graph the trigonometric function.\n\n$y = \\cos(x - \\frac{5\\pi}{6})$ \n\nplot all points corresponding to…

Explanation:

Step1: Find the period

The general form of a cosine function is $y = A\cos(Bx - C)+D$. For $y=\cos(x-\frac{5\pi}{6})$, $B = 1$. The period $T$ of the cosine - function $y=\cos(Bx - C)$ is given by $T=\frac{2\pi}{|B|}$. Since $B = 1$, $T = 2\pi$.

Step2: Find the phase - shift

The phase - shift of the cosine function $y=\cos(Bx - C)$ is $\frac{C}{B}$. Here, $C=\frac{5\pi}{6}$ and $B = 1$, so the phase - shift is $\frac{5\pi}{6}$.

Step3: Find the x - intercepts

Set $y = 0$, so $\cos(x-\frac{5\pi}{6})=0$. Then $x-\frac{5\pi}{6}=\frac{\pi}{2}+k\pi$, $k\in\mathbb{Z}$. Solving for $x$ gives $x=\frac{\pi}{2}+\frac{5\pi}{6}+k\pi=\frac{3\pi + 5\pi}{6}+k\pi=\frac{4\pi}{3}+k\pi$. For one - cycle ($0\leq x\leq2\pi$), when $k = 0$, $x=\frac{4\pi}{3}$.

Step4: Find the maxima

The maximum value of $y = \cos(x-\frac{5\pi}{6})$ is $y = 1$. Set $\cos(x-\frac{5\pi}{6})=1$. Then $x-\frac{5\pi}{6}=2k\pi$, $k\in\mathbb{Z}$. Solving for $x$ gives $x=\frac{5\pi}{6}+2k\pi$. For one - cycle ($0\leq x\leq2\pi$), when $k = 0$, $x=\frac{5\pi}{6}$.

Step5: Find the minima

The minimum value of $y=\cos(x - \frac{5\pi}{6})$ is $y=-1$. Set $\cos(x-\frac{5\pi}{6})=-1$. Then $x-\frac{5\pi}{6}=\pi+2k\pi$, $k\in\mathbb{Z}$. Solving for $x$ gives $x=\pi+\frac{5\pi}{6}+2k\pi=\frac{11\pi}{6}+2k\pi$. For one - cycle ($0\leq x\leq2\pi$), when $k = 0$, $x=\frac{11\pi}{6}$.

Plot the points $(\frac{5\pi}{6},1)$, $(\frac{4\pi}{3},0)$, $(\frac{11\pi}{6}, - 1)$ on the graph within the interval $[0,2\pi]$.

Answer:

Plot the points $(\frac{5\pi}{6},1)$, $(\frac{4\pi}{3},0)$, $(\frac{11\pi}{6}, - 1)$ within one - cycle ($0\leq x\leq2\pi$) and then graph the cosine function.