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Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
Trigonometric function
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Trigonometric function

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  • 1. TRIGONOMETRIC FUNCTION GROUP MEMBERS : NOOR AZURAH ABDULRAZAK WAN NORAZWANI MAHUSIN IRA NUSRAT JAAFAR NUR WAHIDAH SAMI’ON SITI NURHAFIZA HAFINAS
  • 2. OBJECTIVES• To find the angle and convert the angle from degree to radian or vice versa.• To recognize the trigonometric identities, sine and cosine rule.• To solve trigonometric equations.
  • 3. HISTORY• One of the oldest branches of mathematics.• Historical evidence shows that by about 1100 B.C., Chinese were making measurements of distance and height using right- triangle trigonometry.• Greek astronomer Hipparcus, The Father of Trigonometry, is credited with compiling the 1st trigonometric tables.• The trigonometry of Hipparcus and other astronomers was strictly a tool of measurement.
  • 4. USES IN OUR DAILY LIFE• Making measurements of distance and height.• Astronomers field.• Describing physical phenomena that are “periodic”.
  • 5. ANGLES and THEIR MEASURE• An angle is determined by rotating a ray about about its endpoint.• The starting position: initial side• The position after rotation: terminal side• The point connecting the two sides: vertex y terminal side angle initial side x vertex
  • 6. Positive angles are generated Negative angles are generatedwith anticlockwise rotation. with clockwise rotation. y y 135° x -45° x
  • 7. QUADRANT‘A’ represent an angle measure. y Quadrant II Quadrant I angle: 90° <A< 180° angle: 0 <A< 90° x Quadrant III Quadrant IV angle: 180° <A< 270° angle: 270° <A< 360°
  • 8. Angles θ θAcute angle (0°< θ < 90°) Obtuse angle (90° <θ< 180°) 180° 90 °Right angle ( ¼ rotation) Straight angle (1/2 rotation)
  • 9. RADIAN and DEGREE• An angle may be measured in terms of “Radians” rather than degrees.• π radians = 180°• 2 π radians = 360°• Note: π is used to present 3.142
  • 10. CONVERT: DEGREE TO RADIANS and RADIANS TO DEGREE.• Degree Radians• By using formula: Degree x π radians = radians 1 180°• Radians Degree• By using formula : Radians x 180° = degree 1 π radians
  • 11. QUESTIONSConvert to radians.i. 60 °ii. 173°iii. 35°Convert to degree.i. π 4ii. 7 π 8iii. 3 π 5
  • 12. SOLUTIONS• Degree to Radiansi. 60° x π radians = 1.047 radians 1 180°ii. 173° x π radians = 3.019 radians 1 180°iii. 35° x π radians = 0.611 radians 1 180°• Radians to Degreei. π x 180° = 45° 4 π radiansii. 7 π x 180° = 157.5° 8 π radiansiii. 3 π x 180° = 108 ° 5 π radians
  • 13. Graph of y=sin x sin 0° sin 90° sin 180° sin 270° sin 360° 0 +1 0 -1 0
  • 14. Graph y=cos xcos 0° cos 90° cos 180° cos 270° cos 360° +1 0 -1 0 +1
  • 15. Graph y=tan x•The period is π.•Graphs consists repetitions at intervals of π.•The tangent function is undefined at π/2.
