This course will help you in solving traditional Trigonometric, Cartesian and related problems without using the complex formulae that a student has to memorize. This method uses very simple yet seemingly magical ways to solve the same complex problems.

1. “
”
Like the crest of the peacock, like the gem on the head of a snake,
so is mathematics at the head of all knowledge.
Vol 1: Trigon Arithmetic
This course will help you in solving traditional Trigonometric, Cartesian and related
problems without using the complex formulae that a student has to memorize. This
method uses very simple yet seemingly magical ways to solve the same complex
problems.
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2. INTRODUCTION TO TRIGON(S)
• In this course, the term ‘trigon’ refers to a set of 3 numbers
which satisfy the following condition:
• The sum of squares of the first two numbers is equal to the square of
the third number
• e.g. 3,4,5 form a trigon since 32 + 42 = 52
• Hence all trigons satisfy the Pythagoras theorem: 𝑎2
+ 𝑏2
= 𝑐2
• Thus, a set of any 3 numbers satisfying this property is a trigon
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3. TRIGONS AND PYTHAGORAS THEOREM
• The theorem of Pythagoras was known long before Pythagoras (who
lived c540 B.C.), the earliest statement of it being in the Indian
Sulba-sutras dated c800 B.C.
• For example, to quote from Katyanana Sulba-sutra , “the square of
the diagonal of an oblong is equal to the square of both it sides”.
• In fact, the Pythagoras theorem can be obtained from the Vedic
Formula “ “ which means the sum of products is the product of the
sum. Here products means square and the hypotenuse is the sum of
squares of the other two sides.
• The dimensions of altars in ancient India were based on perfect
trigons.
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4. TRIGON BASICS
• A trigon in which all 3 elements are rational numbers is called a
rational trigon
• e.g. 3,4,5; 8,15,17; 1.5,2,2.5; -4,3,5 etc.
• 2,3,√13 is not a rational trigon since √13 is an irrational number.
• Any trigons that are multiples or fractions of other trigons are called
equal trigons.
• e.g. 4,3,5; 8,6,10; 2, 1.5, 2.5; 12,9,15 are equal trigons
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5. TRIGON BASICS (CONTD.)
• A rational trigon is called a primary trigon if all 3 elements are
whole numbers and have no common factor other than 1.
• In every family of equal perfect trigons, there is one and only one
primary trigon.
• For example, among 4,3,5; 8,6,10; 2, 1.5, 2.5; and 12,9,15
4.3,5 represent the primary trigon.
• 3,4,5 is the complimentary trigon for 4,3,5. Hence to obtain a
complementary trigon, we just transpose the first two elements.
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6. TRIGON NOTATION
• Since trigons satisfy Pythagoras theorem, we can
represent right-angled triangles using trigons and vice-
versa.
• The notation 𝜃) 4,3,5 represents the angle 𝜃 (between
sides of length 4 & 5) of the right-angled triangle with
sides 4,3,5 where 4 is the base, 3 is the perpendicular
(height) and 5 is the hypotenuse.
• Thus, Angle) Base, Perpendicular, Hypotenuse is the
general notation for a trigon.
𝜃
3
4 5
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7. BENEFITS OF TRIGON NOTATION
• With this notation – 𝜃) 4,3,5 all the trigonometric ratios are
immediately available.
• For example:
• cos 𝜃 = B/H =4/5
• sin 𝜃 = P/H = 3/5
• tan 𝜃 = P/B = 3/4 𝜃
3
4 5
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8. TRIGON ARITHMETIC
• The Trigon Arithmetic mainly includes:
• Trigon Addition
• Trigon Subtraction
• Other Trigon Arithmetical operations frequently used are:
• Double Angle
• Half Angle
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9. TRIGON ADDITION
• Consider the following two trigons:
• θ) B, P, H
• 𝜙) 𝑏, 𝑝, ℎ
• where B= Base, P=Perpendicular/Height and H=Hypotenuse
• Then the trigon addition is calculated as-
θ B P H
𝜙 b p h
θ+𝜙 Bb-Pp Pb+Bp Hh
• Hence, when we add the trigons for the angles θ and 𝜙,
we get the trigons for the angle (θ + 𝜙)
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10. TRIGON ADDITION – VERTICALLY & CROSSWISE
• The Trigon Addition is actually an application of “Vertically & Crosswise”
formula of Vedic Maths.
B P H B P H B P H
b p h b p h b p h
Bb-Pp Pb+Bp Hh
Base Perpendicular Hypotenuse
𝜃 𝐵 𝑃 𝐻
𝜙 𝑏 𝑝 ℎ
𝜃 + 𝜙 𝐵𝑏 − 𝑃𝑝 𝑃𝑏 + 𝐵𝑝 𝐻ℎ
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11. TRIGON ADDITION – NUMERICAL
• Consider the following two trigons:
• θ) 12,5,13
• 𝜙) 3,4,5
• Then the trigon (θ+𝜙) is calculated as-
θ 12 5 13
𝜙 3 4 5
θ + 𝜙 (12 × 3 − 5 × 4), (5 × 3 + 12 × 4), (13 × 5)
• Hence the trigon for the angle θ+𝜙 is 16, 63, 65
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12. TRIGON ADDITION – GEOMETRICAL ILLUSTRATION
• Consider the following two trigons:
• θ) 4,3,5
• 𝜙) 15,8,17
• Geometrical illustration is on the right
• The addition is done as follows-
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4
35
15
17
8
θ
𝜙
θ 4 3 5
𝜙 15 8 17
𝜃 + 𝜙 36 77 85 36
77
85

