Answer:
[tex]\begin{gathered} a)y=-2\cos\left(\frac{8}{5}x\right)-1 \\ b)y=-3\sin\left(\frac{4}{3}x\right)-2 \end{gathered}[/tex]
Step-by-step explanation:
The standard trigonometric function is represented by the form:
[tex]\begin{gathered} f(x)=A\text{ trig \lparen Bx-C\rparen+D} \\ \text{ where,} \\ \text{ A= amplitude } \\ \text{ B= represents the speed of the cycle} \\ \text{ Period is }\frac{2\pi}{b} \\ \frac{c}{b}\text{ represents the pase shift} \\ d=\text{ represents the vertical shift} \end{gathered}[/tex]
a) Therefore, for the given graph, since it is a cosine function reflected:
[tex]y=-2\cos\left(\frac{8}{5}x\right)-1[/tex]
b) Now, for your own trigonometric function, do a vertical shift down 2 units of the sinusoidal function, amplitude of 3, reflect it, and let's say it has a period of 3pi/2.
[tex]y=-3\sin\left(\frac{4}{3}x\right)\ -2[/tex]