Math 181 - worksheet for fitting trig
functions
Here are two more problems for you to practice finding parameters ω, A, B,
C, so that the function
f (t) = A cos(ω(t − B)) + C
matches a given table of function values.
Recipe
• Identify the maximum and minimum y-value M and m, respectively.
• Choose C = (M + m)/2 (average value).
• Choose A = M − C.
• Look up T as the time interval after which the values in the table
repeat.
• Choose ω = 2π/T .
• Choose B as any value of time t where a maximum occurs.
Here is the example from class.
Problem 1 We are given the data table
0 √ π/4 2π/4
3π/4
π
5π/4
t
√
√
√ 6π/4 7π/4 2π
f (t) 1
2−1
−1 −1 − 2 −3 −1 − 2 −1 2 − 1
1
a) Find C. b) Find A. c) Find ω. d) Find B. e) When will f (t) = 0.3? f)
Find f 0 (t).
Solution. a) M = 1 and m = −3, so C = −1.
b) A = 1 − (−1) = 2.
c) T = 2π (distance between times where a maximum occurs, would work
similar for minimum if you have two of those in your table). So ω = 1.
d) B = 0 or B = 2π are possible choices. We can write our function as
f (t) = 2 cos(t) − 1.
e) We first isolate the cosine:
cos(x) =
1.3
.
2
Then we use the arccos function, aka cos−1 , giving one solution
1.3
x1 = arccos
2
and from looking at the graph, we can see a second solution, using symmetry
about the line x = π.
x2 = 2π − x1 .
Finally, all solutions are then of the form x1 + 2nπ or x2 + 2nπ, with some
(positive or negative) integer n.
Problem 2 Do the same problem, given the data table
−0.4 0.6
1.6 2.6
3.6 4.6
5.6 6.6
7.6 8.6
t
f (t) 40.24 41.7 40.24 36.7 33.17 31.7 33.17 36.7 40.24 41.7
a) Find C. b) Find A. c) Find ω. d) Find B. e) When will f (t) = 40? f)
Find f 0 (t).
Solution. a) M = 41.7 and m = 31.7, so C = 36.7.
b) A = M − C = 5.
c) T = 8.6 − 0.6 = 8 (distance between times where a maximum occurs,
would work similar for minimum if you have two of those in your table). So
ω = 2π/8 = π/4.
d) B = 0.6 or B = 8.6 are possible choices. We can write our function as
π
f (t) = 5 cos
(t − 0.6) + 36.7.
4
e) We first isolate the cosine, putting x = π4 (t − 0.6).
cos(x) =
40 − 36.7
3.3
=
.
5
5
Then we use the arccos function, aka cos − 1, giving one solution
3.3
x1 = arccos
≈ 0.85
5
and from looking at the graph, we can see a second solution, using symmetry
about the line x = π.
x2 = 2π − x1 ≈ 5.43
This means for the t-values
t=
4x
+ 0.6
π
so t1 = 1.68 and t2 = 7.52. Finally, all solutions are then of the form t1 + 8n
or t2 + 8n, with some (positive or negative) integer n (the period of this
function is 8).
f) In sum, our function can be written as
π
(t − 0.6) + 36.7
f (t) = 5 cos
4
and then
f 0 (t) = −
5π
sin(π(t − 0.6)/4).
4