Chapter 4.4, Problem 119E

(a)

To determine

To calculate: The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =-270°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 119E

The value $\left(f+g\right)\left(-270°\right)$ is $1$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Formula Used:

Use the property $\left(f+g\right)\left(\theta \right)=f\left(\theta \right)+g\left(\theta \right)$ , for the given value of $\theta$ .

Use another property $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ .

Calculation:

Substitute the value $\sin \theta$ for $f\left(\theta \right)$ and $\cos \theta$ for $g\left(\theta \right)$ in above identity.

  $\begin{array}{c}\left(f+g\right)\left(\theta \right)=f\left(\theta \right)+g\left(\theta \right)\\ =\sin \theta +\cos \theta \end{array}$

Substitute $-270°$ for $\theta$ in the above equation.

  $\left(f+g\right)\left(-270°\right)=\sin \left(-270°\right)+\cos \left(-270°\right)$

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore

  $\sin \left(-270°\right)=-\sin 270°$ and $\cos \left(-270°\right)=\cos 270°$ .

Therefore,

  $\left(f+g\right)\left(-270°\right)=-\sin \left(270°\right)+\cos \left(270°\right)$

Now substitute the trigonometric value $-1$ for $\sin 270°$ and $0$ for $\cos 270°$ in the above equation.

  $\sin \left(270°\right)=-1$ and $\cos \left(270°\right)=0$ .

Now substitute the values in the formula.

  $\begin{array}{c}\left(f+g\right)\left(-270°\right)=\sin \left(-270°\right)+\cos \left(-270°\right)\\ =-\left(-1\right)+0\\ =1\end{array}$

Therefore the value $\left(f+g\right)\left(-270°\right)$ is $1$ .

(b)

To determine

To calculate: The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =-270°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 119E

The value $\left(g-f\right)\left(-270°\right)$ is $-1$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Formula Used:

Use the property $\left(g-f\right)\left(\theta \right)=g\left(\theta \right)-f\left(\theta \right)$ , for the given value of $\theta$ .

Calculation:

Substitute the values $\sin \theta$ for $f\left(\theta \right)$ and $\cos \theta$ for $g\left(\theta \right)$ in above identity.

  $\begin{array}{c}\left(g-f\right)\left(\theta \right)=g\left(\theta \right)-f\left(\theta \right)\\ =\cos \theta -\sin \theta \end{array}$

Substitute $-270°$ for $\theta$ in the above equation.

  $\left(g-f\right)\left(-270°\right)=\cos \left(-270°\right)-\sin \left(-270°\right)$

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore

  $\sin \left(-270°\right)=-\sin 270°$ and $\cos \left(-270°\right)=\cos 270°$

Therefore,

  $\left(g-f\right)\left(-270°\right)=\cos \left(270°\right)+\sin \left(270°\right)$

Now substitute the trigonometric value $-1$ for $\sin 270°$ and $0$ for $\cos 270°$ in the above equation.

  $\sin \left(270°\right)=-1$ and $\cos \left(270°\right)=0$ .

Now substitute the values in the formula

  $\begin{array}{c}\left(g-f\right)\left(-270°\right)=\cos \left(-270°\right)-\sin \left(-270°\right)\\ =0+\left(-1\right)\\ =-1\end{array}$

Therefore the value $\left(g-f\right)\left(-270°\right)$ is $-1$ .

(c)

To determine

To calculate: The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =-270°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 119E

The value ${\left[g\left(-270°\right)\right]}^2$ is $0$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Formula Used:

Use the value of $g\left(\theta \right)=\cos \theta$ in the required value ${\left[g\left(\theta \right)\right]}^2$ , for the given value of $\theta$ .

Calculation:

Substitute the value $\cos \theta$ for $g\left(\theta \right)$ in above identity.

  ${\left[g\left(\theta \right)\right]}^2={\left(\cos \theta \right)}^2$

Substitute $-270°$ for $\theta$ in the above equation.

  ${\left[g\left(-270°\right)\right]}^2={\left(\cos \left(-270°\right)\right)}^2$

By the trigonometric identity $\cos \left(-\theta \right)=\cos \theta$ , therefore

  $\cos \left(-270°\right)=\cos 270°$

Therefore,

  ${\left[g\left(-270°\right)\right]}^2={\left(\cos \left(270°\right)\right)}^2$

Now substitute the trigonometric value $0$ for $\cos 270°$ in the above equation.

