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Trigonometry Revision

O. A. O. S. C. T. H. H. A. Trigonometry Revision. Hypotenuse. Opposite. θ. Adjacent. Trig Graphs. y = Sin ( x ). For y = a sin (b x ). a = ½ the total height of the graph. b = The number of times the graph repeats over 360 0. y = a sin (b x ).

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Trigonometry Revision

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  1. O A O S C T H H A Trigonometry Revision Hypotenuse Opposite θ Adjacent

  2. Trig Graphs y = Sin (x) For y = a sin (bx) a = ½ the total height of the graph b = The number of times the graph repeats over 3600

  3. y = a sin (bx) Here the total height of the graph is 8. a =4 The graph repeats twice over 3600. b =2 Hence the equation of the graph is y = 4 sin 2x y = a cos (bx) Here the total height of the graph is 10. a = 5 The graph repeats 3 times over 3600. b = 3 Hence the equation of the graph is y = 5 cos 3x

  4. y = a sin (bx) + c Here the total height of the graph is 4. a = 2 The graph repeats twice over 3600. b =2 The centre line has been moved up 3 places. c = 3 Hence the equation of the graph is y = 2 sin 2x +3

  5. Find the equation of the graphs shown below. y = 2 cos(3x) -4 y = 5 sin (½ x)

  6. Sketch the graphs of : • y = 2 sin(3x) • y = cos(x) -3

  7. Trigonometric Equations 180 - θ S A θ T C 360 - θ 180 + θ Once the acute solution to a trig equation is found the other solution is found using the diagram above. For example: (i) Solve sin θ = ½ Since sin θ is + θ lies in the 1st And 2nd quadrant θ = sin-1(½) θ = 300 and 1800 - 300 θ = 300 and 1500

  8. Example 2: Solve cos θ = -½ S A  T C  Since cos θ is negative our solutions will lie in: the 2nd and 3rd Quadrant. To solve all trig equations find the acute value of θ first. Then use it to find the actual solutions. It may be that as in the previous problem, the acute value of θ is one of the actual solutions. Acute value of θ = cos-1 (½) = 600 Actual value of θ = 1800 – 600 and 1800 + 600 θ = 1200 and 2400

  9. (1) Solve for (ii) 3 sin θ + 1 = 0 (i) Cos θ = 0.7 (ii) S A (i) S A  T C T C    3 sin θ + 1 = 0 sin θ = -1/3 Since Cos θ is positive the solutions lies in the 1st and 4th quadrant. Since sin θ is negative the solutions lies in the 3rd and 4th quadrant. Acute value of θ = cos-1 0.7 = 45.60 Acute value of θ = sin-1 (1/3) = 19.50 Actual values of θ are 45.60 and 3600 – 45.60 Actual values of θ are 1800 + 19.50 and 3600 – 19.50 θ = 45.60 and 314.40 θ = 199.50 and 340.50

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