Chapter 4.4, Problem 113E

(a)

To determine

To calculate: The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =-150°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 113E

The value $\left(f+g\right)\left(-150°\right)$ 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 $\theta =-150°$ in the above equation it gives

  $\left(f+g\right)\left(-150°\right)=\sin \left(-150°\right)+\cos \left(-150°\right)$ .

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore $\sin \left(-150°\right)=-\sin 150°$ and $\cos \left(-150°\right)=\cos 150°$ .

Therefore, $\left(f+g\right)\left(-150°\right)=-\sin \left(150°\right)+\cos \left(150°\right)$

By the trigonometric property $\sin \left(180°-\theta \right)=\sin \theta$ and $\cos \left(180°-\theta \right)=-\cos \theta$ ,

Substitute $\theta =30°$ in both the properties $\sin \left(180°-30°\right)=\sin 30°$ , it gives $\sin \left(150°\right)=\sin 30°$ and also $\cos \left(180°-30°\right)=-\cos 30°$ , it gives $\cos \left(150°\right)=-\cos 30°$ .

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

  $\sin \left(150°\right)=\frac{1}{2}$ and $\cos \left(150°\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the values in the formula $\begin{array}{c}\left(f+g\right)\left(-150°\right)=\sin \left(-150°\right)+\cos \left(-150°\right)\\ =-\frac{1}{2}+\frac{\sqrt{3}}{2}\\ =\frac{-1+\sqrt{3}}{2}\end{array}$

Therefore the value $\left(f+g\right)\left(-150°\right)$ is $\frac{-1+\sqrt{3}}{2}$ .

(b)

To determine

To calculate: The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =-150°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 113E

The value $\left(g-f\right)\left(-150°\right)$ 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 $\left(g-f\right)\left(\theta \right)=g\left(\theta \right)-f\left(\theta \right)$ , 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 $\theta =-150°$ in the above equation it gives

  $\left(g-f\right)\left(-150°\right)=\cos \left(-150°\right)-\sin \left(-150°\right)$ .

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore $\sin \left(-150°\right)=-\sin 150°$ and $\cos \left(-150°\right)=\cos 150°$ .

Therefore, $\left(g-f\right)\left(-150°\right)=\cos \left(150°\right)+\sin \left(150°\right)$ .

By the trigonometric property $\sin \left(180°-\theta \right)=\sin \theta$ and $\cos \left(180°-\theta \right)=-\cos \theta$ ,

Substitute $\theta =30°$ in both the properties $\sin \left(180°-30°\right)=\sin 30°$ , it gives $\sin \left(150°\right)=\sin 30°$ and also $\cos \left(180°-30°\right)=-\cos 30°$ , it gives $\cos \left(150°\right)=-\cos 30°$ .

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

  $\sin \left(150°\right)=\frac{1}{2}$ and $\cos \left(150°\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the values in the formula $\begin{array}{c}\left(g-f\right)\left(-150°\right)=\cos \left(-150°\right)-\sin \left(-150°\right)\\ =-\frac{\sqrt{3}}{2}+\frac{1}{2}\\ =\frac{1-\sqrt{3}}{2}\end{array}$

Therefore the value $\left(g-f\right)\left(-150°\right)$ is $\frac{1-\sqrt{3}}{2}$ .

(c)

To determine

To calculate: The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =-150°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 113E

The value ${\left[g\left(-150°\right)\right]}^2$ 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 $\theta =-150°$ in the above equation it gives

  ${\left[g\left(-150°\right)\right]}^2={\left(\cos \left(-150°\right)\right)}^2$ .

By the trigonometric identity $\cos \left(-\theta \right)=\cos \theta$ , therefore $\cos \left(-150°\right)=\cos 150°$ .

Therefore, ${\left[g\left(-150°\right)\right]}^2={\left(\cos \left(150°\right)\right)}^2$ .

By the trigonometric property $\cos \left(180°-\theta \right)=-\cos \theta$ ,

Substitute $\theta =30°$ in both the property $\cos \left(180°-30°\right)=-\cos 30°$ , it gives $\cos \left(150°\right)=-\cos 30°$ .

