How do you write an equation of the cosine function with amplitude of 2, period of 2π/3, phase shift of π/6, and a vertical shift of 1?

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Answer 1 · 2016-09-17T03:38:42

$y=2cos(3x-pi/2)+1$

The general equation of cosine is $y=Acos(Bx-C)+D$ ,
where
$A=$ the amplitude

$(2pi)/absB=$ the period

$C/B=$ the phase shift

$D=$ the vertical shift

In this example, the amplitude is given as 2, the period $(2pi)/3$ , the phase shift is $pi/6$ , and vertical shift is 1.

$A=2$

$(2pi)/3=(2pi)/absB$ so $B=3$

$pi/6=C/B=C/3$

$pi/6=C/3$ so $C=pi/2$

$D=1$

Giving us the equation

$y=2cos(3x-pi/2)+1$