Chapter 4.4, Problem 116E

(a)

To determine

To calculate: The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =\frac{5\pi }{6}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 116E

The value $\left(f+g\right)\left(\frac{5\pi }{6}\right)$ is $\frac{1-\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

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 $\sin \left(\pi -\theta \right)=\sin \theta$ and $\cos \left(\pi -\theta \right)=-\cos \theta$ .

Calculation:

Substitute the values of $\sin \theta$ 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 $\frac{5\pi }{6}$ for $\theta$ in the above equation.

  $\left(f+g\right)\left(\frac{5\pi }{6}\right)=\sin \frac{5\pi }{6}+\cos \frac{5\pi }{6}$

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

Substitute $\frac{\pi }{6}$ for $\theta$ in both the properties

  $\sin \left(\pi -\frac{\pi }{6}\right)=\sin \frac{\pi }{6}$

  $\sin \left(\frac{5\pi }{6}\right)=\sin \frac{\pi }{6}$

And,

  $\cos \left(\frac{5\pi }{6}\right)=-\cos \frac{\pi }{6}$

Now substitute $\frac{1}{2}$ for $\sin \frac{\pi }{6}$ and $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ in the above equation it gives,

  $\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2}$ and $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the values in the formula.

  $\begin{array}{c}\left(f+g\right)\left(\frac{5\pi }{6}\right)=\sin \frac{5\pi }{6}+\cos \frac{5\pi }{6}\\ =\frac{1}{2}-\frac{\sqrt{3}}{2}\\ =\frac{1-\sqrt{3}}{2}\end{array}$

Therefore the value $\left(f+g\right)\left(\frac{5\pi }{6}\right)$ is $\frac{1-\sqrt{3}}{2}$ .

(b)

To determine

To calculate: The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =\frac{5\pi }{6}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 116E

The value $\left(g-f\right)\left(\frac{5\pi }{6}\right)$ is $-\frac{1+\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

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 $\theta$ .

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

Calculation:

Substitute the values of $\sin \theta$ 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 $\frac{5\pi }{6}$ for $\theta$ in the above equation.

  $\left(g-f\right)\left(\frac{5\pi }{6}\right)=\cos \frac{5\pi }{6}-\sin \frac{5\pi }{6}$ .

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

Substitute $\frac{\pi }{6}$ for $\theta$ in both the properties

  $\sin \left(\pi -\frac{\pi }{6}\right)=\sin \frac{\pi }{6}$

  $\sin \left(\frac{5\pi }{6}\right)=\sin \frac{\pi }{6}$

And,

  $\cos \left(\frac{5\pi }{6}\right)=-\cos \frac{\pi }{6}$ .

Now substitute $\frac{1}{2}$ for $\sin \frac{\pi }{6}$ and $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ in the above equation it gives,

  $\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2}$ and $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the values in the formula.

  $\begin{array}{c}\left(g-f\right)\left(\frac{5\pi }{6}\right)=\cos \frac{5\pi }{6}-\sin \frac{5\pi }{6}\\ =-\frac{\sqrt{3}}{2}-\frac{1}{2}\\ =-\frac{1+\sqrt{3}}{2}\end{array}$

Therefore the value $\left(g-f\right)\left(\frac{5\pi }{6}\right)$ is $-\frac{1+\sqrt{3}}{2}$ .

(c)

To determine

To calculate: The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =\frac{5\pi }{6}$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 116E

The value ${\left[g\left(\frac{5\pi }{6}\right)\right]}^2$ is $\frac{3}{4}$ .

Explanation of Solution

Given:

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

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 $\theta$ .

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

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 $\frac{5\pi }{6}$ for $\theta$ in the above equation.

  ${\left[g\left(\frac{5\pi }{6}\right)\right]}^2={\left(\cos \frac{5\pi }{6}\right)}^2$ .

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

Substitute $\frac{\pi }{6}$ for $\theta$ in the above property

  $\cos \left(\pi -\frac{\pi }{6}\right)=-\cos \frac{\pi }{6}$

  $\cos \left(\frac{5\pi }{6}\right)=-\cos \frac{\pi }{6}$

Now substitute $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ in the above equation.

  $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$

Now substitute $-\frac{\sqrt{3}}{2}$ for $\cos \left(\frac{5\pi }{6}\right)$ in the above identity.

