# Find Trigonometric Functions Given Their Graphs With Vertical Shift (1)

Find the amplitude, period, phase shift and vertical shift and the equation of a trigonometric functions given by their graphs. Questions are presented along with detailed solutions. Interactive tutorials on Vertical Shift of trigonometric functions may first be used to understand this concept. More examples on how to find equations of trigonometric graphs are in part (2).

## Question 1

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

-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

## Question 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

## Question 3

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

## Question 4

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 maximum 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.