Trigonometry

Trigonometry
Adjacent
Opposite
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 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
Sideoppositetoangle
Side opposite to C
Hypotenuse
AB
AC
Side adjacent to C
Hypotenuse
BC
AC

 Tangent of C =
=
 Cosecant of C=
=
 Secant of C =
A
CB
Sideoppositetoangle
Side opposite to C
Side adjacent to C
AB
BC
Side adjacent to C
Hypotenuse
Side opposite to 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
Sideoppositetoangle
Side opposite to 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

Trigonometric Ratios of 45°
In Δ ABC, right-angled at B,
if one angle is 45°, then
the other angle is also 45°,
i.e., ∠ A = ∠ C = 45°
So, BC = AB
Now, Suppose BC = AB = a.
Then by Pythagoras Theorem,
AC2 = BC2 + AB2 = a2 + a2
AC2 = 2a2 , or AC = a 2
A
CB
450
a
a
450

 Sin 450 = = = = 1/ 2
 Cos 450 = = = = 1/ 2
 Tan 450 = = = = 1
 Cosec 450 = 1 / sin 450 = 1 / 1/ 2 = 2
 Sec 450 = 1 / cos 450 = 1 / 1/ 2 = 2
 Cot 450 = 1 / tan 450 = 1 / 1 = 1
Side opposite to 450
Hypotenuse
AB
AC
a
a 2
Side adjacent to 450
Hypotenuse
BC
AC
a
Side opposite to 450
Side adjacent to 450
AB
BC
a
a
a 2

 Consider an equilateral triangle ABC.
Since each angle in an equilateral triangle is
60°, therefore,
∠ A = ∠ B = ∠ C = 60°.
Draw the perpendicular AD from A
to the side BC,
Now Δ ABD ≅ Δ ACD --------- (S. A. S)
Therefore, BD = DC
and ∠ BAD = ∠ CAD -----------(CPCT)
Now observe that:
Δ ABD is a right triangle, right-angled at D with ∠ BAD =
30° and ∠ ABD = 60°
600
300
A
B D C

 As you know, for finding the trigonometric ratios, we
need to know the lengths of the sides of the triangle.
So, let us suppose that AB = 2a.
BD = ½ BC = a
AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2
AD = a 3
Now Trigonometric ratios
Sin 300 = =
= = ½
600
300
A
B D C
2a
2a
2a
a aSide opposite to 300
Hypotenuse
BD
AB
a
2a

Cos 300 = = = 3 / 2
Tan 300 = = = 1 / 3
Cosec 300 = 1 / sin 300 = 1 / ½ = 2
Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3
Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3
Now trigonometric ratios of 600
AD
AB
a 3
2a
BD
AD
a
a 3
300
A
B D C
2a
2a
2a
a a

Sin 600 = = = 3 / 2
Cos 600 = = = ½
Tan 600 = = = 3
Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3
Sec 600 = 1 / cos 600 = 1 / ½ = 2
Cot 600 = 1 / tan 600 = 1 / 3
AD
AB
a 3
2a
BD
AB
a
2a
AD
BD
a 3
a
600
A
B D C
2a
2a
2a
a a

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
Cotangent 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

The End
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Trigonometry

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