Chapter 5.2, Problem 79E

Solution

a.

To decide whether $y_1=y_2$ for $y_1=\frac{\left(\sin \left(x+.001\right)-\sin x\right)}{0.001}$ and $y_2=\cos x$ .

Given information:

The functions, $y_1=\frac{\left(\sin \left(x+.001\right)-\sin x\right)}{0.001}$ and $y_2=\cos x$ .

Explanation:

Graph the two functions.

  

From the graph it seems that both the functions equal. That is, $y_1=y_2$ .

Now, find the value of each function for several values of $x$ .

  

The tables show that the functions have different values for the same input. That means, $y_1\ne y_2$ .

Conclusion:

The two functions are not same as the table values are different.

b.

To find the value of $h$ so that the graph of $y_3=y_1-y_2$ appears to be sinusoid in the window $\left[-2\pi ,2\pi \right]$ by $\left[-h,h\right]$ .

Given information:

The functions, $y_1=\frac{\left(\sin \left(x+.001\right)-\sin x\right)}{0.001}$ and $y_2=\cos x$ .

Explanation:

The function $y_3=y_1-y_2$ or $y_3=\frac{\left(\sin \left(x+.001\right)-\sin x\right)}{0.001}-\cos x$ can be written in expanded trigonometric form as $y_3=-1.66667\times {10}^{-7}\cos (x)-0.0005\sin (x)$ .

Find the amplitude of the function.

  $\begin{array}{c}A=\sqrt{{\left(-1.66667\times {10}^{-7}\right)}^2+{\left(-0.0005\right)}^2}\\ =0.0005\end{array}$

Since the amplitude is $0.0005$ , the value of $h$ can be taken to be slightly greater than $0.0005$ , say, $h=0.0006$ .

Now, graph the function to see that it appears to be sinusoid.

  

The function $y_3=-1.66667\times {10}^{-7}\cos (x)-0.0005\sin (x)$ can be rewritten as:

  $\begin{array}{c}{y}_3=\sqrt{{\left(-1.66667\times {10}^{-7}\right)}^2+{\left(-0.0005\right)}^2}\cos \left(x+{\tan }^{-1}\left(\frac{-0.0005}{-1.66667\times {10}^{-7}}\right)\right)\\ y_3\approx 0.0005\cos \left(x+\frac{\pi }{2}\right)\end{array}$

This shows that the function $y_3$ is sinusoid.