left hand side:
Using x2 + y2 = 1 we get: Using cos2θ + sin θ2 = 1 we get:
Comparing the two solutions one realizes that they are identical, and thus
we can repeat the solutions of all the remaining four examples using two col-
umns. The next obvious question is: can we solve trigonometric equations using
this method?
We try a couple of examples:
Example A: Solve tan2 θ = 5 + sec θ. Replace the trigonometric symbols
with coordinates in the unit circle: . Notice that the equation contains
x and y and that all this happens on the unit circle with equation x2 + y2 = 1. So
if we replace y2 by 1 – x2 we get the following equation: .
2 2 2 2
Multiply both sides by x → 1 – x = 5x + x. Or: 6x + x – 1 = 0.
or (3x – 1)(2x + 1) = 0
θ = 1.23 ± 2nπ, where n is any integer or , where n is any integer
Example B: Solve the equation
cos (2 θ) + 3 = sin θ
(1 - 2sin2 θ) + 3 = sin θ
2sin2 θ + sin θ – 4 = 0
2y2 + y – 4 + 0
So y1,2 =
Since both values of y (= sin θ) are outside the interval [-1, + 1] this equation has
no solutions.
At this point, we will show how more trigonometric identities can be de-
rived using the identities we already had, using the “vertically and crosswise”
method discussed above. We think cos 2A is a most versatile trigonometric iden-
tity and it is one of the easiest identities to remember. Most students remember
the first Pythagorean trigonometric identity.
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11. Georgia Journal of Science, Vol. 73 [2015], Art. 3 145
cos2A + sin2A = 1............................................................(1)
Changing the “+” sign into a “–” sign in this identify, we obtain with minor
changes the double angle identify: cos2A – sin2A = cos2A...........................(2)
Adding the two equations together we obtain:
2cos2A – 1 = cos 2A........................................................(3)
Subtracting the second from the first:
1 – 2sin2A = cos 2A.........................................................(4)
thus cos2 A = ...........................................................(5)
and sin2 A = ............................................................(6)
To obtain the half angle formulas (replace 2A by A and A by ) we obtain
cos2 and sin2
hence cos .......................................................(7)
and sin ...............................................................(8)
The (±) signs are used based on the quadrant in which angle lies.
We now obtain the remaining identities using the sum and difference for-
mula or trigonometric identities that we obtained earlier while using the vertical
and crosswise formula to add/subtract Pythagorean triples. All the remaining
identities are obtained by using these four. We have shown earlier how we obtain
the double angle identities by putting B = A, that is by taking the angles to be
equal. Given the double angle identity cos2A, we can obtain the half angle trig
identities. The exciting thing is to see what happens if we replace the positive
by a negative sign in the very first trig identity. This identity provides a professor
with an elegant idea to demonstrate to students that in mathematics signs have
a very important place, as noticed before. Except for the sine and the cosine
formula to solve triangles, the remaining identities can be obtained by using the
following identities. The addition and subtraction formulas are:
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
There are four Product to Sum formulas that can be obtained by taking the
sum or difference of the above identities such as:
sin u sin v = ½(cos(u – v) – cos(u + v))
cos u cos v = ½(cos(u – v) + cos(u + v))
sin u cos v = ½(sin(u + v) + sin(u – v))
cos u sin v = ½(sin(u + v) – sin(u – v))
Substituting (u + v) = c and (u – v) = d to obtain u = ,v = , the previous
identities can then be re-written as: 10
12. 146 Goel and Elstak: Reform of Teaching a Trigonometry Course
cos d – cos c = 2 sin
cos d + cos c = 2 cos
sin c + sin d = 2 sin
sin c – sin d = 2 cos
sin c – sin d = 2 cos
This completes the proof of all the trig identities that are used in trigono-
metric courses except for the sine and cosine formulas to solve a triangle.
The work is interesting and simple. Moreover, the students might realize that
trigonometry is not a monster as many students believe it to be.
In this paper we used the formula “vertically and crosswise” to add or
subtract Pythagorean triples. It is a simple and elegant formula to provide us
with much trigonometry with a small effort. Moreover the sum and difference
formulas for Pythagorean triples may themselves be published along with some
nice applications of them. We reiterate that this paper contains everything that
is taught in a trigonometric course except the exercise sets and trig applications,
but of course it is not a trigonometric textbook. It also does not include angular
and linear velocity formulas, the formulas to find the area of a triangle, sine and
cosine formulas to solve a triangle and also applications that specifically use trig-
onometry. Lastly, it does not contain De Moivre’s theorem.
1. Tirtha, SBK, Vasudeva SA, and Agrawala VS. Vedic mathematics. Vol.
10. Motilal Banarsidass Publ., 1992.
2. Williams KR: Triples. India: Motilal Banarsidass Publ., 2003
3. Nicholas AP, Williams KR and Pickles J: Vertically and Crosswise. India:
Motilal Banarsidass Publ., 2003
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