  $\cos \left(270°\right)=0$ .

Substitute the value in above equation.

  $\begin{array}{c}{\left[g\left(-270°\right)\right]}^2={\left(\cos 270°\right)}^2\\ =0\end{array}$

Therefore the value ${\left[g\left(-270°\right)\right]}^2$ is $0$ .

(d)

To determine

To calculate: The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =-270°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 119E

The value $\left(fg\right)\left(-270°\right)$ is $0$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Formula Used:

Use the property $\left(fg\right)\left(\theta \right)=f\left(\theta \right)g\left(\theta \right)$ , for the given value of $\theta$ .

Use the property $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ .

Calculation:

Substitute the value $\sin \theta$ for $f\left(\theta \right)$ and $\cos \theta$ for $g\left(\theta \right)$ in above identity.

  $\begin{array}{c}\left(fg\right)\left(\theta \right)=f\left(\theta \right)g\left(\theta \right)\\ =\sin \theta \cos \theta \end{array}$

Substitute $-270°$ for $\theta$ in the above equation.

  $\left(fg\right)\left(-270°\right)=\sin \left(-270°\right)\cos \left(-270°\right)$

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore

  $\sin \left(-270°\right)=-\sin 270°$ and $\cos \left(-270°\right)=\cos 270°$

Therefore,

  $\left(fg\right)\left(-270°\right)=-\sin \left(270°\right)\cos \left(270°\right)$

Now substitute the trigonometric value $-1$ for $\sin 270°$ and $0$ for $\cos 270°$ in the formula.

  $\begin{array}{c}\left(fg\right)\left(-270°\right)=-\sin \left(270°\right)\cos \left(270°\right)\\ =-\left(-1\right)\times 0\\ =0\end{array}$

Therefore the value $\left(fg\right)\left(-270°\right)$ for $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ is $0$ .

(e)

To determine

To calculate: The exact value of function $f\left(2\theta \right)$ for $\theta =-270°$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 119E

The value $f\left(2\theta \right)$ for $\theta =-270°$ is $0$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Use another property $\sin \left(-\theta \right)=-\sin \theta$ and $\sin \left(360°+\theta \right)=\sin \theta$ .

Formula Used:

Firstly calculate the angle $2\theta$ for given $\theta$ and then calculate $f\left(2\theta \right)$ .

Calculation:

As $\theta =-270°$ , so $2\theta =-540°$ .

Substitute the value $-540°$ for $2\theta$ in $f\left(2\theta \right)=\sin \left(2\theta \right)$

  $f\left(-540°\right)=\sin \left(-540°\right)$

By the trigonometric property $\sin \left(-\theta \right)=-\sin \theta$ , therefore

  $\sin \left(-540°\right)=-\sin 540°$

Therefore, $f\left(-540°\right)=-\sin \left(540°\right)$

As $\sin \left(360°+\theta \right)=\sin \theta$ .

Substitute $180°$ for $\theta$ in the equation.

  $\sin \left(360°+180°\right)=\sin 180°$

  $\sin 540°=\sin 180°$

Substitute the trigonometric value $0$ for $\sin 180°$ in the above equation

  $\sin \left(540°\right)=0$

Therefore value $f\left(2\theta \right)$ for $\theta =-270°$ is $0$ .

(f)

To determine

To calculate: The exact value of function $g\left(-\theta \right)$ for $\theta =-270°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 119E

The value $g\left(-270°\right)$ is $0$ .

Explanation of Solution

Given:

The functions $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Formula Used:

Use the property $g\left(-\theta \right)=g\left(\theta \right)$ for the given even function $g\left(\theta \right)=\cos \theta$ , for the given value of $\theta$ .

Calculation:

As, $\theta =-270°$ , then $-\theta =270°$ . Therefore, $g\left(-270°\right)=\cos \left(270°\right)$ .

Now, substitute the trigonometric value $0$ for $\cos \left(270°\right)$ .

  $\cos \left(270°\right)=0$ .

Therefore the value of $\cos \left(-\theta \right)$ for $\theta =-270°$ is $0$ .