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Mathematics

Mathematics. Session 1. Trigonometric ratios and Identities. Topics. Measurement of Angles. Definition and Domain and Range of Trigonometric Function . Compound Angles. Transformation of Angles. B. O. A. J001. Measurement of Angles.

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Mathematics

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  1. Mathematics

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

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

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

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

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

  9. 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 ? 1 right angle = 100 grades (=100g) 1 grade = 100 minutes (=100’) 1 minute = 100 Seconds (=100”) System of Measurement of Angles J001 Sexagesimal System (British System) Centesimal System (French System) NO

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

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

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

  15. C B  1c O A Relation Between Angle Subtended by an Arc At The Center of Circle J002 Arc AC = r and Arc ACB = 

  16. B 72o P A Illustrative Problem J002 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 Arc AB = 88 m and AP = ?

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

  18. Some Basic Identities

  19. Illustrative Problem J003 Solution

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

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

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

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

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

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

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

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

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

  29. 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+ cos cos - sin sin sin - sin sin - sin cos cos - sin - cos cos cos sin - cos tan cot -tan - tan - cot tan - tan tan Trigonometric Function For Allied Angles

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

  31. 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  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

  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. We get Proved Trigonometric Ratio of Compound Angle J006

  34. Find the value of (i) sin 75o (ii) tan 105o Illustrative problem 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. If tan (+) = a and tan ( - ) = b Prove that Illustrative problem 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

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

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

  40. Illustrativeproblem 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. Note 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 or

  43. Prove that Illustrativeproblem Solution Proved

  44. 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) Class Exercise - 1

  45. 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) Class Exercise - 1 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. Prove that • tan3A tan2A tanA = tan3A – tan2A – tanA. Þ tan3A = Class Exercise - 2 Solution :- We have 3A = 2A + A Þtan3A = tan(2A + A) Þ tan3A – tan3A tan2A tanA = tan2A + tanA • Þtan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

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

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

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

  50. Prove that Class Exercise - 5 Solution :-

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