text: # 4.5 | Graphs of Sine and Cosine Functions
  
  The graphs of $$y = a \sin(bx-c) +d$$ and $$y = a \cos(bx-c)+d$$ have the following characteristics. (Assume $$b>0.$$)
  
  • The amplitude is given by $$|a|$$ and represents half the distance between the maximum and minimum values of the function.
  • The period is given by $$\displaystyle \frac{2\pi}{b}.$$
  • The phase shift is given by $$c/b$$.
  • The center amplitude is given by $$d.$$ The corresponding midline is given by $$y = d.$$
  • The left and right endpoints of a one-cycle interval can be determined by solving the equations $$bx-c = 0$$ and $$bx -c = 2\pi.$$
  
  Sketch the graph of
  $$y = 4 \sin\left(x – \frac{\pi}{4} \right) +4.$$
  (Include two full periods.) Label $$x$$ and $$y$$ intercepts (if any). Find the amplitude, period, phase shift, and midline.
  
  From the form of the equation we can have that $$a = 4,$$ $$b = 1,$$ $$c = \frac{\pi}{4},$$ and $$d = 4.$$ So that:
  
  • The left and right endpoints of a two cycle interval can be determined by solving the equations $$x – \frac{\pi}{4} = 0$$ and $$x – \frac{\pi}{4} = 4 \pi$$. Thus, our two cycle interval is $$[\pi/4, 17\pi/4]$$.
  • The amplitude is $$4.$$
  • The period is $$\frac{2\pi}{b} = \frac{2\pi}{1} = 2\pi.$$
  • The phase shift is $$\pi/4.$$
  • The midline is $$y = 4.$$
  • The $$y$$-intercept:
  The two cycles we are sketched over the interval $$[\pi/4,17\pi/4]$$ which does not include $$x=0$$. So our graph will not include a $$y$$-intercept.
  • The $$x$$-intercepts:
  Set $$y = 0$$. Then
  \begin{aligned} & y = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & 0 = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & -1 = \sin\left(x – \frac{\pi}{4} \right) \\ \Rightarrow & x- \frac{\pi}{4} = \frac{3\pi}{2} + n 2\pi\\ \Rightarrow & x = \frac{3\pi}{2}+ \frac{\pi}{4} + n 2\pi\\ \Rightarrow & x = \frac{7\pi}{4} + n 2\pi\\ \end{aligned}
  So the $$x$$-intercepts over the interval $$[\pi/4,17\pi/4]$$ are $$\left(\frac{7\pi}{4},0\right)$$ and $$\left(\frac{15\pi}{4},0\right)$$.
  
  Sketch the graph of
  $$y = -1 + 2 \cos \frac{\pi t}{12}.$$
  (Include two full periods.) Label $$t$$ and $$y$$ intercepts. Find the amplitude, period, phase shift, and midline.
  
  First rewrite the equation in standard form
  $$y = 2 \cos \left(\frac{\pi}{12} t \right) -1$$
  So that $$a = 2,$$ $$b = \frac{\pi}{12},$$ $$c = 0,$$ and $$d = -1.$$ Now we have the following:
  
  • The left and right endpoints of a two cycle interval can be determined by solving the equations $$\frac{\pi}{12} t = 0$$ and $$\frac{\pi}{12} t = 4 \pi$$. Thus, our two cycle interval is $$[0, 48]$$.
  • The amplitude is $$2.$$
  • The period is $$\frac{2\pi}{b} = \frac{2\pi}{\frac{\pi}{12}} = 24.$$
  • The phase shift is $$0.$$
  • The midline is $$y = -1.$$
  • The $$y$$-intercept:
  Set $$t = 0$$. Then
  \begin{aligned} y & = -1 + 2 \cos \frac{\pi t}{12} \\ & = -1 + 2 \cos \frac{\pi 0}{12} \\ & = -1 + 2 \cos 0 \\ & = -1 + 2 = -1\\ \end{aligned}
  So the $$y$$-intercept is $$\left( 0,-2 \right)$$
  • The $$t$$-intercepts:
  Set $$y = 0$$. Then
  \begin{aligned} & y = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & 0 = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{1}{2} = \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{\pi t}{12} = \pm\frac{\pi}{3} +n 2\pi \\ \Rightarrow & t = \pm 4 +24 n \\ \end{aligned}
  So the $$t$$-intercepts over the interval $$[0,48]$$ are
  $$(4,0),(20,0),(28,0),(44,0).$$
explanation: The document contains detailed explanations and solutions for two math problems involving the graphs of sine and cosine functions. Each problem includes finding the amplitude, period, phase shift, midline, and intercepts, followed by a step-by-step solution.
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distilabel_metadata.content: # 4.5 | Graphs of Sine and Cosine Functions
  
