Chapter 4.4, Problem 97E

Solution

a.

To explain: Why the value of $b$ is positive.

For any sign change of $a$ , there occur a sign change in $b$ .

Given information:

Sinusoidal equation is given by, $y=a\sin \left[b\left(x-H\right)\right]+k$ .

Calculation:

Considering the equation,

  $y=a\sin \left[b\left(x-H\right)\right]+k$

It is noted that the sine function is odd from the given equation.

  $a\sin (-b(x-h))+k=-a\sin (b(x-h))+k$

Therefore, for any sign change of $a$ , there occur a sign change in $b$ .

b.

To prove: The equation $y=a\cos \left[b(x-h)\right]+k$ has the same graph as $y=a\sin \left[b(x-H)\right]+k$ .

Given information:

The given equations are,

  $\begin{array}{c}y=a\cos \left[b(x-h)\right]+k\\ y=a\sin \left[b(x-H)\right]+k\end{array}$

Proof:

Since a sine curve is identical to a cosine curve offset to the right by $\frac{\pi }{2}$ .

Rewriting the above cosine curve in terms of a sine function,

  $y=a\sin \left[b\left(x-h-\frac{\pi }{2}\right)\right]+k$ , which is equivalent to the second equation $H=h+\frac{\pi }{2}$ .

c.

To graph: The unit circle argument for the identity $\sin (\theta +\pi )=-\sin \theta$ .

Given information:

The identity is given by $\sin (\theta +\pi )=-\sin \theta$ .

Graph:

From the given identity,

  

Interpretation:

The graph indicates the arbitrary angle $\theta$ and $\theta +\pi$ will have the negative side lengths with respect to each other.

d.

To prove: The equation $y=-a\sin \left[b(x-h)\right]+k$ has the same graph as

  $y=a\sin \left[b(x-H)\right]+k$ .

Given information:

The given equations are,

  $\begin{array}{l}y=-a\sin \left[b(x-h)\right]+k\\ y=a\sin \left[b(x-H)\right]+k\end{array}$

Proof:

Consider the two given identities and relating both to get,

  $-a\sin \left[b(x-h)\right]+k=a\sin \left[b(x-H)\right]$

Substitute $H=h+x$ to get,

  $\begin{array}{c}-a\sin \left[b(x-h)\right]+k=a\sin \left[b(x-(h+x)\right]+k\\ -a\sin \left[b(x-h)\right]+k=a\sin \left[b(x-h-x\right]+k\end{array}$

If $H=h+x$ , then the given identity equations are said to be equivalent.

Therefore, both the identity has the same graph for the chosen value $H$ .

e.

To prove: Any sinusoid can be represented $y=a\sin \left[b(x-H)\right]+k$ by combining the results of part (a) to part (d).

Given information:

The given equations are,

  $y=a\sin \left[b(x-H)\right]+k$

Proof:

Comparing all the results from part (a) to (d), it is known that,

The equation includes all requirements for vertical shifting $k$ , horizontal shifting $-H$ , vertical scale $a$ , and horizontal scale $\frac{1}{b}$ to create a particular sinusoid.

Therefore, for any sinusoid it can be represented by the equation $y=a\sin \left[b(x-H)\right]+k$ .