# Find Trigonometric Functions Given Their Graphs With Vertical Shift

Grade 12 trigonometry problems and questions, on how to find the amplitude, period, phase shift, vertical shift and the equation of a trigonometric function given its graph, are presented along with detailed solutions. Interactive tutorials on Vertical Shift of trigonometric functions may first be used to understand this concept.

 Suppose a sine (or cosine) curve is vertically shifted, so it has equation of the form y = a sin(bx + c) + d. a) Find equations for max and min in terms of a and d (assume a > 0) b) Find equations for a and d in terms of max and min. Solution a) Start with the range of function sin(bx +c) which is: -1 ≤ sin(bx +c) ≤ 1 Multiply all terms of the above inequality by a (a >0) to obtain - a ≤ a sin(b x + c) ≤ a Add d to all terms of the above inequality by a to obtain - a + d ≤ a sin(b x + c) + d ≤ a + d The above inequality indicates that y = a sin(b x + c) + d has a maximum value max and a minimum value min given by: max = d + a     and     min = d - a b) If we assume that the formulas max = d + a and min = d - a are equations in two variables d and a, we may easily solve them for d and a to obtain d = (max + min) / 2     and     a = (max - min) / 2 Find the constants a, b, c and d for the curve y = a sin(bx + c) + d graphed below. Solution First the maximum max and the minimum min of the function shown in the graph below are equal to: max = 0     and     min = - 4 Using the formulas obtained in the last problem, we have: d = (max + min) / 2 = - 2     and     a = (max - min) / 2 = 2 We next find the period p from the graph of the function. p = 4 π / 3 Assuming b positive, the period is given by 2π / b. Hence the equation 2π / b = 4 π / 3 which gives b = 3 / 2 We can now write the function as y = a sin(bx + c) + d = 2 sin( 3 x / 2 + c) - 2 One way to determine c is to use a point on the given graph. For example for x = 0, y = 0 according to the graph. Hence 2 sin( 0 + c) - 2 = 0 sin(c) = 1 gives: c = π / 2 + 2 k π , where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... Use c = π / 2 to write an expression to the function as: y = 2 sin( 3 x / 2 + π / 2) - 2 Find the constants a, b, c and d for the curve y = a sin(bx + c) + d graphed below. Solution First the maximum max and the minimum min of the function shown in the graph below are equal to: max = 18 / 5     and     min = - 6 / 5 Using the formulas obtained in the last problem, we have: d = (max + min) / 2 = 6 / 5     and     a = (max - min) / 2 = 12 / 5 We next find the period p from the graph of the function. Since the points (3/5 , 18/5) and (7/5 , - 6/5) delimit half a cycle, the period p is given by twice the difference between the x coordinates of these points.Hence p = 2(7 / 5 - 3 / 5) = 8 / 5 Assuming b positive, the period is given by 2π / b. Hence the equation 2π / b = 8 / 5 which gives b = 5 π / 4 We use the values of a, b and d found above to write the function as y = (12/5) sin( 5 π x / 4 + c) + 6 / 5 Use the point (3 / 5 , 18 / 5) by setting x = 3/5 and y = 18/5 in the equation and simplify. 18 / 5 = (12 / 5) sin( 5 π (3/5) / 4 + c) + 6 / 5 (12 / 5) sin( 3 π / 4 + c) = 18 / 5 - 6 / 5 (12 / 5) sin( 3 π / 4 + c) = 12 / 5 sin( 3 π / 4 + c) = 1 We now solve the above trigonometric equation. 3 π / 4 + c = π / 2 + 2 k π , where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... gives: c = π / 2 - 3 π / 4 = - π / 4 , (we have used the solution for k = 0) We now write an expression to the function as: y = (12 / 5) sin( 5 π x / 4 - π / 4) + 6 / 5 The temperature T, in degree Celsius, during the day is approximated by the function $T(t) = -8\cos(\dfrac{\pi t}{12})+30^{\circ}$ where t is the time in hours and t = 0 corresponds to 12:00am. a) Find the period of T. b) Find the maxiumum value of T. c) Find T at 12 am, 6 am, 12 pm, 6pm and graph T over one period starting from 12 am. Solution a) Period P is given by: P = 2π / (π/12) = 24 hours. b) Maximum value of T is given by: max = 30 + 8 = 38° c) Find values of T at 12 am, 6 am, 12 pm, 6pm 12 am corresponds to t = 0 , hence $T(12 \,\, am) = -8\cos(0)+30^{\circ} = 22^{\circ}$ 6 am corresponds to t = 6 , hence $T(6 \,\, am) = -8\cos(\dfrac{6 \pi}{12})+30^{\circ} = 30^{\circ}$ 12 pm corresponds to t = 12 , hence $T(12 \,\, pm) = -8\cos(\dfrac{12 \pi}{12})+30^{\circ} = 38^{\circ}$ 6 pm corresponds to t = 18 , hence $T(6 \,\, pm) = -8\cos(\dfrac{18 \pi}{12})+30^{\circ} = 30^{\circ}$ Graph shown below.

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