Chapter 6, Problem 70RE

To determine

Determine the amplitude, period and phase shift of each function and graph at least two periods.

Answer to Problem 70RE

Amplitude of $y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)$ is $\boxed{amplitude=1}$

Periodicity of $y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)$ is $\boxed{periodicity=4\pi }$

Phase shift of $y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)$ is $\boxed{\frac{-\frac{\pi }{2}}{\frac{1}{2}}}$

Explanation of Solution

Given information:

  $y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)$

Formula used:

For $f\left(x\right)=a.\cos \left(bx-c\right)+d$ the amplitude is $\left|a\right|$

Periodicity of $a.\cos \left(bx-c\right)+d=\frac{\text{Periodicity of }\cos x}{\left|b\right|}$

For $f\left(x\right)=A\cdot g\left(Bx-C\right)+D$ , where $g\left(x\right)$ is one the basic trigonometric functions, $\frac{C}{B}$ is phase shift and D is vertical shift.

Now calculating the amplitude of the function by using $f\left(x\right)=a.\cos \left(bx-c\right)+d$ the amplitude is $\left|a\right|$ formula:

  $\begin{array}{l}y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)\\ Here,a=-1\\ amplitude=\left|a\right|\\ amplitude=\left|-1\right|\\ \boxed{amplitude=1}\end{array}$

Now calculating periodicity by using $a\cdot \cos \left(bx+c\right)+d=\frac{\text{Periodicity of }\cos x}{\left|b\right|}$ formula:

  $\frac{\text{Periodicity of }\cos x}{\left|b\right|}$

Periodicity of $\cos x$ is $2\pi$ and $b=\frac{1}{2}$ , therefore:

  $\begin{array}{l}=\frac{2\pi }{\left|\frac{1}{2}\right|}\\ \boxed{periodicity=4\pi }\end{array}$

Now calculating the phase shift of $y=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)$ :

  $\begin{array}{l}f\left(x\right)=-\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right)\\ g\left(Bx-C\right)=\cos \left(\frac{1}{2}x+\frac{\pi }{2}\right),B=\frac{1}{2},C=-\frac{\pi }{2},D=0\end{array}$

Therefore, the phasing shift $\frac{C}{B}$ is $\boxed{\frac{-\frac{\pi }{2}}{\frac{1}{2}}}$