Problem 7 - Algebra and Trigonometry

Question

Determine the amplitude and period of each function. Then graph one period of the function. $$y=\sin 2 x$$

Step-by-Step Solution

Answer

The amplitude of the function \(y=\sin 2 x\) is 1 and the period is \(\pi\).

Step 1: Determine Amplitude

The amplitude is the absolute value of the coefficient of the sine function. In this case, the coefficient is 1. So the amplitude of the function \(y=\sin 2 x\) is \(1\).

Step 2: Determine Period

The period of the sine function is calculated by the formula \(\text{Period} = \frac{2\pi}{|B|}\), where B is the coefficient of x in the function. In this case, B is 2. So the period of the function \(y=\sin 2 x\) is \(\frac{2\pi}{2} = \pi\).

Step 3: Sketching the Graph

The graph of \(y=\sin 2 x\) starts at the origin (0,0), reaches its peak at the amplitude (which is 1 in this case) at \(x=\frac{\pi}{4}\), returns back to 0 at \(x=\frac{\pi}{2}\), minimises at \(x=\frac{3\pi}{4}\), and completes one period at \(x = \pi\). This behaviour continues for other values of x, spanned across \(-\infty\) to \(+\infty\). Thus, the wave is compressed horizontally, with a peak and trough occurring more frequently than the standard sine function.