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?

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Answer 1 · 2018-03-30T21:05:21

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

Trigonometric functions can be expressed in the form:

$y = a cos (b x + c) + d$ $y=acos(bx+c)+d$

Where:

$\ \ \bba \ \ \ \ \ \ \ \$ is the amplitude.

$bb((2pi)/b) \ \ \ \ \ \$ is the period. *

$bb((-c)/b) \ \ \ \ \ \$ is the phase shift.

$\ \ \ bbd \ \ \ \ \ \ \ \$ is the vertical shift.

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

We require:

$a=2.4=12/5$

Period of $0.2=1/5$

$:.$

$(2pi)/b=1/5$

$b=10pi$

Phase shift of $pi/3$

$(-c)/(10pi)=pi/3$

$c=-(10pi^2)/3$

Vertical shift of $0.2=1/5$

$d=1/5$

$:.$

$y=12/5cos(10pix-(10pi^2)/3)+1/5$