Chapter 10, Problem 13TP

Solution

To State:

How the graph of the given function will look without actually graphing it. Explain your reasoning.

To Complete:

The given table and verify that the graph will indeed look the way you predicted earlier.

The graph of $y=\sin x\cos x$ is the same as $y=\frac{1}{2}\sin \left(2x\right)$ , that is the graph of $y=\sin x$ shrunk by a factor of $2$ about the origin in both the horizontal and the vertical directions. The values in the following table agree with this prediction.

$x$	$-\pi$	$-\frac{3\pi }{4}$	$-\frac{\pi }{2}$	$-\frac{\pi }{4}$	$0$	$\frac{\pi }{4}$	$\frac{\pi }{2}$	$\frac{3\pi }{4}$	$\pi$
$f\left(x\right)$	$0$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$0$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$0$

Given:

The function $f\left(x\right)=\sin x\cos x$ .

The table to complete:

$x$	$-\pi$	$-\frac{3\pi }{4}$	$-\frac{\pi }{2}$	$-\frac{\pi }{4}$	$0$	$\frac{\pi }{4}$	$\frac{\pi }{2}$	$\frac{3\pi }{4}$	$\pi$
$f\left(x\right)$	?	?	?	?	?	?	?	?	?

Concepts Used:

The trigonometric identity $2\sin x\cos x=\sin 2x$ .

If $a>1$ , then the graph $y=f\left(ax\right)$ is the graph of $y=f\left(x\right)$ shrunk horizontally about the origin by a factor of $a$ .

If $a>1$ , then the graph $y=\frac{1}{a}f\left(x\right)$ is the graph of $y=f\left(x\right)$ shrunk vertically about the origin by a factor of $a$ .

The trigonometric function’s value at standard angles $\sin \left(n\pi \right)=0;n\in \mathbb{Z}$ , $\sin \frac{\pi }{2}=1$ . The identity $\sin \left(-x\right)=-\sin x$ .

Calculations:

Rewrite the equation $y=\sin x\cos x$ :

  $\begin{array}{l}y=\sin x\cos x\\ y=\frac{1}{2}\times 2\sin x\cos x\\ y=\frac{1}{2}\sin \left(2x\right)\end{array}$

The graph of $y=\sin x\cos x$ is the same as $y=\frac{1}{2}\sin \left(2x\right)$ , that is the graph of $y=\sin x$ shrunk by a factor of $2$ about the origin in both the horizontal and the vertical directions.

Complete the table using the definition $f\left(x\right)=\frac{1}{2}\sin \left(2x\right)$ and the value of $\sin x$ at standard values of the angle $x$ .

$x$	$-\pi$	$-\frac{3\pi }{4}$	$-\frac{\pi }{2}$	$-\frac{\pi }{4}$	$0$	$\frac{\pi }{4}$	$\frac{\pi }{2}$	$\frac{3\pi }{4}$	$\pi$
$f\left(x\right)$	$0$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$0$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$0$

The values in the table indicate that the period of $f\left(x\right)$ is indeed $\pi$ , half of that of $\sin x$ . Also the amplitude of $f\left(x\right)$ is $\frac{1}{2}$ , the half of that of $\sin x$ .

Conclusion:

The graph of $y=\sin x\cos x$ is the same as $y=\frac{1}{2}\sin \left(2x\right)$ , that is the graph of $y=\sin x$ shrunk by a factor of $2$ about the origin in both the horizontal and the vertical directions. The values in the table agree with this prediction.