Answer :
[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\[/tex]
[tex]\bf \bullet \textit{ stretches or shrinks}\\ \left. \qquad \right. \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis}[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }{{ D}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\[/tex]
[tex]\bf \left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)[/tex]
now, with that template in mind, let's see
[tex]\bf \begin{array}{llll} y=&\frac{1}{4}cos&\left( \frac{2\pi }{3}\theta \right)\\ &\uparrow &\quad \uparrow\\ &A&\quad B \end{array} \\\\ -------------------------------\\\\ A)\qquad period=\cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{2\pi }{3}}\implies \cfrac{\frac{2\pi}{1} }{\frac{2\pi }{3}}\implies \cfrac{2\pi }{1}\cdot \cfrac{3}{2\pi }[/tex]
and surely you know how much that is
for part B)... well, is right there, from the template what A is
for part C)
well, the equation has no vertical shifting, so the midline is the same as for the parent function cos(θ).
[tex]\bf \bullet \textit{ stretches or shrinks}\\ \left. \qquad \right. \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis}[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }{{ D}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\[/tex]
[tex]\bf \left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)[/tex]
now, with that template in mind, let's see
[tex]\bf \begin{array}{llll} y=&\frac{1}{4}cos&\left( \frac{2\pi }{3}\theta \right)\\ &\uparrow &\quad \uparrow\\ &A&\quad B \end{array} \\\\ -------------------------------\\\\ A)\qquad period=\cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{2\pi }{3}}\implies \cfrac{\frac{2\pi}{1} }{\frac{2\pi }{3}}\implies \cfrac{2\pi }{1}\cdot \cfrac{3}{2\pi }[/tex]
and surely you know how much that is
for part B)... well, is right there, from the template what A is
for part C)
well, the equation has no vertical shifting, so the midline is the same as for the parent function cos(θ).
Answer:
A) Period = 3
B) Amplitude = [tex]\frac{1}{4}[/tex]
C) The equation of midline y=0
Step-by-step explanation:
Given : A sound wave is modeled with the equation [tex]y =\frac{1}{4}cos (\frac{2\pi}{3}) \theta[/tex]
To find :
A) Period
B)Amplitude
C) The equation of Midline
Solution :
The general formula for cosine is:
[tex]y=Acos(Bx)+C[/tex]
Where A is Amplitude
[tex]B=\frac{2\pi}{\text{Period}}[/tex]
C is Mid line
Comparing the given function with general form of cosine we get,
[tex]y =\frac{1}{4}cos (\frac{2\pi}{3}) \theta[/tex]
A) Period - [tex]B=\frac{2\pi}{3}[/tex]
and we know, [tex]B=\frac{2\pi}{\text{Period}}[/tex]
[tex]\frac{2\pi}{3}=\frac{2\pi}{\text{Period}}[/tex]
[tex]\text{Period}=\frac{2\pi\times 3}{2\pi}[/tex]
[tex]\text{Period}=3[/tex]
B) Amplitude- [tex]A=\frac{1}{4}[/tex]
C) The equation of midline
Midline is C=0
The equation of midline is y=0.
