trigonometric (hayati pravita)

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Proj ect Manage ment Trigonomet ric Equatio n Graph Sine Rule Cosine Rule Triangl e’s Area Identit y Ratio

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  • 1. EquationRatioGraphTrigonometricProjectIdentitySine Rule CosineManagement TrianglesRuleArea

2. RATIOS of TRIGONOMETRI 3. RATIO TRIGONOMETRIC onEXTRAORDINARY ANGLES 0 1 1 0 0 1 ~ 4. NOTE: 5. Trigonometric Ration onQuadrantsMenggunakan berkebalikkannya 6. Trigonometric Ration onQuadrantsNOTE: For quadrant I, all ratio of trigonometric is positive (+) For quadrant II, only sinus and cosec are positive (+) For quadrant III, only tan and cotg are positive (+) For quadrant IV, only cosine and sec are positive (+) 7. Trigonometric Ration of Related AngleQuadrant I 8. 9. Trigonometric Ration of Related AngleQuadrant II 10. 11. Trigonometric Ration of RelatedAngleQuadrant III 12. 13. Trigonometric Ration of Related AngleQuadrant IV 14. 15. Trigonometric Ration of Related AngleAngle (n . 90 ) for n real number For n (even number) consist Example : sin (180 ) = sin (2 . 90 ) = sin cos (360 ) = cos (4 . 90 ) = cos tan (360 ) = tan (4 . 90 ) = - tan positive (+) or negative (-) based on their quadrant. 16. Trigonometric Ration of RelatedAngle For n (odd number) Function is changed (complement) The change are :sin cos cos sin tag cotgExample : sin (90 ) = (1 . 90 ) = cos cos (270 ) = (3 . 90 ) = - sin tan (90 ) = (1 . 90 ) = cotg positive (+) or negative (-) based on their quadrant. 17. Trigonometric Ration of RelatedAngleSo, we can conclude that :90 = change trigonometric function with their complement180 = the trigonometric function is consist270 = change trigonometric function with their complement360 = the trigonometric function is consist REMEMBER! positive (+) or negative (-) based on their quadrant. 18. Trigonometric with ANGLE > 360K . 90 + The Method :1. Determine K, its even or odd. If even the trigonometry is consistent. If odd, the trigonometry to be complement of it.2. K divided by 4, and find the remainder.3. Then the remainder is the last quadrant that indicate the quadrant4. Solve it 19. Trigonometric with ANGLE > 360 20. Trigonometric with ANGLE > 360 21. NEGATIVE ANGLE 22. NEGATIVE ANGLE 23. NEGATIVE ANGLE 24. TRIGONOMETRY IDENTITY 25. Trigonometric Equation A trigonometric equation is anyequation that contains a trigonometricfunction. There are 3 kinds of TrigonometricEquation:1. Sinus2. Cosine3. TangentTo solve this problem, we use different formulafor sinus, cosine, and tangent. 26. Problem..1.Find the value of xthatsatisfiestrigonometric equation of,for 0 x 360 Problem solution:x = A + K.360x = 30 + K.360 or x = (180-A) + K.360x = (180-30) + K.360x = 150 + K.360 if: K = 0 x = 30SS = {30, 150} K = 1 x = 390 (not satisfied) or K = 0 x = 150 27. Problem..2.Find the solution set of, for 0 x 360 Problem Solving1. K = 0 x = 60I. K = 1 x = 240 K = 2 x = 420 (NS)2. K = 0 x = -60 (NS) K = 1 x = 120 K = 2 x = 3002. SS = {60, 120, 240, 300}240, 28. Problem..3.Find the solution set of, for 0 x 360 Problem solutionK=0K =1 (NS) SS = { } 29. SINUS In degrees () A is onA is on sin x = sin Aquadrant I Iandquadrant and II, so the sinII, so the sin x = A + K . 