  $\begin{array}{c}{\left[g\left(\frac{5\pi }{6}\right)\right]}^2={\left(\cos \frac{5\pi }{6}\right)}^2\\ ={\left(-\frac{\sqrt{3}}{2}\right)}^2\\ =\frac{3}{4}\end{array}$

Therefore the value ${\left[g\left(\frac{5\pi }{6}\right)\right]}^2$ is $\frac{3}{4}$ .

(d)

To determine

To calculate: The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =\frac{5\pi }{6}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 116E

The value $\left(fg\right)\left(\frac{5\pi }{6}\right)$ is $-\frac{\sqrt{3}}{4}$ .

Explanation of Solution

Given:

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

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 $\sin \left(\pi -\theta \right)=\sin \theta$ and $\cos \left(\pi -\theta \right)=-\cos \theta$ .

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 $\frac{5\pi }{6}$ for $\theta$ in the above equation.

  $\left(fg\right)\left(\frac{5\pi }{6}\right)=\sin \left(\frac{5\pi }{6}\right)\cos \left(\frac{5\pi }{6}\right)$ .

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

Substitute $\frac{\pi }{6}$ for $\theta$ in both the properties

  $\sin \left(\pi -\frac{\pi }{6}\right)=\sin \frac{\pi }{6}$

  $\sin \left(\frac{5\pi }{6}\right)=\sin \frac{\pi }{6}$

And,

  $\cos \left(\frac{5\pi }{6}\right)=-\cos \frac{\pi }{6}$

Now substitute $\frac{1}{2}$ for $\sin \frac{\pi }{6}$ and $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ in the above equation it gives,

  $\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2}$ and $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$ .

Now substitute the above values in the formula

  $\begin{array}{c}\left(fg\right)\left(\frac{5\pi }{6}\right)=\sin \left(\frac{5\pi }{6}\right)\cos \left(\frac{5\pi }{6}\right)\\ =\frac{1}{2}\times -\frac{\sqrt{3}}{2}\\ =-\frac{\sqrt{3}}{4}\end{array}$ .

Therefore the value $\left(fg\right)\left(\frac{5\pi }{6}\right)$ 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 $\theta =\frac{5\pi }{6}$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 116E

The value $f\left(2\theta \right)$ for $\theta =\frac{5\pi }{6}$ is $-\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

Formula Used:

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

Use the property $\sin \left(2\pi -\theta \right)=-\sin \theta$ .

Calculation:

As $\theta =\frac{5\pi }{6}$ , so $2\theta =\frac{5\pi }{3}$ .

Substitute the value $\frac{5\pi }{3}$ for $2\theta$ in $f\left(2\theta \right)=\sin \left(2\theta \right)$

  $f\left(\frac{5\pi }{3}\right)=\sin \frac{5\pi }{3}$ ,

By the trigonometric property $\sin \left(2\pi -\theta \right)=-\sin \theta$ ,

Substitute $\frac{\pi }{3}$ for $\theta$ in both the properties

  $\sin \left(2\pi -\frac{\pi }{3}\right)=-\sin \frac{\pi }{3}$

  $\sin \left(\frac{5\pi }{3}\right)=-\sin \frac{\pi }{3}$

Now substitute $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$ in the above equation it gives,

  $\sin \left(\frac{5\pi }{3}\right)=-\frac{\sqrt{3}}{2}$ .

Therefore value $f\left(2\theta \right)$ for $\theta =\frac{5\pi }{6}$ is $-\frac{\sqrt{3}}{2}$ .

(f)

To determine

To calculate: The exact value of function $g\left(-\theta \right)$ for $\theta =\frac{5\pi }{6}$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 116E

The value $g\left(-\frac{5\pi }{6}\right)$ is $-\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given:

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

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 $\theta$ .

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

Calculation:

As by the property of even function $\cos \left(-\theta \right)=\cos \theta$ .

Substitute $\frac{5\pi }{6}$ for $\theta$ in the equation.

  $\cos \left(-\frac{5\pi }{6}\right)=\cos \left(\frac{5\pi }{6}\right)$

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

Substitute $\frac{\pi }{6}$ for $\theta$ in the above property $\cos \left(\pi -\frac{\pi }{6}\right)=-\cos \frac{\pi }{6}$

  $\cos \left(\frac{5\pi }{6}\right)=-\cos \frac{\pi }{6}$

Now substitute $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ in the above equation.

  $\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$

Substitute $-\frac{\sqrt{3}}{2}$ for $\cos \left(\frac{5\pi }{6}\right)$ in above equation.

  $\cos \left(-\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}$

Therefore the value of $\cos \left(-\frac{5\pi }{6}\right)$ is $-\frac{\sqrt{3}}{2}$ .