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# Trigonometric (hayati pravita)

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### Trigonometric (hayati pravita)

1. 1. Equation Ratio Graph TrigonometricProject Identity Sine Rule CosineManagement Triangle’s Rule Area
2. 2. RATIOS of TRIGONOMETRI
3. 3. RATIO TRIGONOMETRIC onEXTRAORDINARY ANGLES 0 1 1 0 0 1 ~
4. 4. NOTE:
5. 5. Trigonometric Ration on Quadrants Menggunakan berkebalikkannya
6. 6. Trigonometric Ration on QuadrantsNOTE: 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. 7. Trigonometric Ration of Related AngleQuadrant I
8. 8.
9. 9. Trigonometric Ration of Related AngleQuadrant II
10. 10.
11. 11. Trigonometric Ration of Related AngleQuadrant III
12. 12.
13. 13. Trigonometric Ration of Related AngleQuadrant IV
14. 14.
15. 15. Trigonometric Ration of Related AngleAngle (n . 90° ± α) for n ϵ real number For n (even number) consist Example : sin (180 – α)° = sin (2 . 90 – α)° = sin α cos (360 – α)° = cos (4 . 90 – α)° = cos α tan (360 – α)° = tan (4 . 90 – α)° = - tan α positive (+) or negative (-) based on their quadrant.
16. 16. Trigonometric Ration of Related Angle For n (odd number) Function is changed (complement) The change are : sin  cos cos sin tag  cotgExample : sin (90 – α)° = (1 . 90 – α)° = cos α° cos (270 – α)° = (3 . 90 – α)° = - sin α° tan (90 – α)° = (1 . 90 – α)° = cotg α°positive (+) or negative (-) based on their quadrant.
17. 17. Trigonometric Ration of Related AngleSo, we can conclude that :90 ± α = change trigonometric function with their complement180 ± α = the trigonometric function is consist270 ± α = change trigonometric function with their complement360 ± α = the trigonometric function is consist REMEMBER! positive (+) or negative (-) based on their quadrant.
18. 18. Trigonometric with ANGLE > 360° K . 90 + αThe Method :1. Determine K, it’s 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. 19. Trigonometric with ANGLE > 360°
20. 20. Trigonometric with ANGLE > 360°
21. 21. NEGATIVE ANGLE
22. 22. NEGATIVE ANGLE
23. 23. NEGATIVE ANGLE
24. 24. TRIGONOMETRY IDENTITY
25. 25. Trigonometric Equation A trigonometric equation is any equation that contains a trigonometric function. There are 3 kinds of Trigonometric Equation: 1. Sinus 2. Cosine 3. Tangent To solve this problem, we use different formula for sinus, cosine, and tangent.
26. 26. Problem..1. Find the value of x that satisfies trigonometric equation of , for 0° ≤ x ≥ 360° ◦ Problem solution: x = A + K.360 x = 30 + K.360 or x = (180-A) + K.360 x = (180-30) + K.360 x = 150 + K.360 if: K = 0  x = 30 SS = {30,° 150°} K = 1  x = 390 (not satisfied) or K = 0  x = 150
27. 27. Problem..2. Find the solution set of , for 0° ≤ x ≥ 360° Problem Solving 1. K = 0  x = 60 I. K = 1  x = 240 K = 2  x = 420 (NS) 2. K = 0  x = -60 (NS) K = 1  x = 120 K = 2  x = 300 2. SS = {60°, 120°, 240°, 300°} 240°,
28. 28. Problem..3. Find the solution set of , for 0° ≤ x ≥ 360° Problem solution K=0 K =1  (NS) SS = { }
29. 29. SINUS In degrees (°) A is on A is on ◦ sin x = sin A quadrant I Iand quadrant and II, so the sin II, so the sin x = A + K . 360 always (+) always (+) ◦ or x = (180-A) + K . 360 In radian (π) ◦ sin x = sin A x = A + K . 2π x = (π-A) + K. 2π K is integer(K = + 1, + 2, + 3, + 4, ….)
30. 30. COSINE In degrees (°) ◦ cos x = cos A x = A + K . 360 ◦ or x = -A + K . 360 In radian (π) ◦ cos x = sin A x = A + K . 2π x = -A + K. 2π K is integer(K = + 1, + 2, + 3, + 4, ….)
31. 31. TANGENT In degrees (°) ◦ tan x = sin A x = A + K . 180 In radian (π) ◦ tan x = sin A x=A+K.π K is integer(K = + 1, + 2, + 3, + 4, ….)
32. 32. Trigonometric Function Real Number set or its section set. Trigonometric function value is value from the function for each given value x. e.g: f(x) = sin x f(x) = cos x f(x) = tan x Based on this function, we can make a simple graph easily.
33. 33. Trigonometric Function Example: If function f is defined by f(x) = sin2 x – cos2 x, determine value x causing function f intersecting X-axis!• Problem Solving: function f intersects X-axis b) 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 + cos 1 x) =0 a) sin x – cos x = 0 x = 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 225° zero value
34. 34. Trigonometric Function Graph Simple Trigonometric Function Graph is a geometric description of a trigonometric function. This graph can make us easily to analyze the value of the function, types of functions, etc.