13. APPLICATIONS OF TRIGON ADDITION
• Given sin 𝛼=3/5 and tan 𝛽=7/24 find-
• sin(𝛼 + 𝛽)
• sec(𝛼 + 𝛽)
• tan(𝛼 + 𝛽)
• cot(𝛼 + 𝛽)
In modern mathematics, to solve these kinds of equations we have to
memorize lots of formulae. However, it is not needed when we apply
the Vedic formulae “Sum of Products = Product of the Sum” [SoP=PoS],
and “Vertically & Crosswise” together in trigon form.
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14. APPLICATIONS OF TRIGON ADDITION (CONTD.)
• Clearly, trigon and trigonometric ratio are related. Here
we have been given sin 𝛼=3/5, i.e. P=3 and H=5 and tan 𝛽=7/24,
i.e. p=7 and b=24.
• Using SoP=PoS and V&C, we get
The primary trigon for 75,100,125 is 3,4,5. Hence we have
𝛼 + 𝛽) 3,4,5
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𝛼 4 3 5
𝛽 24 7 25
𝛼 + 𝛽 75 100 125
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15. APPLICATIONS OF TRIGON ADDITION (CONTD.)
• Now that we have 𝛼 + 𝛽) 3,4,5, the following can be very easily
calculated.
• sin(𝛼 + 𝛽) = P/H = 4/5
• sec(𝛼 + 𝛽) = H/B = 5/3
• tan(𝛼 + 𝛽) = P/B = 4/3
• cot(𝛼 + 𝛽) = B/P = 3/4
Hence, given any two or more trigonometric ratio, we can very
easily obtain any trigonometric ration of the form m𝛼 ± 𝑛𝛽.
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16. TRIGON ADDITION – COMPLEMENTARY TRIGONS
• Given sin 𝛼=3/5 and sec 𝛽=5/4 find sin(𝛼 + 𝛽)
• Using SoP=PoS, we have 𝛼) 4,3,5 and 𝛽) 3,4,5
• Since we need to calculate only sin 𝛼 + 𝛽 we don’t need to
calculate the base/1st element of the trigon. Similarly, if we
needed to find tan 𝛼 + 𝛽, we wouldn’t need to evaluate the 3rd
element of the trigon.
• Here we clearly have sin(𝛼 + 𝛽) = 25/25 = 1.
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𝛼 4 3 5
𝛽 3 4 5
𝛼 + 𝛽 − 25 25

