1 Answer

Julien Lafont · Answer 1 · 2022-11-20T20:58:23+0000

Answer:

$\displaystyle 1(1)/(4)\pi$

Explanation:

$\displaystyle \boxed{y = 4cos\:(1(3)/(5)x - 1(3)/(10)\pi)} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(13)/(16)\pi} \hookrightarrow (1(3)/(10)\pi)/(1(3)/(5)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{-(3)/(4)\pi} \hookrightarrow (2)/(1(3)/(5))\pi \\ Amplitude \hookrightarrow 4$

OR

$\displaystyle \boxed{y = 4sin\:(1(3)/(5)x + 1(1)/(5)\pi)} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{-(3)/(4)} \hookrightarrow (-1(1)/(5)\pi)/(1(3)/(5)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{1(1)/(4)\pi} \hookrightarrow (2)/(1(3)/(5))\pi \\ Amplitude \hookrightarrow 4$

Keep in mind that although this IS a sine function, if you plan on writing your equation as a function of cosine, then by all means, go right ahead. As you can see, the photograph farthest to the right displays the trigonometric graph of
$\displaystyle y = 4cos\:1(3)/(5)x,$ in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the sine graph [photograph farthest to the left], accourding to the horisontal shift formula above. Also keep in mind that the −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 cosine graph [photograph farthest to the right] is shifted
$\displaystyle (13)/(16)\pi\:units$ to the left, which means that in order to match the sine graph [photograph farthest to the left], we need to shift the graph FORWARD
$\displaystyle (13)/(16)\pi\:units,$ which means the C-term will be positive, and by perfourming your calculations, you will arrive at
$\displaystyle \boxed{(13)/(16)\pi} = (1(3)/(10)\pi)/(1(3)/(5)).$ So, the cosine graph of the sine graph, accourding to the horisontal shift, is
$\displaystyle y = 4cos\:(1(3)/(5)x - 1(3)/(10)\pi).$ Intermediate, the centre photograph displays the trigonometric graph of
$\displaystyle y = 4sin\:1(3)/(5)x.$ Again, just like before, we must figure out the appropriate C-term that will make this sine graph horisontally shift and map onto the sine graph in the leftward photograph. So, between the two photographs, we can tell that the centre sine graph is shifted
$\displaystyle (3)/(4)\pi\:units$ to the right, which means that in order to match the leftward sine graph, we need to shift the graph BACKWARD
$\displaystyle (3)/(4)\pi\:units,$ which means the C-term will be negative, and by perfourming your calculations, you will arrive at
$\displaystyle \boxed{-(3)/(4)\pi} = (-1(1)/(5)\pi)/(1(3)/(5)).$ So, the sine graph is
$\displaystyle y = 4sin\:(1(3)/(5)x + 1(1)/(5)\pi).$ Now, the amplitude is obvious to figure out because it is the A-term. Moreover, 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. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.