  • 16. RIGHT ANGLE TRIANGLE TRIGONOMETRY Sine θ = Opposite side = yHypotenuse Hypotenuse r r Opposite θ side ,y Cosine θ = Adjacent side = x Adjacent Hypotenuse r side , x Tangent θ = Opposite side = y Adjacent side x
  • 17. • Tan θ = sin θ cos θ• Sec θ = 1 = r cos θ x• Cosec θ = 1 = r sin θ y• Cot θ = 1 = x tan θ y
  • 18. TRIGONOMETRY RATIOS FOR SPECIAL ANGLES 30° 2 45° 2 1 45°3 1 60° 1
  • 19. 0° 30° 45° 60° 90°sin θ 0 1 1 1 3 2 2 2cos θ 1 1 1 0 3 2 2 2tan θ 0 1 1 UNDEFINED 3 3
  • 20. TRIGONOMETRIC IDENTITIES• sin²θ + cos²θ = 1• 1 + cot²θ = cosec²θ• Tan²θ + 1 = sec²θ
  • 21. How to proven??y p(x,y) phytagoras theorem: r² = x² + y²…….① r y θ x
  • 22. From graph…cos θ = x sin θ = y r rDivided ① by r² gives : r² = x² + y² r² r² r² 1 = x ² + y² r r
  • 23. 1 = cos² θ + sin ² θ……..②Divide ② by cos ² gives : 1 = cos² θ + sin²θ cos² θ cos² θ cos² θ 1 ²= 1 + sin θ ² cos θ cos θ
  • 24. sec²θ = 1 + tan²θ………③Divide ② by sin²θ 1 = cos²θ + sin²θ sin²θ sin²θ sin²θ 1 ² = cos θ ² + 1 sin θ sin θ cosec ²θ = cot²θ + 1
  • 25. Negative angles• sin (-θ )= - sin θ• cos (- θ ) = cos θ• tan (-θ ) = - tan θ
  • 26. Prove the following identitiesa) ( 1 + sin θ)² = 1 + sin θ cos²θ 1- sin θSolution( 1 + sin θ)² = ( 1 + sin θ ) ( 1 + sin θ ) cos²θ 1- sin²θ = ( 1 + sin θ ) ( 1 + sin θ ) ( 1 - sin θ ) ( 1 + sin θ )
  • 27. = 1 + sin θ 1- sin θb) ( 1 + tan² θ )² = sec ⁵ θ cos θ
  • 28. solution( 1 + tan² θ )² = ( sec²θ )² cos θ cos θ = sec⁴ θ cos θ = 1 x sec⁴ θ cos θ = sec θ x sec⁴ θ = sec ⁵ θ
  • 29. c) ( sin θ + cos θ )² + ( sin θ - cos θ )² = 2SolutionLHS= ( sin θ + cos θ )² + ( sin θ – cos θ) ( sin θ – cos θ)= sin²θ + 2 sin θ cos θ + cos²θ + sin²θ – 2 sin θ cos θ + cos²θ= sin² θ + cos² θ + sin² θ + cos² θ= 1+ 1=2LHS = RHS SO, PROVEN.
  • 30. d) sec θ – tan θ = cos θ 1 + sin θSolutionRHS, cos θ = cos θ 1 – sin θ 1 + sin θ 1 – sin θ = cos θ –cos θ sin θ 1- sin ²θ
  • 31. = cos θ –cos θ sin θ cos²θ = cos θ - cos θ sin θ cos²θ cos²θ = 1 - sin θ cos θ cos θ = sec θ - tan θRHS = LHS , SO PROVEN
  • 32. Trigonometric Equation• A trigonometric equation is an equations that contains a trigonometric expression with a variable, such as sin x
  • 33. Step in solving trigonometric equations• Step 1 : Identify the range for the given angle• Step 2 : identify the quadrant for the basic angle• Step 3 : Find the basic angle (α )• List all the answers in radian or degree ( depends on the given range )
  • 34. Solve the following equations for angles in the given rangea) tan θ = 1 , 0 ̊ ≤ θ ≤ 360b) tan 2x = 1 0 ̊ ≤ x ≤ 360 ̊
  • 35. solutionsa) Step 1 : 0 ̊ ≤ θ ≤ 360 Step 2 : quadrant 1 and 3 Step 3 : tan α = 1 1 α = tan 1 = 45 ̊ Step 4 : θ = 45 ̊ , 225 ̊
