Find the fundamental period of a sine function with a horizontal shift

Expert intro: This kind of multiple-choice trig question is really about spotting what matters and what does not. The vertical shift and horizontal shift can look important, but the period only depends on the coefficient attached to $x$ inside the sine.

Detailed walkthrough

We are given

$y=2-3\sin\left(\frac{\pi}{4}(x-1)\right).$

To find the fundamental period, look only at the coefficient of $x$ inside the sine function.

Step 1: Identify $B$

The inside is

$\frac{\pi}{4}(x-1).$

So the coefficient multiplying $x$ is

$B=\frac{\pi}{4}.$

Step 2: Apply the period formula

For sine, the fundamental period is

$\frac{2\pi}{|B|}.$

Substitute $B=\frac{\pi}{4}$:

$\frac{2\pi}{\pi/4}=2\pi\cdot\frac{4}{\pi}=8.$

Step 3: Choose the answer

The correct choice is

$\boxed{8}$

So the answer is C.

💡 Pitfall guide

Do not let the $2$, the $-3$, or the $(x-1)$ distract you. The constant outside the sine changes amplitude and vertical position; the $-1$ changes horizontal shift. Neither one changes the period. The only part that controls period is the coefficient of $x$ inside the sine.

🔄 Real-world variant

If the function had been $y=2-3\sin\left(\frac{\pi}{4}(x+1)\right)$, the period would still be $8$. Changing $(x-1)$ to $(x+1)$ only moves the graph left instead of right. If the coefficient inside were doubled, say $\frac{\pi}{2}(x-1)$, then the period would drop to $4$.

🔍 Related terms

fundamental period, amplitude, phase shift