Download Formula for Representing Sinusoidal Functions with Horizontal Shift and more Study Guides, Projects, Research Advanced Calculus in PDF only on Docsity!
Formulas for sinusoidal functions
In Section 1.5, we reviewed the functions sin x and cos x, their relation to points on a unit circle, and their graphs. For reference, the graphs of y = sin x and y = cos x are shown below.
x
y
− 2 π −π^ π 2 π 3 π 4 π
− 1
y = sin x
x
y
− 2 π −π^ π 2 π 3 π 4 π
− 1
y = cos x
Our textbook has many examples and problems involving sinusoidal functions. In most of these examples, however, the sinusoidal function f (x) is at its minimum, maximum, or midline when x = 0, so no horizontal shift of sin or cos is necessary. When a horizontal shift is necessary, it is often convenient to write a formula for the sinusoidal function as a horizontal shift of cos since its often easy to determine a value of x when f (x) is maximized. In particular, we can write any sinusoidal function in the following form.
Suppose that f (x) is a sinusoidal function with amplitude A, midline M , and period P , and suppose that f (x) takes its maximum value at x = c. Then f (x) = A cos
2 π P
(x − c)
+ M.
Here are some examples using this way of writing sinusoidal functions.
Example 0.1. Find a formula for the sinusoidal function graphed below.
x
y
y = F (x)
Solution. The fuction F (x) oscillates between a minimum value of −20 and a maximum value of 40. The midline is the number halfway between the max and min, so
midline =
(max + min) =
The amplitude is half the distance between the max and the min, so
amplitude =
(max − min) =
Check that these make sense. If the midline is 10 and the amplitude is 30, then the max would be 10 + 30 = 40, which is correct, and the min would be 10 − 30 = −20, which is correct. To find the period of F (x), note that F (x) completes a full cycle betwen 150 (where its at a max) and 1150 (where its at a max again), so the period is 1000. At x = 0, F (x) isn’t at its max, its min, or its midline, so we will need a horizontal shift. Note that F (x) is at a maximum at x = 150, so using the boxed formula above, we have
F (x) = 30 cos
2 π 1000
(x − 150)
Example 0.2. Find a formula for the sinusoidal function graphed below.
θ
r
− 3 π π 5 π 9 π
r = G(θ)
Solution. The fuction G(θ) oscillates between a minimum value of 0.1 and a maximum value of 0.7. The midline is the number halfway between the max and min, so
midline =
(max + min) =
The amplitude is half the distance between the max and the min, so
amplitude =
(max − min) =
Check that these make sense. If the midline is 0.4 and the amplitude is 0.3, then the max would be 0 .4 + 0.3 = 0.7, which is correct, and the min would be 0. 4 − 0 .3 = 0.1, which is correct. To find the period of G(θ), note that G(θ) completes a full cycle betwen π (where its at a min) and 9π (where its at a min again), so the period is 8π. At θ = 0, G(θ) isn’t at its max, its min, or its midline, so we will need a horizontal shift. Note that G(θ) is at a maximum at θ = − 3 π and at θ = 5π, so using the boxed formula above, we have
G(θ) = 0.3 cos
2 π 8 π
θ − (− 3 π)
(θ + 3π)
or
G(θ) = 0.3 cos
2 π 8 π
(θ − 5 π)
(θ − 5 π)
