Question

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

Answer 1

`\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} \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-(\pi )/(6)) \right]-2 \\\\\\ y=5cos(9\theta-(3\pi )/(2))-2\\\\ -------------------------------\\\\ \begin{array}{lllcclllll} y=&5cos(&9\theta&-(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{-(3\pi )/(2)}{9}\implies \cfrac{-(3\pi )/(2)}{(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}`