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lubasha [3.4K]
3 years ago
10

What are the phase shift and period for the function y = 5cos[9(θ − 30°)] − 2?

Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0
\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}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}

so, with that template in mind, let's see

\bf y=5cos[9(\theta-30^o)]-2\implies y=5cos\left[9(\theta-\frac{\pi }{6})  \right]-2
\\\\\\
y=5cos(9\theta-\frac{3\pi }{2})-2\\\\
-------------------------------\\\\
\begin{array}{lllcclllll}
y=&5cos(&9\theta&-\frac{3\pi }{2} )&-2\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-------------------------------\\\\

\bf \textit{horizontal/phase shift}\implies \cfrac{C}{B}\implies \cfrac{-\frac{3\pi }{2}}{9}\implies \cfrac{-\frac{3\pi }{2}}{\frac{9}{1}}
\\\\\\
-\cfrac{3\pi }{2}\cdot \cfrac{1}{9}\implies -\cfrac{\pi }{6}\\\\
-------------------------------\\\\
period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{9}
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