graph the following equation (y = \frac{1}{2}sin(2\theta+\frac{5pi}{3}) + 2).

Answer

Explanation:

Step1: Identify the amplitude

The general form of a sinusoidal function is $y = A\sin(B\theta - C)+D$. For the function $y=\frac{1}{2}\sin(2\theta+\frac{5\pi}{3}) + 2$, the amplitude $A=\frac{1}{2}$. This means the graph will oscillate between values that are $\frac{1}{2}$ unit above and below the mid - line.

Step2: Identify the period

The period of a sinusoidal function $y = A\sin(B\theta - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Here, $B = 2$, so $T=\frac{2\pi}{2}=\pi$. This means the graph will complete one full cycle in an interval of length $\pi$.

Step3: Identify the phase shift

The phase - shift is given by $\frac{C}{B}$. For $y=\frac{1}{2}\sin(2\theta+\frac{5\pi}{3})+2$, we can rewrite it as $y=\frac{1}{2}\sin(2(\theta+\frac{5\pi}{6})) + 2$. So the phase - shift is $-\frac{5\pi}{6}$ (a shift to the left).

Step4: Identify the vertical shift

The vertical shift is $D = 2$. So the mid - line of the graph is $y = 2$.

The graph of $y=\frac{1}{2}\sin(2\theta+\frac{5\pi}{3})+2$ has an amplitude of $\frac{1}{2}$, a period of $\pi$, a phase - shift of $-\frac{5\pi}{6}$ and a vertical shift of 2. The correct graph is one that oscillates with these characteristics. Without seeing all the options in full detail, we can note that a graph with a period of $\pi$ will have a more frequent wave pattern compared to a standard sine function with period $2\pi$. The mid - line at $y = 2$ and amplitude of $\frac{1}{2}$ means the function will range from $y=2-\frac{1}{2}=\frac{3}{2}$ to $y=2 + \frac{1}{2}=\frac{5}{2}$.

Since the period is $\pi$, option B is more likely as it shows a more frequently repeating wave pattern compared to option A which seems to have a longer period.