1. Mathematics

2. Trigonometric ratios and Identities
Session 1
Trigonometric ratios and Identities

3. Topics Measurement of Angles Definition and Domain and Range
of Trigonometric Function
Compound Angles
Transformation of Angles

4. Measurement of Angles J001 B O A
Angle is considered as the figure obtained by rotating initial ray about its end point.

5. Measure and Sign of an Angle
J001
Measure of an Angle :-
Amount of rotation from initial side to terminal side.
Sign of an Angle :- B
Rotation anticlockwise –
Angle positive O A B’
Rotation clockwise –
Angle negative

6. Right Angle O Y X
J001
Revolving ray describes one – quarter of a circle then we say that measure of angle is right angle
Angle < Right angle  Acute Angle
Angle > Right angle  Obtuse Angle

7. Quadrants J001 Y O X’ X X’OX – x - axis Y’OY – y - axis Y’ II Quadrant
III Quadrant
IV Quadrant
X’OX – x - axis
Y’OY – y - axis

8. System of Measurement of Angle
J001
Measurement of Angle
Sexagesimal System or
British System
Centesimal System or
French System
Circular System or
Radian Measure

9. System of Measurement of Angles
J001
Sexagesimal System (British System)
1 right angle = 90 degrees (=90o)
1 degree = 60 minutes (=60’)
1 minute = 60 seconds (=60”)
Is 1 minute of sexagesimal
1 minute of centesimal ? =
Centesimal System (French System)
1 right angle = 100 grades (=100g)
1 grade = 100 minutes (=100’)
1 minute = 100 Seconds (=100”) NO

10. System of Measurement of Angle
J001
Circular System O r A B 1c
If OA = OB = arc AB

11. System of Measurement of Angle
J001
Circular System O A C B 1c

12. Relation Between Degree Grade And Radian Measure of An Angle
J002 OR

13. Illustrative Problem J002
Find the grade and radian measures of the angle 5o37’30”
Solution

14. Illustrative Problem J002 Solution
Find the grade and radian measures of the angle 5o37’30”
Solution

15. Relation Between Angle Subtended by an Arc At The Center of Circle B
J002
Arc AC = r and Arc ACB = 

16. Illustrative Problem J002 Solution B Arc AB = 88 m and AP = ? 72o P A
A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72o at the center. Find the length of rope. [ Take  = 22/7 approx.].
Solution P A B
72o
Arc AB = 88 m and AP = ?

17. Definition of Trigonometric Ratiosx O Y X
P (x,y) M  y r
J003

18. Some Basic Identities

19. Illustrative Problem
J003
Solution

20. Signs of Trigonometric Function In All Quadrants
J004
In First Quadrant x O Y X
P (x,y) M  y r
Here x >0, y>0,
>0

21. Signs of Trigonometric Function In All Quadrants
J004
In Second Quadrant  X X’ Y Y’
P (x,y) x y r
Here x <0, y>0,
>0

22. Signs of Trigonometric Function In All Quadrants
J004  X’ X
P (x,y) O Y’ Y M
In Third Quadrant
Here x <0, y<0,
>0

23. Signs of Trigonometric Function In All Quadrants
J004
In Fourth Quadrant  X O
P (x,y) Y’ M
Here x >0, y<0,
>0

24. Signs of Trigonometric Function In All Quadrants
J004 X Y’ X’ Y O
II Quadrant
sin & cosec are Positive
I Quadrant
All Positive
III Quadrant
tan & cot are Positive
IV Quadrant
cos & sec are Positive
ASTC :- All Sin Tan Cos

25. Illustrative Problem J004  lies in second If cot  =
quadrant, find the values of other five trigonometric function
Solution
Method : 1

26. Illustrative Problem J004 Solution Y P (-12,5) r 5  -12 X X’ Y’
 lies in second
If cot  =
quadrant, find the values of other five trigonometric function
Solution
Method : 2 Y  X X’ Y’
P (-12,5)
-12 5 r
Here x = -12, y = 5 and r = 13

27. Domain and Range of Trigonometric Function
J005
Functions
Domain
Range
sin R
[-1,1]
cos R
[-1,1]
tan R
cot R
Faculty should explain domain of any one of trigonometric function in the class with the help off funda book.
sec
R-(-1,1)
cosec
R-(-1,1)

28. Illustrative problem Prove that J005
is possible for real values of x and y only when x=y
J005
Solution
But for real values of x and y
is not less than zero

29. Trigonometric Function For Allied Angles
If angle is multiple of 900 then
sin  cos;tan  cot; sec  cosec
If angle is multiple of 1800 then
sin  sin;cos  cos; tan  tan etc.
Trig. ratio -
90o-
90o+
180o-
180o+
360o-
360o+
sin
- sin
cos
cos
sin
- sin
- cos
tan
- tan
cot
- cot
-tan

30. Trigonometric Function For Allied Angles
Trig. ratio -
90o-
90o+
180o-
180o+
360o-
360o+
cot
- cot
tan
-tan
-cot
sec
cosec
- cosec
- sec
cosec
- cosec
sec
-cosec

31. Periodicity of Trigonometric Function
J005
If f(x+T) = f(x)  x,then T is called period of f(x) if T is the smallest possible positive number
Periodicity : After certain value of x the functional values repeats itself
Period of basic trigonometric functions
sin (360o+) = sin  period of sin is 360o or 2
cos (360o+) = cos  period of cos is 360o or 2
tan (180o+) = tan  period of tan is 180o or 

