

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 , ± 1 , ± 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 , ± 1 , ± 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.
