Chapter 5.4, Problem 42E

To determine

To complete: the table and graph the two functions in the same viewing window.Then verify the identity algebraically.

Answer to Problem 42E

$x$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$	$1.2$	$1.4$
$y_1$	$0.04$	$0.15$	$0.32$	$0.51$	$0.71$	$0.87$	$0.97$
$y_2$	$0.04$	$0.15$	$0.32$	$0.51$	$0.71$	$0.87$	$0.03$

Explanation of Solution

Given information:

The table and the graph as evidence that $y_1=y_2$ ..

$x$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$	$1.2$	$1.4$
$y_1$
$y_2$

Where $y_1=\sin \left(x+\pi \right)\sin \left(x-\pi \right),y_2={\sin }^{2}x$

Calculation:

Complete the table using the table feature of the graphing utility.

$x$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$	$1.2$	$1.4$
$y_1$	$0.04$	$0.15$	$0.32$	$0.51$	$0.71$	$0.87$	$0.97$
$y_2$	$0.04$	$0.15$	$0.32$	$0.51$	$0.71$	$0.87$	$0.03$

Graph the two functions on the same window.

  

The functions $y_1$ and $y_2$ have the same graph.

Based on the sum, and the difference formula of cosine function, we have

  $\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v$ , and $\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v$

Replace $u$ with $x$ , and $v$ with $\pi$ .

  $\sin \left(x+\pi \right)\sin \left(x-\pi \right)=\left(\sin x\cos \pi -\cos x\sin \pi \right)\left(\sin x\cos \pi +\cos x\sin \pi \right)$

Use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$

  $\begin{array}{c}\left(\sin x\cos \pi -\cos x\sin \pi \right)\left(\sin x\cos \pi +\cos x\sin \pi \right)={\left(\sin x\cos \pi \right)}^2-{\left(\cos x\sin \pi \right)}^2\\ ={\sin }^{2}x{\cos }^2\pi -{\cos }^{2}x{\sin }^2\pi \\ ={\cos }^{2}x\left(1\right)-{\sin }^{2}x\left(0\right)\\ ={\sin }^{2}x\end{array}$

The results are verified algebraically.