35. 35. Trigonometric Function Graph The simple trigonometric function graph: 1. The graph of function f(x) = sin x, for 0 ≤ x ≥ 360 2. The graph of function f(x) = cos x, for 0 ≤ x ≥ 360 3. The graph of function f(x) = tan x, for 0 ≤ x ≥ 360
36. 36. Drawing a Simple Graph How to make it? 1. Determine the intersect point of x-axis 2. Determine the intersect point of y 3. Find vertex point (max and min) 4. Draw that graph
37. 37. Problem.. If the function of trigonometric is f(x) = sin 3x, for 0° ≤ x ≥ 360°. Draw the graph of that function! 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
38. 38. Problem.. i. f(x) = sin 3x y = sin 3x K=0x=0 K = 1  x = 120 0 = sin 3x K = 2  x = 240 sin 3x= sin 0 K = 3  x = 360 3x = 0 + K.360 K = 0  x = 60 x = 0 + K.120 K = 1  x = or 3x = (180-0) + 180 K.360 K = 2  x = 3x = 180 + K.360 300 x = 60 + K.120 Intersect point x-axis are {(0°,0), (60°, 0), (120°, 0), (180°, 0), (240°, 0), (300°,0), (360°, 0)}
39. 39. Problem.. ii. y = sin 3x K = 0  x = 90° Intersect point K = 1  x = y = sin 0 y-axis = (0°,0) 210° K = 2  x = y=0 330° Min iii. Vertex point for y = sin 3x the max point is Max 1 -1 = sin 3x for y = sin 3x the max point is 1 sin 3x = sin 270° 1 = sin 3x K = 0  x = 30° 3x = 270° + K.360 K = 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. 40. Y = sin 3x Y = 3sin ( x + 30 0 )1 SHIFT GRAPH TRIGONOMETRIC FUNCTIONS 90 0 210 0 330 0 30 0 60 0 1200 1500 1800 240 0270 0300 0 360 0 0 -1 a = sum of wave If sin was replaced by cos Y = a sin (x+ α) Sin = shape of graph or tan it can change the α = shifting shape of graph.
41. 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. 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 90 x-30 = 90 + K.360 K = 0  x = 120 x = 120 + K.360 or K=0 x-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. 43. Problem.. Intersect point y-axis = (0°, √3) ii. y = 2cos (x-30) y = 2cos (-30) *cosmin-cosplus K = 0  x = y = 2cos 30 210° y = 2. ½√3  y = √3 Min for y = 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.360 x = 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. 44. 2 Y = 2 Cos (x-30)√3 300 0 210 00 60 0 120 0 360 0-2 4
45. 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 ½ 1 ½ 0 -1/2 -1 -1/2 0
46. 46. y Y = sin x1 270 0 x 0 0 180 0 360 0 90 -1 50
47. 47. Cosine Function Graph The graph of function f(x) = cos x, for 0 ≤ x ≥ 360 x 0 60 90 120 180 240 270 300 360 y 1 1/2 0 1/2 -1 - 1/2 -1 1/2 1
48. 48. 1 Y = Cos x 270 00 90 0 180 0 360 0-1 52
49. 49. Tan Function Graph The graph of function f(x) = tan x, for 0 ≤ x ≥ 360x 0 45 90 135 180 225 270 315 360 y 0 1 ∞ -1 0 1 ∞ 1 0
50. 50. Y = Tg x1 135 0 270 0 315 0 0 45 0 0 225 0 360 0 90 0 180-1
51. 51. Sinus Rules C b a A B D It can be used if there’re given 2 angles, and a side
52. 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 2 2 b 6 6 3 2 b Sin600 Sin450 2 2 b 6 6 6 1 b 3 6 2 3 1 2 2 2
53. 53. Cosine Rules C b a A B D
54. 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 C c2 = (6)2 + (4)2 – 2.(6).(4).cos 1200 c2 = 36 + 16 – 2.(6).(4).( – ½ ) C c2 = 52 + 24 120 0 b=4 a c 2 = 76 c = √76 = 2√19 A c= B 6
55. 55. C Triangle’s Area b a A B 2 Sides and Measure 1 of c D its Angel are Determined.  2 Angles and Measure 1 of its side is Determined. 3 Sides
56. 56. Problem..  In triangle ABC given A = 1500 b = 12 cm and c = 5 cm, so the area of triangle ABC is.. Problem Solution : C Area of ABC = ½ b c sin A b=12 a = ½ (12) (5) sin 1500A 1500 = ½(12) (5) sin (1800 – 300)c= B 5 = ½ (12) (5) sin 300 = ½ (12) (5) ½
57. 57. Problem..Find the area of ABC Triangle if given a = 10, b = 12, and c = 14Problem Solution: C b=12 a=10 A c=14 B
58. 58. Problem..Find the area of triangle ABC if given a = 5, A = 60°, C = 75°.Problem Solution: C b 75° a=5 60° 45° A c B