Chapter 13.8, Problem 60PPSE

Solution

i.

To graph: The given function in the same viewing window.

Given:

The functions are $y=-\cos x$ and $y=-\sec x$ .

Calculation:

The graph of the function is given below in the same viewing window.

  

Here the blue line indicates the function $y=-\cos x$ and the green line indicates the function $y=-\sec x$ .

Conclusion:

Hence, the graph of the functions is given above.

ii.

To find: The domain the domain range and period of each function.

The domain, range, and period of the $y=-\cos x$ is $\left(-\infty ,\infty \right)$ , $\left[-1,1\right]$ and $2\pi$ respectively. And the domain, range, and period of the $y=-\sec x$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ , $\left\{x|x\ne \frac{\pi }{2}+\pi n\right\}$ and $2\pi$ .

Given:

The functions are $y=-\cos x$ and $y=-\sec x$ .

Calculation:

Consider the following graph to obtain the domain, range and amplitude.

  

The blue line in the ranges from $y=-1$ to $y=+1$ so the range of the function $y=-\cos x$ is $\left[-1,1\right]$ . The domain of the function is $\left(-\infty ,\infty \right)$ because it starts from negative infinity and goes to positive infinity. The period of the function is $3\pi -\pi =2\pi$ .

According to the graph, the range of the function $y=-\sec x$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ . The domain of the function is all real numbers except $x=\frac{\pi }{2}+\pi n$ because at this point $\cos x=0$ it means the function is undefined. The period of the function is $3\pi -\pi =2\pi$ .

Conclusion:

Hence, the domain, range, and period of the $y=-\cos x$ is $\left(-\infty ,\infty \right)$ , $\left[-1,1\right]$ and $2\pi$ respectively. And the domain, range, and period of the $y=-\sec x$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$ , $\left\{x|x\ne \frac{\pi }{2}+\pi n\right\}$ and $2\pi$ .

iii.

To find: The value of $x$ for which the given function is equivalent.

  $\pi n$

Given:

The functions are $y=-\cos x$ and $y=-\sec x$ .

Calculation:

It is understood that $-\cos x=-\sec x$ .

Rewrite the function as follows:

  $\begin{array}{c}-\cos x=-\frac{1}{\cos x}\\ {\cos }^{2}x=1\\ \cos x=\pm 1\end{array}$

So, the value of $x$ is,

  $\begin{array}{c}x={\cos }^{-1}\left(\pm 1\right)\\ x=0^{\circ },{180}^{\circ }\\ x=\pi n\end{array}$

Here, $n$ is positive integer.

Conclusion:

Hence, the value of $x$ is $\pi n$ .

iv.

To compare: The graph of the function and obtain the similarities and differences.

  $12\text{ hours }22\text{ min}$

Given:

The functions are $y=-\cos x$ and $y=-\sec x$ .

Calculation:

Consider the following graphs for comparison:

  

The shape of both the graphs are completely different and their domain and range are also different. The period of both the graphs are same. Both the curves are above the $x-\text{axis}$ together and vice versa.

Conclusion:

Hence, the shape, domain and range are different while the period and the behaviour are same.

v.

To justify: The given statement.

Both the curves are above $x-\text{axis}$ together so when one is positive then the other is also positive. Similarly, both the curves are below $x-\text{axis}$ together so when one is negative then the other is also negative.

Given:

The functions are $y=-\cos x$ and $y=-\sec x$ .

Calculation:

Consider the following graph:

  

From the graph, it is understood that when the curve of one function is above $x-\text{axis}$ then the curve of another function is also above $x-\text{axis}$ . It means when one function is positive then another function is also positive. Similarly, when the curve of one function is below $x-\text{axis}$ then the curve of another function is also below $x-\text{axis}$ . It means when one function is negative then another function is also negative.

To understand this, consider $x={180}^{\circ }$ .

It is understood that $\cos {180}^{\circ }=-1$ . So,

  $\begin{array}{c}-\cos x=-\cos \left({180}^{\circ }\right)\\ =1\end{array}$

Now substitute $x={180}^{\circ }$ in $-\sec x$ to check this is positive or not.

  $\begin{array}{c}-\sec x=-\frac{1}{\cos x}\\ =-\frac{1}{\cos {180}^{°}}\\ =1\end{array}$

So, it can be concluded that when $-\cos x$ is positive then $-\sec x$ is also positive.

Similarly, consider $x=0^{\circ }$ then,

  $\begin{array}{c}-\cos 0^{\circ }=-1\\ -\sec 0^{\circ }=-\frac{1}{\cos 0^{\circ }}\\ =-\frac{1}{1}\\ =-1\end{array}$

So, when the $-\cos x$ is negative then $-\sec x$ is also negative.

Conclusion:

Hence, both the curves are above $x-\text{axis}$ together so when one is positive then the other is also positive. Similarly, both the curves are below $x-\text{axis}$ together so when one is negative then the other is also negative.