Trignometry

Trigonome
try

Adjacent
Opposite

 The word ‘trigonometry’ is derived from the Greek words
‘tri’(meaning three), ‘gon’ (meaning sides) and ‘metron’
(meaning measure).
 Trigonometry is the study of relationships between the sides
and angles of a triangle.
 Early astronomers used it to find out the distances of the
stars and planets from the Earth.
 Even today, most of the technologically advanced
methods used in Engineering and Physical Sciences are
based on trigonometrical concepts.
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 A triangle in which one angle
is equal to 90 is called right
triangle.
 The side opposite to the right
angle is known as
hypotenuse.
AC is the hypotenuse
 The other two sides are
known as legs.
AB and BC are the legs
Trigonometry deals with Right Triangles
A
CB


 In any right triangle, the area of the square whose side is the hypotenuse is
equal to the sum of areas of the squares whose sides are the two legs.
A
CB

(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AC2 = BC2 + AB2

Pythagoras Theorem Proof:
 Given: Δ ABC is a right angled triangle where  B =
900 And AB = P, BC= b and AC = h.
 To Prove: h2 = p2 + b2
 Construction : Draw a  BD from
B to AC , where AD = x and CB = h-x ,
 Proof : In Δ ABC and Δ ABD,
Δ ABC  Δ ABD --------(AA)
In Δ ABC and Δ BDC both are similar
So by these similarity,
p
b
h
A
B
C

Or P2 = x × h And b2 = h (h – x)
Adding both L.H.S. and R.H. S. Then
p2 + b2 = (x × h) + h (h – x)
Or p2 + b2 = xh + h2 – hx
Hence the Pythagoras theorem
p2 + b2 = h2
b
xh
h
b
p
x
h
p 
 And
p
b
h
A
B
C

 Let us take a right triangle ABC
 Here, ∠ ACB () is an acute angle.
 The position of the side AB with
respect to angle  . We call it the
side opposite to angle .
 AC is the hypotenuse of the right
triangle and the side BC is a part of
 . So, we call it the side adjacent
to angle .
A
CB

Sideoppositetoangle
Side adjacent to angle ‘’

 The trigonometric ratios of the
angle C in right  ABC as follows
:
 Sine of  C =
=
 Cosine of C=
=
A
CB

Side opposite to 
C Hypotenuse
AB
AC
Side adjacent to
 CHypotenuse
BC
AC

 Tangent of C =
=
 Cosecant of C=
=
 Secant of C =
A
CB

CSide adjacent to
 CAB
BC
Side adjacent to
 C
Hypotenuse
C
Hypotenuse
AC
AB
AC
AB
=

 Cotangent of C
 Above Trigonometric Ratio
arbitrates as sin C, cos C, tan
C , cosec C , sec C, Cot C .
 If the measure of angle C is ‘’
then the ratios are :
sin , cos , tan , cosec , sec 
and cot 
A
CB

C
Side adjacent to
 C AB
BC
= =

 Tan  =
 Cosec  = 1 / Sin 
 Sec  = 1 / Cos 
 Cot  = Cos  / Sin 
= 1 / Tan 
A
CB

p
b
h


cos
sin

1. Sin  = p / h
2. Cos  = b / h
3. Tan  = p / b
4. Cosec = h / p
5. Sec  = h / b
6. Cot  = b / p
A
CB

p
b
h

T. Ratios 0 30 45 60 90
Sine 0 ½ 1/2 3/2 1
Cosine 1 3/2 1/2 ½ 0
Tangent
0 1/ 3 1 3
Not
defined
Cosecant Not
defined
2 2 2/ 3 1
Secant
1 2/ 3 2 2
Not
defined
Cotangen
t
Not
defined
3 1 1/ 3 0

 Relation of  with Sin when 00    900
 The greater the value of ‘’, the
greater is the value of Sin.
Smallest value of Sin  = 0
Greatest value of Sin  = 1
 Relation of  with Cos when 00    900
 The greater the value of ‘’, the smaller
is the value of Cos.
Smallest value of Cos  = 0
Greatest value of Cos  = 1

 Relation of  with tan when 00    900
 Tan  increases as ‘’ increases
 But ,tan  is not defined at ‘’ = 900
 Smallest value of tan  = 0

 If 00    900
1. Sin(900- ) = Cos 
2. Cos(900- ) = Sin 
 If 00<   900
1. Tan(900- ) = Cot 
2. Sec(900- ) = Cosec 
 If 00   < 900
1. Cot(900- )= Tan 
2. Cosec(900- ) = Sec 
A
CB

p
b
h

 Sin2 +Cos2 = 1
 Sec2  -Tan2 = 1
 Cosec2 - Cot2 = 1

Trignometry

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