How do you write an equation of a cosine function with Amplitude=2.4, Period=0.2, Phase Shift=pi/3, and Vertical shift=.2?

Mar 30, 2018

$y = \frac{12}{5} \cos \left(10 \pi x - \frac{10 {\pi}^{2}}{3}\right) + \frac{1}{5}$

Explanation:

Trigonometric functions can be expressed in the form:

$y = a \cos \left(b x + c\right) + d$

Where:

$\setminus \setminus \setminus \boldsymbol{a} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ is the amplitude.

$\boldsymbol{\frac{2 \pi}{b}} \setminus \setminus \setminus \setminus \setminus \setminus$ is the period. *

$\boldsymbol{\frac{- c}{b}} \setminus \setminus \setminus \setminus \setminus \setminus$ is the phase shift.

$\setminus \setminus \setminus \boldsymbol{d} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ is the vertical shift.

(where $2 \pi$ is the normal period of the cosine function ) *

We require:

$a = 2.4 = \frac{12}{5}$

Period of $0.2 = \frac{1}{5}$

$\therefore$

$\frac{2 \pi}{b} = \frac{1}{5}$

$b = 10 \pi$

Phase shift of $\frac{\pi}{3}$

$\frac{- c}{10 \pi} = \frac{\pi}{3}$

$c = - \frac{10 {\pi}^{2}}{3}$

Vertical shift of $0.2 = \frac{1}{5}$

$d = \frac{1}{5}$

$\therefore$

$y = \frac{12}{5} \cos \left(10 \pi x - \frac{10 {\pi}^{2}}{3}\right) + \frac{1}{5}$