graph the trigonometric function.\n\n$y = \\sin(x - \\frac{3\\pi}{4})$ \n\nplot all points corresponding to…

Explanation:

Step1: Find the period

The general form of a sine - function is $y = A\sin(Bx - C)+D$. For $y=\sin(x-\frac{3\pi}{4})$, $B = 1$. The period of the sine - function $y=\sin(Bx - C)$ is $T=\frac{2\pi}{|B|}$. Since $B = 1$, the period $T = 2\pi$.

Step2: Find the x - intercepts

Set $y = 0$, so $\sin(x-\frac{3\pi}{4})=0$. Then $x-\frac{3\pi}{4}=k\pi$, where $k\in\mathbb{Z}$. For one - cycle, when $k = 0$, $x=\frac{3\pi}{4}$; when $k = 1$, $x=\frac{3\pi}{4}+\pi=\frac{7\pi}{4}$.

Step3: Find the maxima

The maximum value of $y = \sin(u)$ is 1 when $u=\frac{\pi}{2}+2k\pi$. Set $x-\frac{3\pi}{4}=\frac{\pi}{2}+2k\pi$. For one - cycle ($k = 0$), $x=\frac{\pi}{2}+\frac{3\pi}{4}=\frac{5\pi}{4}$.

Step4: Find the minima

The minimum value of $y=\sin(u)$ is - 1 when $u=\frac{3\pi}{2}+2k\pi$. Set $x-\frac{3\pi}{4}=\frac{3\pi}{2}+2k\pi$. For one - cycle ($k = 0$), $x=\frac{3\pi}{2}+\frac{3\pi}{4}=\frac{9\pi}{4}$.

To graph:

• Plot the x - intercepts at $x=\frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.
• Plot the maximum point at $(\frac{5\pi}{4},1)$.
• Plot the minimum point at $(\frac{9\pi}{4}, - 1)$.
• Then connect these points with a smooth sine - curve over the interval of one cycle (for example, from $\frac{3\pi}{4}$ to $\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}$).

Answer:

Plot the points $(\frac{3\pi}{4},0),(\frac{5\pi}{4},1),(\frac{7\pi}{4},0),(\frac{9\pi}{4}, - 1)$ and draw a sine - curve through them for one cycle.