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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
• 11. QUESTIONSConvert to radians.i. 60 °ii. 173°iii. 35°Convert to degree.i. π 4ii. 7 π 8iii. 3 π 5
• 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