360 always (+) always (+) or x = (180-A) + K . 360 In radian () sin x = sin A x = A + K . 2 x = (-A) + K. 2K is integer(K = + 1, + 2, + 3, + 4, .) 30. COSINE In degrees () cos x = cos A x = A + K . 360 or x = -A + K . 360 In radian () cos x = sin A x = A + K . 2 x = -A + K. 2K is integer(K = + 1, + 2, + 3, + 4, .) 31. TANGENT In degrees () tan x = sin Ax = A + K . 180 In radian () tan x = sin Ax=A+K.K is integer(K = + 1, + 2, + 3, + 4, .) 32. Trigonometric Function Real Number set or its section set. Trigonometric function value is valuefrom the function for each given valuex.e.g: f(x) = sin xf(x) = cos xf(x) = tan x Based on this function, we can make asimple graph easily. 33. Trigonometric Function Example:If function f is defined by f(x) = sin2 x cos2 x, determinevalue x causing function f intersecting X-axis! Problem Solving: function f intersects X-axisb) sin x + cos x = 0 if f(x) = 0 sin x = -cos x sin2x cos2x =0 = -1 tan x = - (sin x cos x)(sin x + cos1 x) =0 a) sin x cos x = 0x = 135 and 315 sin x = cos x So, the solution set is:= 1 tan x = 1{45, 135, 225, 315} can causing function f has x = 45 and 225zero value 34. Trigonometric Function Graph Simple Trigonometric Function Graphis a geometric description of atrigonometric function. This graph can make us easily toanalyzethe value ofthefunction, types of functions, etc. 35. Trigonometric Function Graph The simple trigonometric functiongraph:1. The graph of function f(x) = sin x, for 0 x 3602. The graph of function f(x) = cos x, for 0 x 3603. The graph of function f(x) = tan x, for 0 x 360 36. Drawing a Simple Graph How to make it?1. Determine the intersect point of x-axis2. Determine the intersect point of y3. Find vertex point (max and min)4. Draw that graph 37. Problem.. If the function of trigonometric is f(x) =sin 3x, for 0 x 360. Draw thegraph of that function!i. Find the intersect point of x-axis y = 0ii.Find the intersect point of y-axis x = 0iii. Find the vertex point (max-min)iv.Draw that graph 38. Problem.. i. f(x) = sin 3xy= sin 3xK=0x=0 K = 1 x = 1200= sin 3xK = 2 x = 240sin 3x= sin 0K = 3 x = 3603x = 0 + K.360 K = 0 x = 60 x = 0 + K.120 K = 1 x =or3x = (180-0) + 180 K.360 K = 2 x =3x = 180 + K.360 300 x = 60 + K.120Intersect point x-axis are {(0,0), (60, 0), (120, 0), (180, 0), (240, 0), (300,0), (360, 0)} 39. Problem..ii.y = sin 3x K = 0 x = 90Intersect point K = 1 x = y = sin 0y-axis = (0,0) 210K = 2 x = y=0330Miniii. Vertex point fory = sin 3x the max point is Max1-1 = sin 3x for y = sin 3x the max point is 1sin 3x = sin 270 1 = sin 3x K = 0 x = 30 3x = 270 + K.360K = 1 x =x = 90 + K.120 sin 3x = sin 90 Min = {(90,-150 3x = 90 + K.360 K = 2 x =1),(210, -1),x = 30 + K.120 270(330, -1)}Max = {(30,1),(150,1), (270,1)} 40. Y = sin 3x Y = 3sin ( x + 30 0 )1 SHIFT GRAPHTRIGONOMETRICFUNCTIONS90 0210 0 330 0 30 0 60 01200 1500 1800 240 0270 0300 0360 00-1a = sum of waveIf sin was replaced by cosY = a sin (x+ )Sin = shape of graph or tan it can change the = shifting shape of graph. 