  • 36. b) tan 2x = 1 step 1 : 0 ̊ ≤ θ ≤ 360 0 ̊ ≤ 2x ≤ 720 step 2 : quadrant 1 and 3 step 3 : tan α = 1 α= tan 1 1 α = 45 ̊ step 4 : 2x = 45 ̊, 225 ̊, 405 ̊,585 ̊ x = 22.5 ̊, 112.5 ̊, 202.5 ̊, 292.5 ̊
  • 37. TRIGONOMETRIC EQUATION1. Solution of trigonometric equation such as sin = k , cos = k ,tan = k2. Solve equations in quadratic form3. Express sin , cos and tan in term of t where t tan 2
  • 38. Express sin , cos & tan in term of t where t tan 2 2 tantan 2 1 tan 2 2 tan 2t tan 2 1 tan 2 1 t2 2 2 2 x (2t ) (1 t 2 ) 2 2t 1+t² x2 4t 2 1 2t 2 t4 x2 t4 2t 2 1 2 2 2 1-t² x t 1 t 1 2 2 2 x t 1 x t2 1
  • 39. 2t tan 1 t2 2t sin 1 t2 1 t2 cos 2t 1+t² 1 t2Equation in the form a cos Ɵ + b sin Ɵ =k 1 t2Can be solved using these expression
  • 40. Example• Solve the equation 3cos x -8sin x= -2, 0°≤Ɵ≤360° 3cos x 8sin x 2 1 t2 2t 3 8 2 1 t2 1 t2 3 1 t2 8 2t 2 1 t2 3 3t 2 16t 2 2t 2 t 2 16t 5 0 b b 2 4ac t 2a 16 162 4 1 5 2 16 236 2 x t 0.3066 t tan 2 t 16.3066
  • 41. 0 x 360 x 0 180 2Tan positive in quadrant 1 and 3 tan negative in quadrant 2 and 4 x tan 16.3066 x 2 tan 0.3066 x 1 2 tan 16.3066 x 2 tan 1 0.3066 86.49 2 x x 17.05 180 86.49 2 2 x 34.1 x 93.51 2 x 187.02 x 34.9 ,187.02
  • 42. Express a cos Ɵ ± b sin Ɵ as R cos (Ɵ±α) or R sin (Ɵ±α)R cos a cos b sinR(cos cos sin sin ) a cos b sinR cos cos R sin sin a cos b sinEquating the coefficient of cos Ɵ: R cos α = a …………….(1)Equating the coefficient of sin Ɵ: R sin α = b …………….(2) R 2 cos 2 R 2 sin 2 a 2 b2 (1)²+(2)² R 2 cos 2 sin 2 a 2 b2 R2 a 2 b2 R a 2 b2
  • 43. R sin b (1)÷(2) R cos a b tan aa cos b sin R cos  a sin b cos R sin( )where where R a 2 b2 R a 2 b2 b b tan tan a a
  • 44. ExampleExpress 4 cos Ɵ – 3 sin Ɵ = 1 in the form of R cos (Ɵ + α) andsolve for Ɵ. 4 cos 3sin R cos( ) R(cos cos sin sin ) R cos cos R sin sin R cos 4 R sin 3 R 42 32 3 tan 4 R 25 1 3 R 5 tan 4 R cos( 36.87 ) 36.87
  • 45. 4 cos 3sin 5cos 36.87 4 cos 3sin 15cos 36.87 1 1 cos 36.87 5 1 1 ( 36.87 ) cos 5 36.87 78.46 36.87 78.46 ,360 78.46 41.59 , 244.67
  • 46. Equation in linear formExample 1Solve 4 sin θ – 3 cos θ = 0 for angles in therangeSolution 4 sin θ = 3 cos θ sin 3 = 4 cos
  • 47. 3tan = 4 3tan α = 4 1 3 α = tan 4 α = 36.9 ̊ θ = 36.9 ̊, 216.9 ̊
  • 48. Equation in quadratic formSolve the following trigonometric equations1. 2 sin ² x+ 5 cos x + 1 for -180 ̊≤ x ≤ 180 ̊Solution sin ² x + cos ² x = 1 sin ² x = 1- cos ² x 2(1- cos ² x) + 5 cos x + 1 = 0 2 - 2 cos ² x + 5 cos x + 1 = 0 - 2 cos ² x + 5 cos x + 3 = 0
  • 49. 2 cos ² x – 5 cos ² x – 3 = 0let y = cos x2y²- 5y – 3 = 0( y-3 )( 2y+1 ) = 0 1y = 3 and y = 2cos x = 1 2 1cos α = 2 1 1 α = cos 2 α = 60 ̊
  • 50. x = 120 ̊, -120 ̊ x = -120 ̊ , 120 ̊cos x = 3 ( no solution )2) 3 cot ²θ + 5 cosec θ + 1 for -2 ≤θ≤ 2solution 3 ( cosec ²θ-1) + 5 cosec θ + 1 = 0 3 cosec ²θ – 3 + 5 cosec θ + 1 = 0 3 cosec ²θ + 5 cosec θ – 2 = 0
