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Mathematics. Trigonometric ratios and Identities Session 1.

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Presentation on theme: "Mathematics. Trigonometric ratios and Identities Session 1."— Presentation transcript:

1 Mathematics

2 Trigonometric ratios and Identities Session 1

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

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

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

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

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

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

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

10 System of Measurement of Angle Circular System J001 O r r r A B 1c1c If OA = OB = arc AB

11 System of Measurement of Angle Circular System OA C B 1c1c J001

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

13 Illustrative Problem Find the grade and radian measures of the angle 5 o 37’30” Solution J002

14 Illustrative Problem Find the grade and radian measures of the angle 5 o 37’30” Solution J002

15 Relation Between Angle Subtended by an Arc At The Center of Circle OAC 1c1c  B Arc AC = r and Arc ACB = J002

16 Illustrative Problem 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 72 o at the center. Find the length of rope. [ Take  = 22/7 approx.]. Solution P A B 72 o Arc AB = 88 m and AP = ? J002

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

18 Some Basic Identities

19 Illustrative Problem Solution J003

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

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

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

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

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

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

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

27 FunctionsDomainRange sin R[-1,1] cos R [-1,1] sec R-(-1,1) cosec R-(-1,1) tan R cot R Domain and Range of Trigonometric Function J005

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

29 Trigonometric Function For Allied Angles Trig. ratio -- 90 o -90 o +180 o -180 o +360 o -360 o + cos sin - sin - cos cos tan - tan cot - cot -tan tan- tantansin- sin cos sin- sin sin If angle is multiple of 90 0 then sin  cos;tan  cot; sec  cosec If angle is multiple of then sin  sin;cos  cos; tan  tan etc.

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

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

32 Trigonometric Ratio of Compound Angle 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 J006

33 Trigonometric Ratio of Compound Angle J006 We get Proved

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

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) = sin 2 A - sin 2 B = cos 2 B - cos 2 A  cos (A+B) cos (A-B) = cos 2 A - sin 2 B = cos 2 B - sin 2 A

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

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

40 Illustrative problem Find the maximum and minimum value of 7cos + 24sin Solution  Max. value =25, Min. value = -25 Ans.

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  2 sinAsinB = cos(A - B) - cos(A+B) [Note]

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

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 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 Solution :- Prove that tan3A tan2A tanA = tan3A – tan2A – tanA. We have 3A = 2A + A tan3A = tan(2A + A)  tan3A =  tan3A – tan3A tan2A tanA = tan2A + tanA tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

47 Class Exercise - 3 If sin  = sin  and cos  = cos , then (c) (d) (a) (b) Solution :- and

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

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

50 Class Exercise - 5 Prove that Solution :-

51 Class Exercise - 5 Solution :- Prove that

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 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 cos + b sin = c, then show that Solution :- We have … (i) are roots of equatoin (i),

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

56 Class Exercise - 7 If a and b are the solutions of acos + bsin = 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) a 2 + b 2 = c 2 + d 2 + cd (c) a 2 + b 2 = c 2 + d 2 (d) ab = cd (b)

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

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

60 Class Exercise -9 Solution :- The value of

61 Class Exercise -10 If,,  and  lie between0 and, then value  of tan2  is (a) 1 (c) 0 (b) (d) Solution :-  and between 0 and, Consequently, cos and sin are positive.

62 Class Exercise -10 Solution :- If,,  and  lie between0 and, then value  of tan2  is


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