17. TRIGON ADDITION – COMPLEMENTARY TRIGONS
• In the last example, if we calculate the first element we get-
• The primary trigon for 0,25,25 is 0,1,1 which implies that the
base of the triangle is Zero. This is because 0,1,1 represent the
angle 90º (or 𝜋/2c)
• Any two angles - 𝛼 and 𝛽 are called complementary angles if
𝛼 + 𝛽 = 90º = 𝜋/2c
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𝛼 4 3 5
𝛽 3 4 5
𝛼 + 𝛽 0 25 25

18. TRIGON ADDITION – COMPLEMENTARY TRIGONS
• Here we have 𝛼 + 𝛽 = 90º = 𝜋/2c
• The trigon for 𝛼 is 4,3,5 and trigon for 𝛽 is 3,4,5.
• Here 4,3,5 is complementary trigon of 3,4,5.
• In terms of trigon addition if sum of two trigon is 0,1,1
they are complementary trigons.
• 3,4,5 is the complimentary trigon for 4,3,5. Hence to
obtain a complementary trigon, we just transpose the
first two elements.
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19. TRIGON ADDITION – MORE EXAMPLES
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θ 3 4 5
𝜙 15 8 17
𝜃 + 𝜙 13 84 85
36
𝛼 24 7 25
𝛽 3 4 5
𝛼 + 𝛽 44 117 125
A 40 9 41
B 4 3 5
A+B 133 156 205
C 12 5 13
D 8 15 17
C + D 21 220 221

20. TRIGON ADDITION – DOUBLE ANGLE
• Consider the angle θ) B,P,H then the double angle 2θ can be
obtained as-
• For example,
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θ 𝐵 𝑃 𝐻
θ 𝐵 𝑃 𝐻
2𝜃 𝐵2 − 𝑃2 2𝐵𝑃 𝐻2
36
C 12 5 13
C 12 5 13
2C 119 120 169

21. DOUBLE ANGLE – MORE EXAMPLES
Using the formula we can compute trigons for double angles very
easily-
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∅ 4 3 5
2∅ 7 24 25
𝛼 15 8 17
2𝛼 161 240 289

22. TRIGONS AND COORDINATE GEOMETRY
Consider the double angle for the trigon 3,4,5 -
Here we see that the first element, i.e., the base is negative. This
is because the resultant angle is obtuse.
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∅ 3 4 5
2∅ − 7 24 25
-7
24 Y
X
2∅ ∅
∅

23. TRIGONS AND COORDINATE GEOMETRY
Similarly, the same trigon values can be in any of the four
quadrants each representing an angle in that quadrant-
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-7,24,25 Y
X
7,24,25
7,-24,25-7,-24,25

24. TRIGON SUBTRACTION
• Consider the following two trigons:
• θ) B, P, H
• 𝜙) 𝑏, 𝑝, ℎ
• where B= Base, P=Perpendicular/Height and H=Hypotenuse
• Then the trigon subtraction is calculated as-
θ B P H
𝜙 b p h
θ − 𝜙 Bb+Pp Pb-Bp Hh
• Hence, when we subtract the trigon for the angle 𝜙 from
the angle θ, we get the trigons for the angle (θ − 𝜙)
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25. TRIGON SUBTRACTION – V&C
• The Trigon Subtraction is another application of “Vertically & Crosswise
(V&C)” formula of Vedic Maths.
B P H B P H B P H
b p h b p h b p h
Bb+Pp Pb-Bp Hh
Base Perpendicular Hypotenuse
𝜃 𝐵 𝑃 𝐻
𝜙 𝑏 𝑝 ℎ
𝜃 − 𝜙 𝐵𝑏 + 𝑃𝑝 𝑃𝑏 − 𝐵𝑝 𝐻ℎ
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26. TRIGON SUBTRACTION – NUMERICAL
• Consider the following two trigons:
• θ) 12,5,13
• 𝜙) 3,4,5
• Then the trigon θ − 𝜙 is calculated as-
θ 12 5 13
𝜙 3 4 5
θ − 𝜙 (12 × 3 + 5 × 4), (5 × 3 − 12 × 4), (13 × 5)
• Hence the trigon for the angle (θ − 𝜙) is 56, -33, 65
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27. TRIGON SUBTRACTION – GEOMETRICAL ILLUSTRATION
• Consider the following two trigons:
• θ) 4,3,5
• 𝜙) 15,8,17
• Geometrical illustration is on the right
• The addition is done as follows-
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4
5
15
17
8
θ 4 3 5
𝜙 15 8 17
𝜃 − 𝜙 84 13 85 36
85
13
3
84
𝜙
𝜃-𝜙θ

