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INTRODUCTION AND APPLICATIONS OF TRIGONOMETRY 1.
Trigonometric ratios of angle A in right triangle ABC right angled at B (i) sine A (sin A) = Side opposite to angle A BC = Hypotenuse AC
C
(ii) cosine A (cos A) = Side adjacent to angle A AB = Hypotenuse AC (iii) tangent A (tan A) = Side opposite to angle A BC = Side adjacent to angle A A B (iv) cosecant A (cosec A) Hypotenuse AC = Side opposite to angle A BC
A
=
(v) secant A (sec A)
=
Hypotenuse AC = Side adjacent to angle A AB
(vi) cotangent A (cot A)
=
Side opposite to angle A AB = Side adjacent to angle A BC
B
Above formulae can be memorized as below: Pandit
Badri
Prasad
Har
Har
Bole
Sona
Chandi
Tole
sin = 2.
prepinducular( p ) base( b ) cos = hypotenuse(h ) hypotenuse(h )
tan =
prepinducular( p ) base( b )
Reciprocal relation 1 1 1 cosec A = , sec A = , cot A = . cosA sin A tan A For example: In a ABC, right angled at B, tan A =
24 , find the other trigonometric ratios of the angle 7
A. In ABC tan A =
BC 24 24 = 7 AB 7
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BC : AB = 24 : 7
Let the positive real number k be the constant of proportionality.
C
BC = 24 k and AB = 7 k
Now, using Pythagoras theorem, we have AC 2
24 k
= AB 2 + BC 2 = (7k )2 + (24k )2 = 49k 2 + 576 k 2 = 625 k 2
AC
=
A
B
625k 2
AC
= 25 k
sin A
=
BC 24k 24 = = , AC 25k 25
i.e.,
sin A =
24 25
cos A =
AB 7k 7 = = , AC 25k 25
i.e.,
cos A =
7 25
cosec A =
7 cm
1 25 1 1 7 25 = , sec A = = = and cot A = sin A 24 tan A 24 cos A 7
3.
Quotient relation of trigonometric ratios cos A sin A tan A = , cot A = sin A cos A
4.
Values of all the trigonometric ratios of 0, 30, 45, 60 and 90. A
0°
30°
45°
60°
90°
sin A
0
1 2
1
3 2
1
cos A
1
3 2
1 2
1 2
0
tan A
0
1
3
Not defined
cosec A
Not defined
sec A
1
cot A
Not defined
2
1 3
2
2
2
2
2
2
3
Not defined
1
1
3
1
3
0
3
There is an easy way to remember the values of sin for = 0, 30, 45, 60 and 90.
0
30
45
60
90
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cos tan
Write the five numbers in the sequence of 0, 1, 2, 3, 4. Divide by 4 and take their square root. Write the values of sin by cos in reverse order
0
Dividing values of sin by cos
0
1
sin i.e., tan = cos
Note: (i)
1 2
1
3 2 1
1
2
2
3 2
1
Increasing order
1 2
0
Decreasing order
Not defined
Increasing order
1
3
3
The values of sin increases from 0 to 1 as increases from 0 to 90 and value of cos decreases from 1 to 0 as increases from 0 to 90. The value of tan also increases from 0 to a bigger number as increases from 0 to 90.
(ii) If A and B are acute angles such that A B, then sin A sin B, cos A cos B, tan A tan B and cosec A cosec B, sec A sec B, cot A cot B. For example: 1 1 Evaluate: sin 2 30º cos 2 45º + 4 tan 2 30º + sin 2 90º + cot 2 60º 2 8 1 1 sin 2 30º cos 2 45º + 4tan 2 30º + sin 2 90º + cot 2 60º 2 8 2
2
2
1 1 1 1 1 1 + (1)2 + + 4 = 2 8 3 2 2 3
2
1 1 1 1 1 4 1 1 1 1 = + 4 + 1+ = + + + 3 2 8 3 8 3 2 24 4 2 =
5.
