Answer:
[tex]\displaystyle y = 3sin\:1\frac{3}{5}\pi{x} + 5[/tex]
Step-by-step explanation:
[tex]\displaystyle \boxed{y = 3cos\:(1\frac{3}{5}\pi{x} - \frac{\pi}{2}) + 5} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 5 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{5}{16}} \hookrightarrow \frac{\frac{\pi}{2}}{1\frac{3}{5}\pi} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{4}} \hookrightarrow \frac{2}{1\frac{3}{5}\pi}\pi \\ Amplitude \hookrightarrow 3[/tex]
OR
[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{4}} \hookrightarrow \frac{2}{1\frac{3}{5}\pi}\pi \\ Amplitude \hookrightarrow 3[/tex]
From the above information, you should now have an ideya of how to interpret trigonometric equations like this.
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