32. Trigonometric Ratio of Compound Angle
J006
Angles of the form of A+B, A-B, A+B+C, A-B+C etc. are called compound angles
(I) The Addition Formula
 sin (A+B) = sinAcosB + cosAsinB
 cos (A+B) = cosAcosB - sinAsinB

33. Trigonometric Ratio of Compound Angle
J006
We get
Proved
Ask the students to attempt cot(A+B) themselves

34. Illustrative problem Find the value of (i) sin 75o (ii) tan 105o
Solution
(i) Sin 75o = sin (45o + 30o)
= sin 45o cos 30o + cos 45o sin 30o

35. Trigonometric Ratio of Compound Angle
(I) The Difference Formula
 sin (A - B) = sinAcosB - cosAsinB
 cos (A - B) = cosAcosB + sinAsinB
Note :- by replacing B to -B in addition formula we get difference formula

36. Illustrative problem If tan (+) = a and tan ( - ) = b Prove that
Solution

37. Some Important Deductions
 sin (A+B) sin (A-B) = sin2A - sin2B = cos2B - cos2A
 cos (A+B) cos (A-B) = cos2A - sin2B = cos2B - sin2A
Ask the students to find the expressions for Sin ( A+B+C) and Cos ( A+B+C)

38. To Express acos + bsin in the form kcos or sin
Similarly we get acos + bsin = sin

39. Illustrative problem
Find the maximum and minimum values of 7cos + 24sin
Solution
7cos +24sin

40. Illustrative problem Solution  Max. value =25, Min. value = -25 Ans.
Find the maximum and minimum value of 7cos + 24sin
Solution
 Max. value =25, Min. value = Ans.
Ask them to try using the Sin (A+B ) form

41. Transformation Formulae
 Transformation of product into sum and difference
 2 sinAcosB = sin(A+B) + sin(A - B)
 2 cosAsinB = sin(A+B) - sin(A - B)
 2 cosAcosB = cos(A+B) + cos(A - B)
Proof :- R.H.S = cos(A+B) + cos(A - B)
= cosAcosB - sinAsinB+cosAcosB+sinAsinB
= 2cosAcosB =L.H.S
Ask student to do 2 sinAsinB = cos(A - B) - cos(A+B)
 2 sinAsinB = cos(A - B) - cos(A+B) [Note]

42. Transformation Formulae
 Transformation of sums or difference into products
By putting A+B = C and A-B = D in the previous formula we get this result
Note or

43. Illustrative problem
Prove that
Solution
Proved

44. Class Exercise - 1
If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)

45. Class Exercise - 1 Solution :-
If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)
Solution :-
Let the coin is kept at a distance r from the eye to hide the moon completely. Let AB = Diameter of the coin. Then arc AB = Diameter AB = 2.2 cm

46. Class Exercise - 2 Prove that tan3A tan2A tanA = tan3A – tan2A – tanA.
Solution :-
We have 3A = 2A + A
Þ tan3A = tan(2A + A)
Þ tan3A =
Þ tan3A – tan3A tan2A tanA = tan2A + tanA
Þ tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

47. Class Exercise - 3 If sina = sinb and cosa = cosb, then (b) (a) (d)
Solution :-
and

48. Class Exercise - 4 Prove that Solution:-
LHS = sin20° sin40° sin60° sin80°

49. Class Exercise - 4
Prove that
Solution:-
Proved.

50. Class Exercise - 5
Prove that
Solution :-

51. Class Exercise - 5
Prove that
Solution :-

52. Class Exercise - 6 The maximum value of 3 cosx + 4 sinx + 5 is
(d) None of these
(a) 5
(b) 9
(c) 7
Solution :-

53. Class Exercise - 6 Solution :-
The maximum value of 3 cosx + 4 sinx + 5 is
Solution :-
\ Maximum value of the given expression = 10.

54. Class Exercise - 7 If a and b are the solutions of a
cosq + b sinq = c, then show that
Solution :-
We have … (i)
are roots of equatoin (i),

55. Class Exercise - 7 Solution :- sin and sin are roots of equ. (ii).
If a and b are the solutions of acosq + bsinq
= c, then show that
Solution :-
sin and sin are roots of equ. (ii).
Hence
Again from (i),

56. Class Exercise - 7 Solution :- (iv)
If a and b are the solutions of acosq + bsinq
= c, then show that
Solution :-
(iv)
 and  be the roots of equation (i),
cos and cos are the roots of equation (iv).
Now

57. Class Exercise - 8
If a seca – c tana = d and b seca + d tana = c, then
(a) a2 + b2 = c2 + d2 + cd
(b)
(c) a2 + b2 = c2 + d2
(d) ab = cd

58. Class Exercise - 8
If a seca – c tana = d and b seca + d tana = c, then
Solution :-
….. (I)
Again
Squaring and adding (i) and (ii), we get
….. ii

59. Class Exercise -9
The value of
(a) 2 sinA
(c) 2 cosA
(b)
(d)

60. Class Exercise -9
The value of
Solution :-

61. Class Exercise -10 If , , and b lie between0 and , then value
of tan2a is
(a) 1
(c) 0
(b)
(d)
 and between 0 and ,
Solution :-
Consequently, cos(a - b) and sin(a + b) are positive.

62. Class Exercise -10 Solution :- If , ,
and b lie between0 and , then value
of tan2a is
Solution :-