Question

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

Answer 1

Hint: For writing the equation of the cosine function as stated in the above problem, we need to use the standard form. Since we have been given to write the equation of the cosine function, so we use the standard form $\left( y-Y \right)=A\cos \left( B\left( x-X \right)+C \right)$. Here $Y$ and $X$ are respectively the vertical and the horizontal shifts, $B$ is the constant related to the period $T$ by the relation $B=\dfrac{2\pi }{T}$, and $C$ is the phase shift. From the information given in the question, we can substitute the values of these constants in the standard form and get the required equation of the cosine function.

Complete step-by-step solution:
We know that the standard form of the equation of the cosine function is given as
$\left( y-Y \right)=A\cos \left( B\left( x-X \right)+C \right).........(i)$
Now, as given in the question the amplitude of the cosine function is equal to $2.4$. So we can write
$\Rightarrow A=2.4.......(ii)$
The period is given to be equal to $0.2$. So we have
$\Rightarrow T=0.2.......(iii)$
We know that the constant $B$ is related to the period $T$ by the relation
$B=\dfrac{2\pi }{T}$
Substituting (iii) in the above relation, we have
$\begin{align}
  & \Rightarrow B=\dfrac{2\pi }{0.2} \\
 & \Rightarrow B=10\pi .......(iv) \\
\end{align}$
The phase shift of the cosine function is given to be equal to pi/3. So we can write
$C=\dfrac{\pi }{3}........(v)$
Lastly, the vertical shift is given to be equal to $0.2$. So we have
$Y=0.2........(vi)$
Now, substituting the equations (ii), (iii), (iv), (v) and (vi) in (i) to get
$\left( y-0.2 \right)=2.4\cos \left( 10\pi \left( x-X \right)+\dfrac{\pi }{3} \right)$
Since there is no value of the horizontal shift given in the question, we assume it to be zero. Therefore substituting $X=0$ in the above equation we get
\[\begin{align}
& \Rightarrow \left( y-0.2 \right)=2.4\cos \left( 10\pi \left( x-0 \right)+\dfrac{\pi }{3} \right) \\
& \Rightarrow \left( y-0.2 \right)=2.4\cos \left( 10\pi x+\dfrac{\pi }{3} \right) \\
\end{align}\]
Finally adding $0.2$ both the sides, we get
$\Rightarrow y=2.4\cos \left( 10\pi x+ \dfrac{\pi }{3} \right)+0.2$
Hence, the above equation represents the required equation of the cosine function given in the question.

Note: The sign of the vertical shift can be either positive or negative. It need not necessarily be only positive. In the above case, we have taken it to be positive since its sign was not mentioned.