1 Answer

Watchme · Answer 1 · 2019-06-23T01:24:58+0000

Answer(s):

$\displaystyle y = 4cos\:(4\theta \pm \pi) \\ y = -4cos\:4\theta$

Explanation:

$\displaystyle y = Acos(B\theta - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{\pm(\pi)/(4)} \hookrightarrow (\pm\pi)/(4) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{(\pi)/(2)} \hookrightarrow (2)/(4)\pi \\ Amplitude \hookrightarrow 4$

OR

$\displaystyle y = -Acos(B\theta - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{(\pi)/(2)} \hookrightarrow (2)/(4)\pi \\ Amplitude \hookrightarrow 4$

You will need the above information to help you interpret the graph. First off, keep in mind that this cosine graph will have TWO equations because the curvature begins upward from
$\displaystyle [0, -4]$ instead of downward from
$\displaystyle [0, 4],$ telling you that one equation will have a “negative” symbol inserted in the beginning of the equation. Before we go any further though, we must figure the period of the graph out. So, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits
$\displaystyle [0, -4],$ from there to
$\displaystyle [-(\pi)/(2), -4],$ they are obviously
$\displaystyle (\pi)/(2)\:unit$ apart, telling you that the period of the graph is
$\displaystyle (\pi)/(2).$ Now, as you can see, the photograph on the right displays the trigonometric graph of
$\displaystyle y = 4cos\:4\theta.$ Now, if you look hard enough, you will see that both graphs are “mirror reflections” of one another, meaning you can figure the rest of the terms out one of two ways. The first way is to figure the appropriate C-term out that will make the graph horisontally shift and map onto the original cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also, keep in mind that −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the rightward graph is shifted
$\displaystyle (\pi)/(4)\:unit$ on both sides of the y-axis, which means that in order to match the original graph, we need to shift the graph back, which means the C-term will be both negative and positive; and by perfourming your calculations, you will arrive at
$\displaystyle \boxed{\pm(\pi)/(4)} = (\pm\pi)/(4).$ So, one equation of the cosine graph, accourding to the horisontal shift, is
$\displaystyle y = 4cos\:(4\theta \pm \pi).$ Now that we got this out the way, we can focuss on finding the second equation. Another way is to write an equation with a “negative” symbol inserted in the beginning [like I mentioned earlier]. Now, sinse we are writing an equation with the negative, the graph will not have a horisontal shift; so, C will be zero. With this said, the second equation is
$\displaystyle y = -4cos\:4\theta.$ Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at
$\displaystyle y = 0,$ in which each crest is extended four units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.