  • 51. let y = cosec θ3y² + 5y – 2 = 0( 3y – 1 )( y + 2 ) = 0 1y= and y = - 2 3 1cosec θ = 1 3 =1sin 3sin = 3 ( no solution )cosec θ = -2
  • 52. 1 = -2sinsin = - 2 1sin α = 2sin α = 12 α = 30 ̊ α=6 7 5 θ= , , 6 6 6
  • 53. COMPOUND ANGLEusing substitution, it is clear to see that; sin x y sin x sin y cos x y cos x cos y tan x y tan x tan y example 3 sin(30 30 ) sin 60 2 1 1 3 sin 30 sin 30 1 2 2 2
  • 54. SUM & DIFFERENCE OF SINE• Replacing y with –y and nothing that• Cos(-y)=cos y since cosine is even function• Sin(-y)=-sin y since sine is odd function sin( x y) sin x cos y cos x sin y sin x y sin sin x y sin x cos y cos x sin y sin x cos y cos x sin y
  • 55. Example• Find the exact value of sin105 sin105 sin 60 45 sin 60 cos 45 cos 60 sin 45 3 2 1 2   2 2 2 2 6 2 4
  • 56. SUM & DIFFERENCE OF COSINEcos( x y) cos x cos y sin x sin ycos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y
  • 57. Example• Find the exact value of cos15º cos15 cos 60 45 cos 60 cos 45 sin 60 sin 45 1 2 3 2   2 2 2 2 2 6 4
  • 58. SUM & DIFFERENCE OF TANGENT• As we know… sin tan cos sin x y tan x y cos x y sin x cos y cos x sin y cos x cos y  sin x sin y
  • 59. sin x cos y cos x sin y cos x cos y cos x cos ytan( x y) cos x cos y sin x sin y  cos x cos y cos x cos y sin x sin y cos x cos y sin x sin y 1  cos x cos y tan x tan y 1  tan x tan y
  • 60. Example• Find the value of 75º in exact radical form.Solution… 75º=45º+30º let x=45º y=30º tan x tan y tan( x y) 1 tan x tan y tan 45 tan 30 tan(45 30 ) 1 tan 45 tan 30 1 1 3 1 1 1 3
  • 61. 3 1 3 12 3
  • 62. COFUNCTION FORMULAS•In a right triangle, the two acute angles arecomplementary.•Thus, if one acute angle of a right triangleis x, the other is 90 x
  • 63. cos x y cos x cos y sin x sin ycos y cos cos y sin sin y 2 2 2 0 cos y 1 sin y sin y The cofunction identity for cosine cos y sin y 2
  • 64. let y 2 x cos x sin x 2 2 2 cos x sin x 2• The cofunction identity for cosine sin x cos x 2
  • 65. sin x cos x 2 cos x• Divide all equation with 2 sin x 2 cos x cos x cos x 2 2 sin x 2 cos x sin x cos x 2 The cofunction identity tan x cot x for tangent 2
  • 66. B A B 90 a a sin A= cos B = c c a a tan A = cot A =a c b b c c sec A = csc B = a aC A b
  • 67. Write in term of its cofunction• Sin11º = cos (90º-11º) • Sec 52º = cos79º =csc (90º-52º) =csc 38º• Cot 87º = tan (90º-87º) = tan 3º
  • 68. • sin 2x = 2 sin(x) cos(x) 2• cos 2x =cos x sin 2 x 2 = 2cos x 1 = 1 2sin 2 x 2 tan x• tan 2x = 1 tan 2 x
  • 69. sin 2 A 2sin Acos AWe know that, sin(A+B)=sinAcosB+sinBcosAIf we let B=A,then sin(A+A)=sinAcosA+cosAsinAHence, sin2A=2sinAcosA