28. APPLICATIONS OF TRIGON SUBTRACTION
• Given cos 𝛼=7/25 and 𝑠𝑖𝑛 𝛽=4/5 where 𝛼 is acute and 𝛽 is obtuse,
find cos(𝛼 − 𝛽)
• It is to be noted that for obtuse angles the base is
negative.
Hence cos(𝛼 − 𝛽) = 75/125 = 3/5
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𝛼 7 24 25
𝛽 − 3 4 5
𝛼 − 𝛽 75 − 125
36

29. APPLICATIONS OF TRIGON SUBTRACTION (CONTD.)
• Given tan 𝛼=24/7 and 𝛼 is acute find sec(𝛼 − 30)
Hence sec(𝛼 − 30) =
50
7 3+34
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𝛼 7 24 25
30 √3 1 2
𝛼 − 30 7 3 + 34 − 50
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30. APPLICATIONS OF TRIGON SUBTRACTION (CONTD.)
•Given cos 𝛼 = −3/5 and 𝑠𝑖𝑛 𝛽 = 5/13,find 𝑡𝑎𝑛(2𝛼 − 𝛽)
Hence 𝑡𝑎𝑛(2𝛼 − 𝛽) =
253
204
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𝛼 − 3 4 5
2𝛼 − 7 − 24 25
𝛽 12 5 13
2𝛼 − 𝛽 − 204 − 253 325
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31. TRIGON ARITHMETIC – HALF ANGLE
• Consider the angle θ) B,P,H then the half angle θ/2 can be
obtained as-
Where, the 3rd element, i.e., the hypotenuse can be obtained by
the first two elements. The sign of the half angle is decided based
on the quadrant of the angle.
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θ 𝐵 𝑃 𝐻
θ
2
𝐵 + 𝐻 𝑃 −
OR
𝜃
2
− 𝐵 + 𝐻 − 𝑃 −

32. HALF ANGLE –EXAMPLE
Using the formula we can compute trigons for half angles very
easily.
Given that θ)-3-4,5 the half angle can be found as given below:
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θ − 3 − 4 5
θ
2
1 − 2 √5
OR
𝜃
2
− 1 2 √5

33. HALF ANGLE –EXAMPLE
If cot θ = 4/3 find cos 2θ tan θ/2.
Solution:
Here we have cos 2θ=7/25 and tan θ/2 = 1/3
Hence cos 2θ tan θ/2 = 7/75
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𝜃 4 3 5
2θ 7 24 25
𝜃
2
9 3 −

34. APPLICATIONS OF TRIGON:
• This presentation has only covered the mere basics of trigon using
Vedic Mathematics formulae V&C and SoP=PoS.
• This basic arithmetic can be used to solve all kinds of Trigonometric
equations, Inverse and Hyperbolic Trigonometric functions and
Solution to Triangles (e.g. Sine Rule) and much more.
• Solutions to many of the questions asked in JEE, JEE Advanced,
Quizzes, GATE etc. can be done very quickly using these methods.
• The same thing can be used in Coordinate Geometry and various
other branches of mathematics.
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35. TRIGON APLLICATIONS IN COORDINATE GEOMETRY
• Rotate point P(4,3) anti-clockwise through an angle of 90o about the origin.
• Using Trigon Addition, we have -
So the position of the rotated point Q is (-3,4).
• It is to be noted that if the rotation is anti-clockwise we add the triple and
if it is clockwise we subtract the triple.
• Starting with the simplest of problems, even some of the most complex
ones can be easily solved by applying the formulae of Vedic Mathematics.
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𝑃 4 3 5
90 𝑜 0 1 1
Q −3 4 5