  The graphs of $$y = a \sin(bx-c) +d$$ and $$y = a \cos(bx-c)+d$$ have the following characteristics. (Assume $$b>0.$$)
  
  • The amplitude is given by $$|a|$$ and represents half the distance between the maximum and minimum values of the function.
  • The period is given by $$\displaystyle \frac{2\pi}{b}.$$
  • The phase shift is given by $$c/b$$.
  • The center amplitude is given by $$d.$$ The corresponding midline is given by $$y = d.$$
  • The left and right endpoints of a one-cycle interval can be determined by solving the equations $$bx-c = 0$$ and $$bx -c = 2\pi.$$
  
  Sketch the graph of
  $$y = 4 \sin\left(x – \frac{\pi}{4} \right) +4.$$
  (Include two full periods.) Label $$x$$ and $$y$$ intercepts (if any). Find the amplitude, period, phase shift, and midline.
  
  From the form of the equation we can have that $$a = 4,$$ $$b = 1,$$ $$c = \frac{\pi}{4},$$ and $$d = 4.$$ So that:
  
  • The left and right endpoints of a two cycle interval can be determined by solving the equations $$x – \frac{\pi}{4} = 0$$ and $$x – \frac{\pi}{4} = 4 \pi$$. Thus, our two cycle interval is $$[\pi/4, 17\pi/4]$$.
  • The amplitude is $$4.$$
  • The period is $$\frac{2\pi}{b} = \frac{2\pi}{1} = 2\pi.$$
  • The phase shift is $$\pi/4.$$
  • The midline is $$y = 4.$$
  • The $$y$$-intercept:
  The two cycles we are sketched over the interval $$[\pi/4,17\pi/4]$$ which does not include $$x=0$$. So our graph will not include a $$y$$-intercept.
  • The $$x$$-intercepts:
  Set $$y = 0$$. Then
  \begin{aligned} & y = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & 0 = 4 \sin\left(x – \frac{\pi}{4} \right) +4 \\ \Rightarrow & -1 = \sin\left(x – \frac{\pi}{4} \right) \\ \Rightarrow & x- \frac{\pi}{4} = \frac{3\pi}{2} + n 2\pi\\ \Rightarrow & x = \frac{3\pi}{2}+ \frac{\pi}{4} + n 2\pi\\ \Rightarrow & x = \frac{7\pi}{4} + n 2\pi\\ \end{aligned}
  So the $$x$$-intercepts over the interval $$[\pi/4,17\pi/4]$$ are $$\left(\frac{7\pi}{4},0\right)$$ and $$\left(\frac{15\pi}{4},0\right)$$.
  
  Sketch the graph of
  $$y = -1 + 2 \cos \frac{\pi t}{12}.$$
  (Include two full periods.) Label $$t$$ and $$y$$ intercepts. Find the amplitude, period, phase shift, and midline.
  
  First rewrite the equation in standard form
  $$y = 2 \cos \left(\frac{\pi}{12} t \right) -1$$
  So that $$a = 2,$$ $$b = \frac{\pi}{12},$$ $$c = 0,$$ and $$d = -1.$$ Now we have the following:
  
  • The left and right endpoints of a two cycle interval can be determined by solving the equations $$\frac{\pi}{12} t = 0$$ and $$\frac{\pi}{12} t = 4 \pi$$. Thus, our two cycle interval is $$[0, 48]$$.
  • The amplitude is $$2.$$
  • The period is $$\frac{2\pi}{b} = \frac{2\pi}{\frac{\pi}{12}} = 24.$$
  • The phase shift is $$0.$$
  • The midline is $$y = -1.$$
  • The $$y$$-intercept:
  Set $$t = 0$$. Then
  \begin{aligned} y & = -1 + 2 \cos \frac{\pi t}{12} \\ & = -1 + 2 \cos \frac{\pi 0}{12} \\ & = -1 + 2 \cos 0 \\ & = -1 + 2 = -1\\ \end{aligned}
  So the $$y$$-intercept is $$\left( 0,-2 \right)$$
  • The $$t$$-intercepts:
  Set $$y = 0$$. Then
  \begin{aligned} & y = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & 0 = -1 + 2 \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{1}{2} = \cos \frac{\pi t}{12} \\ \Rightarrow & \frac{\pi t}{12} = \pm\frac{\pi}{3} +n 2\pi \\ \Rightarrow & t = \pm 4 +24 n \\ \end{aligned}
  So the $$t$$-intercepts over the interval $$[0,48]$$ are
  $$(4,0),(20,0),(28,0),(44,0).$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains detailed explanations and solutions for two math problems involving the graphs of sine and cosine functions. Each problem includes finding the amplitude, period, phase shift, midline, and intercepts, followed by a step-by-step solution.
  
  Label: multiple-question-and-answer