Chapter 4.5, Problem 23E

To determine

To calculate:An interval that corresponds to one cycle of the given function.

Answer to Problem 23E

An interval that corresponds to one cycle( $T=2\pi$ )of the given function is $x\in \left[0,2\pi \right]$ .

Explanation of Solution

Given information:The given function is $y=\sin \left(x-1\right)$ .

Formula used:

The generalized equation of sine function is

  $y=A\sin \left(Bx+C\right)+D$

Here

  $\begin{array}{l}\text{Amplitude}=Amp=\text{|}A\text{|}\\ \text{Period}\text{of}\text{function}=T=\frac{2\pi }{B}\\ \text{Horizontal/Phase}\text{shift}=Bx+C=0\to x=-\frac{C}{B}\\ \text{Vertical}\text{shift}=D\end{array}$

Calculation:

The generalized equation of sine function is

  $y=A\sin \left(Bx+C\right)+D$.........(1)

The given function is $y=\sin \left(x-1\right)$.........(2)

By comparing equation (1) & (2)

  $B=1$

The period of the function is

  $\begin{array}{l}T=\frac{2\pi }{B}=\frac{2\pi }{1}\\ T=2\pi \end{array}$

An interval that corresponds to $T=2\pi$ is $x\in \left[0,2\pi \right]$ .

Hence an interval that corresponds to one cycle( $T=2\pi$ )of the given function is $x\in \left[0,2\pi \right]$ .