Chapter 5.5, Problem 1E

1. (a) To define the inverse sine function, we restrict the domain of sine to the interval __________. On this interval the sine function is one-to-one, and its inverse function sin⁻¹ is defined by sin⁻¹ x = y ⇔ sin _____ = _____. For example, ${\sin }^{-1}\frac{1}{2}=\_\_\_\_\_$ because sin _____ = _____.
2. (b) To define the inverse cosine function, we restrict the domain of cosine to the interval __________. On this interval the cosine function is one-to-one and its inverse function cos⁻¹ is defined by cos⁻¹ x = y ⇔ cos _____ = _____. For example, ${\cos }^{-1}\frac{1}{2}=\_\_\_\_\_$ because cos _____ = _____.

To determine

To fill: The statement “To define the inverse function , we restrict the domain of sine to the interval _______.On this interval the sine function is one to one, and its inverse function ${\sin }^{-1}x=y\iff \sin$_______=_______. For example, ${\sin }^{-1}\frac{1}{2}=$_______.because $\sin$_______=_______.”

Answer to Problem 1E

To define the inverse function, we restrict the domain of sine to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. On this interval the sine function is one to one, and its inverse function ${\sin }^{-1}x=y\iff \sin y=x$. For example, ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$ because $\sin \frac{\pi }{6}=\frac{1}{2}$.

Explanation of Solution

To prove ${\sin }^{-1}x=y\iff \sin y=x$ as shown below,

Theorem used:

“If a function f is continuous as well as one to one on the interval, then its inverse is also one to one and continuous on that interval.”

Calculation:

To obtain the domain of the sine function, take the help of the graph as shown in below.

From the Figure 1, it is observed that the domain of the sine function is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

To prove the if and only if statement, first assume that $y={\sin }^{-1}x$ is given.

Then, the claim to prove is $x=\sin y$.

That is,

Use the above theorem mentioned above to conclude about the inverse function $y={\sin }^{-1}x$ is continuous.

Since the above function is continuous and one to one, that is rewrite as follows,

$\begin{array}{l}{\sin }^{-1}x=y\\ x=\sin y\end{array}$

Hence, the statement is obtained ${\sin }^{-1}x=y\iff \sin y=x$.

By the above statement the example is prove as follow.

That is,

$\begin{array}{l}{\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}\\ \frac{1}{2}=\sin \frac{\pi }{6}\end{array}$

Therefore, the example is $\sin \frac{\pi }{6}=\frac{1}{2}$.

To determine

To fill: The statement “To define the inverse function , we restrict the domain of cosine to the interval $$. On this interval the cosine function is one to one, and its inverse function ${\cos }^{-1}x=y\iff \cos$ $$= $$. For example, ${\cos }^{-1}\frac{1}{2}=$ $$ because $\cos$ $$= $$.”

Answer to Problem 1E

To define the inverse function, we restrict the domain of cosine to the interval $\left[0,\pi \right]$. On this interval the cosine function is one to one, and its inverse function ${\cos }^{-1}x=y\iff \cos y=x$. For example, ${\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$ because $\cos \frac{\pi }{3}=\frac{1}{2}$.

Explanation of Solution

To prove ${\sin }^{-1}x=y\iff \sin y=x$ as shown below,

Theorem used:

“If a function f is continuous as well as one to one on the interval, then its inverse is also one to one and continuous on that interval.”

Calculation:

To obtain the domain of the cosine function, take the help of the graph as shown in below.

From the Figure 2, it is observed that the domain of the sine function is $\left[0,\pi \right]$.

To prove above the if and only if statement, first assume that $y={\cos }^{-1}x$ is given.

Then, the claim to prove is $x=\cos y$.

That is,

Use the above theorem mentioned above to conclude about the inverse function $y={\cos }^{-1}x$ is continuous.

Since the above function is continuous and one to one, that is rewrite as follows,

$\begin{array}{l}{\cos }^{-1}x=y\\ x=\cos y\end{array}$

Hence, the statement is obtained ${\cos }^{-1}x=y\iff \cos y=x$.

By the above statement the example is prove as follows.

That is,

$\begin{array}{l}{\cos }^{-1}\frac{1}{2}=\frac{\pi }{6}\\ \frac{1}{2}=\cos \frac{\pi }{6}\end{array}$

Therefore, the example is $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$.