Now substitute $\frac{\sqrt{3}}{2}$ for $\cos 30°$ in the above equation it gives $\cos \left(150°\right)=-\frac{\sqrt{3}}{2}$ .

  $\begin{array}{c}{\left[g\left(-150°\right)\right]}^2={\left(\cos 150°\right)}^2\\ ={\left(-\frac{\sqrt{3}}{2}\right)}^2\\ =\frac{3}{4}\end{array}$ .

Therefore the value ${\left[g\left(-150°\right)\right]}^2$ is $\frac{3}{4}$ .

(d)

To determine

To calculate: The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =-150°$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 113E

The value $\left(fg\right)\left(-150°\right)$ 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 $\theta =-150°$ in the above equation it gives

  $\left(fg\right)\left(-150°\right)=\sin \left(-150°\right)\cos \left(-150°\right)$ .

By the trigonometric identity $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ , therefore $\sin \left(-150°\right)=-\sin 150°$ and $\cos \left(-150°\right)=\cos 150°$ .

Therefore, $\left(fg\right)\left(-150°\right)=-\sin \left(150°\right)\cos \left(150°\right)$ .

By the trigonometric property $\sin \left(180°-\theta \right)=\sin \theta$ and $\cos \left(180°-\theta \right)=-\cos \theta$ ,

Substitute $\theta =30°$ in both the properties $\sin \left(180°-30°\right)=\sin 30°$ , it gives $\sin \left(150°\right)=\sin 30°$ and also $\cos \left(180°-30°\right)=-\cos 30°$ , it gives $\cos \left(150°\right)=-\cos 30°$ .

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

  $\sin \left(150°\right)=\frac{1}{2}$ and $\cos \left(150°\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the above values in the formula

  $\begin{array}{l}\left(fg\right)\left(-150°\right)=-\sin \left(150°\right)\cos \left(150°\right)\\ =-\frac{1}{2}\times -\frac{\sqrt{3}}{2}\\ =\frac{\sqrt{3}}{4}\end{array}$ .

Therefore the value $\left(fg\right)\left(-150°\right)$ 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 $\theta =-150°$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 113E

The value $f\left(2\theta \right)$ for $\theta =-150°$ 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 $\theta =-150°$ , so $2\theta =-300°$ .

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

  $f\left(-300°\right)=\sin \left(-300°\right)$ ,

By the trigonometric property $\sin \left(-\theta \right)=-\sin \theta$ , therefore $\sin \left(-300°\right)=-\sin 300°$ .

Therefore, $f\left(-300°\right)=-\sin \left(300°\right)$

As $\sin \left(360°-\theta \right)=-\sin \theta$ , substitute $\theta =60°$ in the equation it gives $\sin \left(360°-60°\right)=-\sin 60°$ , it gives $\sin 300°=-\sin 60°$ .

Now substitute $\frac{\sqrt{3}}{2}$ for $\sin 60°$ it gives $\sin \left(300°\right)=-\frac{\sqrt{3}}{2}$

Therefore value $f\left(2\theta \right)$ for $\theta =-150°$ is $\frac{\sqrt{3}}{2}$ .

(f)

To determine

To calculate: The exact value of function $g\left(-\theta \right)$ for $\theta =-150°$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 113E

The value $g\left(150°\right)$ 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:

Substitute $\theta =-150°$ in the equation it gives $\begin{array}{c}g\left(-\theta \right)=g\left(150°\right)\\ =\cos \left(150°\right)\end{array}$ .

As per trigonometric identity $\cos \left(180°-\theta \right)=-\cos \theta$ , where $\theta$ is any acute angle.

Now substitute $\theta =30°$ in the above equation it gives $\cos \left(180°-30°\right)=-\cos 30°$ , which gives $\cos \left(150°\right)=-\cos 30°$

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

  $\cos \left(150°\right)=-\frac{\sqrt{3}}{2}$ .

Therefore the value of $\cos \left(-\theta \right)$ for $\theta =-150°$ is $-\frac{\sqrt{3}}{2}$ .