3 + 32 + 12 + 1 48 = =2 24 24
Trigonometric ratios of complementary angles. (i) sin (90° - A) = cos A (ii) cos (90° - A ) = sin A (iii) tan (90° - A) = cot A (iv) cot (90° - A) = tan A (v) cosec (90° - A) = sec A (vi) sec (90° - A) = cosec A For example: If A, B and C are the interior angles of a triangle ABC, show that sin
B +C A = cos . 2 2
Since A, B and C are angles of a ABC, A + B + C = 180º. Now,
6.
B +C A + B +C − A sin = sin 2 2
A A 180 º − A = sin = sin 90 º − = cos . 2 2 2 Fundamental trigonometric identities. (Square relation)
(i)
cos2 A + sin2 A = 1
(ii) sec2 A = 1 + tan2 A
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For example: Prove that: (tan 2 A − tan 2B) =
(sin 2 A − sin 2B) cos 2 Acos 2B
=
(cos 2B− cos 2 A) cos 2B cos 2 A
We have
(tan2 A − tan2B)
=
=
=
=
=
Also, (tan2 A − tan2 B )
=
=
=
=
= Hence, (tan2 A − tan2 B ) =
sin 2 A cos 2 A
−
sin 2 B cos 2 B
sin 2 A cos 2 B − cos 2 A sin 2 B cos 2 A cos 2 B sin 2 A(1 − sin 2 B ) − (1 − sin 2 A) sin 2 B cos 2 A cos 2 B
sin 2 A − sin 2 A sin 2 B − sin 2 B + sin 2 A sin 2 B cos 2 A cos 2 B (sin 2 A − sin 2 B ) cos 2 A cos 2 B sin 2 A cos 2 A
−
sin 2 B cos 2 B
sin 2 A cos 2 B − cos 2 A sin 2 B cos 2 A cos 2 B
(1 − cos 2 A) cos 2 B − cos 2 A(1 − cos 2 B ) cos 2 A cos 2 B cos 2 B − cos 2 A cos 2 B − cos 2 A + cos 2 A cos 2 B cos 2 A cos 2 B
(cos 2 B − cos 2 A) cos 2 B cos 2 A (sin2 A − sin 2 B ) cos 2 A cos 2 B
=
(cos 2 B − cos 2 A) cos 2 B cos 2 A
APPLICATIONS OF TRIGONOMETRY 1.
Line of sight: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
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Object
Angle of elevation:
Li ne
of
si gh t
It is the angle formed by the line of sight with horizontal through the eyes of observer when the object is above the horizontal level.
Angle of elevation
Eye
3.
Angle of depression:
Eye
It is the angle formed by the line of sight with the horizontal when the object is below the horizontal level.
Horizontal level
Horizontal level Angle of depression
ne Li t gh si of Object
4.
The height of length of an object or the distance between distant objects can be determined with the help of trigonometric ratios.
5.
Solving right triangle in applications of trigonometry Step: 1
Make all the possible right angles.
Step: 2
Mark all given quantities or angles.
Step: 3
If the figure involves more than one triangle always assume the common side (say x) connecting these right triangles to proceed.
Step 4:
Write trigonometric ratios involving given quantity, assumed quantity and the quantity to be calculated using given angles.
Step 5:
Solve the simultaneous expressions obtained in step 4 to get the solution.
For example: A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60º and from the same point, the angle of elevation of the top of the pedestal is 45º. Find the height of the pedestal.
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https://www.entrancei.com/ Let BD is a pedestal and AD is a statue with height 1.6 m.
A
BCD = 45º, ACB = 60º BC = x m Height of statue = 1.6 m
1.6m
In ABC, we have tan 60 º = 3=
x=
AB BC
D
1 .6 + h x
h 60º
1.6 + h
..... (i)
3
45º
C
x
In BCD, we have tan 45 º = 1=
BD BC
h x
x=h
..... (ii)
From (i) and (ii), h=
1.6 + h 3
h 3 = 1 .6 + h h 3 − h = 1 .6
h ( 3 − 1) = 1.6 h=
h= =
1 .6 3 −1
3 +1 3 +1
1.6 ( 3 + 1) 3 −1 1.6 (1.732 + 1) 2
h = 0.8 (2.732) h = 2.19 m (approx)
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B
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