41. Problem..If given trigonometric function f(x) = 2 cos (x-30), draw the graph of that function!Problem solution: i. Find the intersect point of x-axis y = 0 ii. Find the intersect point of y-axis x = 0 iii. Find the vertex point (max-min) iv. Draw that graph 42. Problem..i.f(x) = 2 cos (x-30)y= 2 cos (x-30)0= 2 cos (x-30)0= cos (x-30) cos (x-30) = sin 90x-30 = 90 + K.360 K = 0 x = 120 x = 120 + K.360or K=0x-30 = -90 + K.360 x = -60 K = 1 x = x = -60 + K.360 300 Intersect point x-axis are{(-60,0), (120, 0), (300, 0)} 43. Problem.. Intersect pointy-axis = (0, 3) ii.y = 2cos (x-30)y = 2cos (-30) *cosmin-cosplus K = 0 x =y = 2cos 30210y = 2. 3 y = 3 Minfory = sin 3x the max point is -2 iii. Vertex point-2 = 2cos (x-30) -1 = cos (x-30)Max cos (x-30) = cos 180 for y = 2cos (x-30)x-30 = 180 + K.360x = 210 + K.360 the max point is 2K = 0 x = 2 = 2cos (x-30) Min =30 1 = cos (x-30) {(210, -2), cos (x-30) = cos 0 Max =x-30 = 0 + K.360{(30,2)}x = 30 + K.360 44. 2 Y = 2 Cos (x-30)3 300 0 210 00 60 0 120 0 360 0-2 4 45. Sin Function Graph The graph of function f(x) = sin x, for 0 x 360 x 0 30 90 150 180 210 270 330 360 y 0 1 0 -1/2 -1 -1/2 0 46. yY = sin x1270 0 x0 0 180 0 360 0 90-150 47. Cosine Function Graph The graph of function f(x) = cos x, for 0 x 360x 060 90 120 180 240 270 300 360y 1 1/2 0 1/2 -1 - 1/2 -1 1/2 1 48. 1 Y = Cos x 270 0090 0 180 0360 0-1 52 49. Tan Function Graph The graph of function f(x) = tan x, for 0 x 360x 0 45 90 135 180 225 270 315 360y 0 1-10 1 10 50. Y = Tg x1 135 0 270 0 315 0 0 45 00 225 0 360 090 0 180-1 51. Sinus Rules C ba ABDIt can be used iftherere given 2 angles,and a side 52. Problem..In Triangle ABC, given c = 6cm, B =600 and C = 450. Find the length of b! Problem Solving:0 1 b c 2 3 6 b 1 SinB SinC 22b6 6 3 2 b Sin600 Sin450 2 2 b 6 6 6 1 b3 6 2 3 1 2 22 53. Cosine Rules C b a A B D 54. Problem.. In Triangle ABC, given a = 6, b = 4 and C = 1200. Find the length of c!Problem Solution:c2 = a2 + b2 2.a.b.cos Cc2 = (6)2 + (4)2 2.(6).(4).cos 1200c2 = 36 + 16 2.(6).(4).( ) Cc2 = 52 + 24 120 0 b=4ac 2 = 76c = 76 = 219 A c= B 6 55. CTriangles Areab a A B 2 Sides and Measure 1 ofc Dits Angel are Determined. 2 Angles and Measure 1of its side is Determined. 3 Sides 56. Problem.. In triangle ABC given A = 1500 b = 12 cm and c = 5 cm, so the area of triangle ABCis.. Problem Solution : C Area of ABC = b c sin Ab=12a= (12) (5) sin 1500A 1500= (12) (5) sin (1800 300)c= B 5= (12) (5) sin 300= (12) (5) 57. Problem..Find the area of ABC Triangle if given a = 10, b = 12, and c = 14Problem Solution: Cb=12 a=10 Ac=14B 58. Problem..Find the area of triangle ABC if given a = 5, A = 60, C = 75.Problem Solution:C b 75 a=560 45 AcB