  • 70. 2 2 cos2 A cos A sin AWe know that, cos( A B) cos A cos B sin A sin BIf we let B=A,thencos( A A) cos A cos A sin Asin A 2 2Hence, cos2 A cos A sin A 2 cos 2 A 2cos A 1 2 cos 2 A 1 2sin A
  • 71. 2 tan A tan 2 A 1 tan 2 AWe know that tan A tan Btan( A B) 1 tan A tan BIf we let B=A, tan A tan A tan( A A) 1 tan A tan A 2 tan AHence, tan 2 A 1 tan 2 A
  • 72. Example 1… 5If sin andlies in quadrant II, find the exact 3value of sin 2 .Solution: 5 y sin 3 r 2 2 2 x 5 13 2 x 25 169 2 x 144 x 144 12
  • 73. So, x 12 cos r 13 sin2 2sin cos 5 12 120 sin 2 2 13 13 169
  • 74. EXAMPLE 2…. 3if tan with is acute angle, find the 4exact value of: a) tan 2 b) tan 4 2 tanSolution:a) tan 2 2 1 tan 3 2 4 3 3 1 4 4 24 7
  • 75. b) tan 4Solution:tan 4 tan(2 2 ) tan 2 tan 2 1 tan 2 tan 2 2 tan 2 2 1 tan 2 24 2 7 2 24 1 7 336 527
  • 76. HALF-ANGLE FORMULAE…. sin 2sin cos 2 2 cos cos 2 2 sin 2 2 2cos 2 1 2 1 2 sin 2 2 2 tan tan 2 1 tan 2 2
  • 77. HALF-ANGLE…. sin 2sin cos 2 2We know that, sin 2 A 2sin Acos ALet A , 2 sin 2 2sin cos 2 2 2Hence, sin 2sin cos 2 2
  • 78. cos cos 2 sin 2 2 2We know that, cos 2 A cos 2 A sin 2 A cos 2 A 2cos 2 A 1 cos 2 A 1 2sin 2 ALet A , Hence, 2 cos 2 cos 2 sin 2 cos cos 2 sin 2 2 2 2 2 2 2 cos 2 1 2cos 2 1 2 2 1 2sin 2 1 2 sin 2 2 2
  • 79. 2 tan 2 tan 1 tan 2 2 2 tan AWe know that,tan 2 A 1 tan 2 ALetting A , 2 2 tan 2 2 tan 2 2 1 tan 2 2Hence, 2 tan 2 tan 1 tan 2 2
  • 80. ExAMPLE….Without using calculator,compute the exact valueof cos 112.5⁰.Solution: ocos 112.5⁰= cos 225 2112.5⁰ lies in quadrant II,where only the sine andcosecant are (+)Thus, - sign is used in the half-angle formulae
  • 81. cos 112.5⁰= cos 225o 2 1 cos2250 2 2 1 2 2 2 2 4 2 2 2
  • 82. THE LAW OF SINESIf A, B, and C are the measures of the angles of a triangle,and a, b, and c are the lengths of the sides opposite theseangles, then a b c = = sin A sin B sin CThe ratio of the length of the side of any triangle to the sineof the angle opposite that side is the same for all three sidesof the triangle.
  • 83. EXAMPLE Solve triangle ABC if A = 50°, C = 33.5°, and b = 76. C 33.5° b = 76 a = 50°, 50° A B c
  • 84. THE LAW OF COSINESIf A, B and C are the measures of the angles of a triangle, anda, b and c are the lengths of the sides opposite these angles,then a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos CThe square of a side of a triangle equals the sum of the squaresof the other two sides minus twice their product times the cosineof their included angle
  • 85. EXAMPLE Solve the triangle with A = 60°, b = 20, and c = 30. C b = 20 • = 50°, A B c = 30
  • 86. AREA OF TRIANGLE Area = 1/2(a)(b)(SinC)
  • 87. EXAMPLEFind the area of this triangle 6cm 52° 14cm

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