Trigonometry Revision θ Hypotenuse Opposite Adjacent O SH A CH O TA

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 360 0

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

y = a sin (bx) + c Here the total height of the graph is 4. a = 2 The graph repeats twice over 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

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

Sketch the graphs of : (i)y = 2 sin(3x) (ii)y = cos(x) -3

Trigonometric Equations AS TC θ Once the acute solution to a trig equation is found the other solution is found using the diagram above θ θ θ For example: (i) Solve sin θ = ½ θ = sin -1 (½) Since sin θ is + θ lies in the 1 st And 2 nd quadrant θ = 30 0 and θ = 30 0 and 150 0

Example 2: Solve cos θ = -½ AS TC Since cos θ is negative our solutions will lie in: the 2 nd and 3 rd 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) = 60 0 Actual value of θ = – 60 0 and θ = and 240 0

(1) Solve for (i) Cos θ = 0.7 (ii) 3 sin θ + 1 = 0 (i) AS TC AS TC Since Cos θ is positive the solutions lies in the 1 st and 4 th quadrant. Acute value of θ = cos = Actual values of θ are and – θ = and sin θ + 1 = 0 sin θ = - 1 / 3 Since sin θ is negative the solutions lies in the 3 rd and 4 th quadrant. Acute value of θ = sin -1 ( 1 / 3 ) = Actual values of θ are and – θ = and (ii)