Chapter 4.4, Problem 111E

(a)

To determine

To calculate: The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =315°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 111E

The value $\left(f+g\right)\left(315°\right)$ is $0$ .

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 $\theta =315°$ in the above equation it gives

  $\left(f+g\right)\left(315°\right)=\sin 315°+\cos 315°$ .

By the trigonometric property $\sin \left(360°-\theta \right)=-\sin \theta$ and $\cos \left(360°-\theta \right)=\cos \theta$ ,

Substitute $\theta =45°$ in both the properties $\sin \left(360°-45°\right)=-\sin 45°$ , it gives $\sin \left(315°\right)=-\sin 45°$ and also $\cos \left(360°-45°\right)=\cos 45°$ , it gives $\cos \left(315°\right)=\cos 45°$ .

Now substitute $\frac{1}{\sqrt{2}}$ for $\sin 45°$ and $\frac{1}{\sqrt{2}}$ for $\cos 45°$ in the above equation it gives,

  $\sin \left(315°\right)=-\frac{1}{\sqrt{2}}$ and $\cos \left(315°\right)=\frac{1}{\sqrt{2}}$ .

Now substitute the values in the formula $\begin{array}{c}\left(f+g\right)\left(315°\right)=\sin 315°+\cos 315°\\ =-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ =0\end{array}$

Therefore the value $\left(f+g\right)\left(315°\right)$ is $0$ .

(b)

To determine

To calculate: The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =315°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 111E

The value $\left(g-f\right)\left(315°\right)$ is $\sqrt{2}$ .

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 $\theta =315°$ in the above equation it gives

  $\left(g-f\right)\left(315°\right)=\cos 315°-\sin 315°$ .

By the trigonometric property $\sin \left(360°-\theta \right)=-\sin \theta$ and $\cos \left(360°-\theta \right)=\cos \theta$ ,

Substitute $\theta =45°$ in both the properties $\sin \left(360°-45°\right)=-\sin 45°$ , it gives $\sin \left(315°\right)=-\sin 45°$ and also $\cos \left(360°-45°\right)=\cos 45°$ , it gives $\cos \left(315°\right)=\cos 45°$ .

Now substitute $\frac{1}{\sqrt{2}}$ for $\sin 45°$ and $\frac{1}{\sqrt{2}}$ for $\cos 45°$ in the above equation it gives,

  $\sin \left(315°\right)=-\frac{1}{\sqrt{2}}$ and $\cos \left(315°\right)=\frac{1}{\sqrt{2}}$ .

Now substitute the values in the formula $\begin{array}{c}\left(g-f\right)\left(315°\right)=\cos 315°-\sin 315°\\ =\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\\ =\frac{2}{\sqrt{2}}\\ =\sqrt{2}\end{array}$

Therefore the value $\left(g-f\right)\left(315°\right)$ is $\sqrt{2}$ .

(c)

To determine

To calculate: The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =315°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 111E

The value ${\left[g\left(315°\right)\right]}^2$ is $\frac{1}{2}$ .

Explanation of Solution

Given:

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

Concept Used:

Use the 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 $\theta =315°$ in the above equation it gives

  ${\left[g\left(315°\right)\right]}^2={\left(\cos 315°\right)}^2$ .

By the trigonometric property $\cos \left(360°-\theta \right)=\cos \theta$ ,

Substitute $\theta =45°$ in the above property it gives $\cos \left(360°-45°\right)=\cos 45°$ , it gives $\cos \left(315°\right)=\cos 45°$ .

Now substitute $\frac{1}{\sqrt{2}}$ for $\cos 315°$ in the above identity,

  $\begin{array}{c}{\left[g\left(315°\right)\right]}^2={\left(\cos 315°\right)}^2\\ ={\left(\frac{1}{\sqrt{2}}\right)}^2\\ =\frac{1}{2}\end{array}$ .

Therefore the value ${\left[g\left(315°\right)\right]}^2$ is $\frac{1}{2}$ .

(d)

To determine

To calculate: The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =315°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 111E

The value $\left(fg\right)\left(315°\right)$ is $-\frac{1}{2}$ .

Explanation of Solution

Given:

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

Concept Used:

Use the property $\left(fg\right)\left(\theta \right)=f\left(\theta \right)g\left(\theta \right)$ , 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 $\theta =315°$ in the above equation it gives

  $\left(fg\right)\left(315°\right)=\sin \left(315°\right)\cos \left(315°\right)$ .

By the trigonometric property $\sin \left(360°-\theta \right)=-\sin \theta$ and $\cos \left(360°-\theta \right)=\cos \theta$ ,

Substitute $\theta =45°$ in both the properties $\sin \left(360°-45°\right)=-\sin 45°$ , it gives $\sin \left(315°\right)=-\sin 45°$ and also $\cos \left(360°-45°\right)=\cos 45°$ , it gives $\cos \left(315°\right)=\cos 45°$ .

Now substitute $\frac{1}{\sqrt{2}}$ for $\sin 45°$ and $\frac{1}{\sqrt{2}}$ for $\cos 45°$ in the above equation it gives,

  $\sin \left(315°\right)=-\frac{1}{\sqrt{2}}$ and $\cos \left(315°\right)=\frac{1}{\sqrt{2}}$ .

Now substitute the above values in the formula

  $\begin{array}{l}\left(fg\right)\left(315°\right)=\sin \left(315°\right)\cos \left(315°\right)\\ =-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\\ =-\frac{1}{2}\end{array}$ .

Therefore the value $\left(fg\right)\left(315°\right)$ for $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ is $-\frac{1}{2}$ .

(e)

To determine

To calculate: The exact value of function $f\left(2\theta \right)$ for $\theta =315°$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 111E

The value $f\left(2\theta \right)$ for $\theta =315°$ is $-1$ .

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 $\theta =315°$ , so $2\theta =630°$ .

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

  $f\left(630°\right)=\sin 630°$ ,

As $\sin \left(360°+\theta \right)=\sin \theta$ , substitute $\theta =270°$ in the equation it gives $\sin \left(360°+270°\right)=\sin 270°$ , it gives $\sin 630°=\sin 270°$ .

Now $\sin \left(180°+\theta \right)=-\sin \theta$ , substitute $\theta =90°$ in the equation it gives $\sin \left(270°\right)=-\sin 90°$ .

Now substitute $1$ for $\sin 90°$ it gives $\sin \left(630°\right)=-1$

Therefore value $f\left(2\theta \right)$ for $\theta =315°$ is $-1$ .

(f)

To determine

To calculate: The exact value of function $g\left(-\theta \right)$ for $\theta =315°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 111E

The value $g\left(-315°\right)$ is $\frac{1}{\sqrt{2}}$ .

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 $\theta =315°$ in the equation it gives $\cos \left(-315°\right)=\cos \left(315°\right)$ .

As per trigonometric identity $\cos \left(360°-\theta \right)=\cos \theta$ , where $\theta$ is any acute angle.

Now substitute $\theta =45°$ in the above equation it gives $\cos \left(360°-45°\right)=\cos 45°$ , which gives $\cos \left(315°\right)=\cos 45°$

Substitute $\frac{1}{\sqrt{2}}$ for $\cos \left(45°\right)$ it gives

  $\cos \left(315°\right)=\frac{1}{\sqrt{2}}$ .

Therefore the value of $\cos \left(-\theta \right)$ for $\theta =315°$ is $\frac{1}{\sqrt{2}}$ .