Question

1. graph the function $y = 3 - 6(\sin x)^2$ on the interval $0, 2\pi$. include the work for the period and the work on how to find the distance between the marks.

Explanation:

Step1: Simplify the function using trigonometric identity

We know the double - angle identity \(\cos(2x)=1 - 2\sin^{2}x\), so \(\sin^{2}x=\frac{1 - \cos(2x)}{2}\).
Substitute \(\sin^{2}x\) into the function \(y = 3-6\sin^{2}x\):

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Step2: Find the period of the function

For a cosine function of the form \(y = A\cos(Bx + C)+D\), the period \(T=\frac{2\pi}{|B|}\).
In the function \(y = 3\cos(2x)\), \(A = 3\), \(B = 2\), \(C = 0\), \(D = 0\).
So the period \(T=\frac{2\pi}{|2|}=\pi\).
On the interval \([0,2\pi]\), the number of periods of \(y = 3\cos(2x)\) is \(\frac{2\pi}{T}=\frac{2\pi}{\pi}=2\) periods.

Step3: Analyze the key points for graphing

For the cosine function \(y = \cos(x)\), the key points in one period \([0,2\pi]\) are \((0,1)\), \((\frac{\pi}{2},0)\), \((\pi, - 1)\), \((\frac{3\pi}{2},0)\), \((2\pi,1)\).

For \(y = 3\cos(2x)\), we use the substitution \(u = 2x\).

• When \(x = 0\), \(u=0\), \(y = 3\cos(0)=3\times1 = 3\).
• When \(x=\frac{\pi}{4}\), \(u = 2\times\frac{\pi}{4}=\frac{\pi}{2}\), \(y = 3\cos(\frac{\pi}{2})=3\times0 = 0\).
• When \(x=\frac{\pi}{2}\), \(u=\pi\), \(y = 3\cos(\pi)=3\times(- 1)=-3\).
• When \(x=\frac{3\pi}{4}\), \(u=\frac{3\pi}{2}\), \(y = 3\cos(\frac{3\pi}{2})=3\times0 = 0\).
• When \(x=\pi\), \(u = 2\pi\), \(y = 3\cos(2\pi)=3\times1 = 3\).
• When \(x=\frac{5\pi}{4}\), \(u=\frac{5\pi}{2}\), \(y = 3\cos(\frac{5\pi}{2})=3\times0 = 0\).
• When \(x=\frac{3\pi}{2}\), \(u = 3\pi\), \(y = 3\cos(3\pi)=3\times(-1)=-3\).
• When \(x=\frac{7\pi}{4}\), \(u=\frac{7\pi}{2}\), \(y = 3\cos(\frac{7\pi}{2})=3\times0 = 0\).
• When \(x = 2\pi\), \(u = 4\pi\), \(y = 3\cos(4\pi)=3\times1 = 3\).

Step4: Determine the distance between the marks (x - axis intervals)

The interval is \([0,2\pi]\). We can divide the interval \([0,2\pi]\) into sub - intervals based on the period and the key points.
The period of \(y = 3\cos(2x)\) is \(\pi\). If we want to mark the x - axis with intervals that match the key points of the function, we can use the x - values of the key points we found above. The distance between consecutive key points (e.g., between \(x = 0\) and \(x=\frac{\pi}{4}\)) can be determined by the step in \(x\) that gives us the key points. Since the argument of the cosine function is \(2x\), to get the standard cosine key points (at \(0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\) for the argument \(u = 2x\)), we solve for \(x\) as \(x=\frac{u}{2}\). So the distance between marks (the step in \(x\)) can be \(\frac{\pi}{4}\) (since when \(u\) increases by \(\frac{\pi}{2}\), \(x\) increases by \(\frac{\pi}{4}\)).

To graph the function:

• Plot the key points \((0,3)\), \((\frac{\pi}{4},0)\), \((\frac{\pi}{2}, - 3)\), \((\frac{3\pi}{4},0)\), \((\pi,3)\), \((\frac{5\pi}{4},0)\), \((\frac{3\pi}{2}, - 3)\), \((\frac{7\pi}{4},0)\), \((2\pi,3)\).
• Then connect these points with a smooth curve, since the cosine function is continuous.

Answer:

The function \(y = 3 - 6\sin^{2}x\) simplifies to \(y = 3\cos(2x)\) with period \(\pi\). The key points on \([0,2\pi]\) are \((0,3)\), \((\frac{\pi}{4},0)\), \((\frac{\pi}{2}, - 3)\), \((\frac{3\pi}{4},0)\), \((\pi,3)\), \((\frac{5\pi}{4},0)\), \((\frac{3\pi}{2}, - 3)\), \((\frac{7\pi}{4},0)\), \((2\pi,3)\) and the distance between marks (x - axis intervals) can be \(\frac{\pi}{4}\) (or other intervals based on the desired level of detail). The graph is a cosine - shaped curve (since \(y = 3\cos(2x)\)) with amplitude \(3\), period \(\pi\), passing through the plotted key points.