Chapter 4.4, Problem 109E

(a)

To determine

To calculate: The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =30°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 109E

The value $\left(f+g\right)\left(30°\right)$ is $\frac{1+\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

Concept Used:

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

Calculation:

Substitute the values of $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ 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 $\theta =30°$ in the above equation it gives

  $\left(f+g\right)\left(30°\right)=\sin 30°+\cos 30°$ .

Now substitute $\frac{1}{2}$ for $\sin 30°$ and $\frac{\sqrt{3}}{2}$ for $\cos 30°$ in the above identity,

  $\begin{array}{c}\left(f+g\right)\left(30°\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}\\ =\frac{1+\sqrt{3}}{2}\end{array}$ .

Therefore the value $\left(f+g\right)\left(30°\right)$ is $\frac{1+\sqrt{3}}{2}$ .

(b)

To determine

To calculate: The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =30°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 109E

The value $\left(g-f\right)\left(30°\right)$ is $\frac{\sqrt{3}-1}{2}$ .

Explanation of Solution

Given:

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

Concept 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 of $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ 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 $\theta =30°$ in the above equation it gives

  $\left(g-f\right)\left(30°\right)=\cos 30°-\sin 30°$ .

Now substitute $\frac{1}{2}$ for $\sin 30°$ and $\frac{\sqrt{3}}{2}$ for $\cos 30°$ in the above identity,

  $\begin{array}{c}\left(g-f\right)\left(30°\right)=\frac{\sqrt{3}}{2}-\frac{1}{2}\\ =\frac{\sqrt{3}-1}{2}\end{array}$ .

Therefore the value $\left(g-f\right)\left(30°\right)$ is $\frac{\sqrt{3}-1}{2}$ .

(c)

To determine

To calculate: The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =30°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 109E

The value ${\left[g\left(30°\right)\right]}^2$ is $\frac{3}{4}$ .

Explanation of Solution

Given:

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

Concept 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 values of $g\left(\theta \right)=\cos \theta$ in above identity

  ${\left[g\left(\theta \right)\right]}^2={\left(\cos \theta \right)}^2$ , substitute $\theta =30°$ in the above equation it gives

  ${\left[g\left(30°\right)\right]}^2={\left(\cos 30°\right)}^2$ .

Now substitute $\frac{\sqrt{3}}{2}$ for $\cos 30°$ in the above identity,

  $\begin{array}{c}{\left[g\left(30°\right)\right]}^2={\left(\cos 30°\right)}^2\\ ={\left(\frac{\sqrt{3}}{2}\right)}^2\\ =\frac{3}{4}\end{array}$ .

Therefore the value ${\left[g\left(30°\right)\right]}^2$ is $\frac{3}{4}$ .

(d)

To determine

To calculate: The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =30°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 109E

The value $\left(fg\right)\left(30°\right)$ is $\frac{\sqrt{3}}{4}$ .

Explanation of Solution

Given:

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

Concept Used:

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

Calculation:

Substitute the values of $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ 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 $\theta =30°$ in the above equation it gives

  $\left(fg\right)\left(30°\right)=\sin \left(30°\right)\cos \left(30°\right)$ .

Now substitute $\frac{1}{2}$ for $\sin 30°$ and $\frac{\sqrt{3}}{2}$ for $\cos 30°$ in the above identity,

  $\begin{array}{c}\left(fg\right)\left(\theta \right)=\frac{1}{2}\times \frac{\sqrt{3}}{2}\\ =\frac{\sqrt{3}}{4}\end{array}$ .

Therefore the value $\left(fg\right)\left(30°\right)$ for $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ is $\frac{\sqrt{3}}{4}$ .

(e)

To determine

To calculate: The exact value of function $f\left(2\theta \right)$ for $\theta =30°$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 109E

The value $f\left(2\theta \right)$ for $\theta =30°$ is $\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

Concept Used:

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

Calculation:

As $\theta =30°$ , so $2\theta =60°$ .

Substitute the value of $2\theta$ in $f\left(2\theta \right)=\sin \left(2\theta \right)$

  $f\left(60°\right)=\sin 60°$ ,

Now substitute $\frac{\sqrt{3}}{2}$ for $\sin 60°$ in the above equation,

  $f\left(60°\right)=\frac{\sqrt{3}}{2}$ .

Therefore value $f\left(2\theta \right)$ for $\theta =30°$ is $\frac{\sqrt{3}}{2}$ .

(f)

To determine

To calculate: The exact value of function $g\left(-\theta \right)$ for $\theta =30°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 109E

The value $g\left(-30°\right)$ is $\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

Concept 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 by the property of even function $\cos \left(-\theta \right)=\cos \theta$ , now substitute $\theta =30°$ in the equation it gives $\cos \left(-30°\right)=\cos \left(30°\right)$ .

Substitute $\frac{\sqrt{3}}{2}$ for $\cos \left(30°\right)$ it gives

  $\cos \left(-30°\right)=\frac{\sqrt{3}}{2}$ .

Therefore the value of $\cos \left(-\theta \right)$ for $\theta =30°$ is $\frac{\sqrt{3}}{2}$ .