Chapter 4.4, Problem 118E

(a)

To determine

To calculate:The exact value of function $\left(f+g\right)\left(\theta \right)$ for $\theta =-\frac{5\pi }{3}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 118E

The value $\left(f+g\right)\left(-\frac{5\pi }{3}\right)$ 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 $\sin \left(-\theta \right)=-\sin \theta$ and $\cos \left(-\theta \right)=\cos \theta$ .

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 $-\frac{5\pi }{3}$ for $\theta$ in the above equation.

  $\left(f+g\right)\left(-\frac{5\pi }{3}\right)=\sin \left(-\frac{5\pi }{3}\right)+\cos \left(-\frac{5\pi }{3}\right)$

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

Substitute $\frac{5\pi }{3}$ for $\theta$ in above identities.

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

And,

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

Therefore,

  $\left(f+g\right)\left(-\frac{5\pi }{3}\right)=-\sin \left(\frac{5\pi }{3}\right)+\cos \left(\frac{5\pi }{3}\right)$

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

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

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

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

And,

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

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

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

Now substitute the values in the formula.

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

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

(b)

To determine

To calculate:The exact value of function $\left(g-f\right)\left(\theta \right)$ for $\theta =-\frac{5\pi }{3}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 118E

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

Explanation of Solution

Given:

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

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

Use another property $\sin \left(2\pi -\theta \right)=-\sin \theta$ and $\cos \left(2\pi -\theta \right)=\cos \theta$ .

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 $-\frac{5\pi }{3}$ for $\theta$ in the above equation.

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

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

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

Substitute the values in above equation.

  $\left(g-f\right)\left(-\frac{5\pi }{3}\right)=\cos \left(\frac{5\pi }{3}\right)+\sin \left(\frac{5\pi }{3}\right)$

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

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

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

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

And,

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

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

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

Now substitute the values in the formula.

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

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

(c)

To determine

To calculate:The exact value of function ${\left[g\left(\theta \right)\right]}^2$ for $\theta =-\frac{5\pi }{3}$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 118E

The value ${\left[g\left(-\frac{5\pi }{3}\right)\right]}^2$ is $\frac{1}{4}$ .

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 the property $\cos \left(-\theta \right)=\cos \theta$ and $\cos \left(2\pi -\theta \right)=\cos \theta$ .

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 $-\frac{5\pi }{3}$ for $\theta$ in the above equation.

  ${\left[g\left(-\frac{5\pi }{3}\right)\right]}^2={\left(\cos \left(-\frac{5\pi }{3}\right)\right)}^2$

By the trigonometric identity $\cos \left(-\theta \right)=\cos \theta$ , therefore $\cos \left(-\frac{5\pi }{3}\right)=\cos \frac{5\pi }{3}$ .

Therefore,

  ${\left[g\left(-\frac{5\pi }{3}\right)\right]}^2={\left(\cos \left(\frac{5\pi }{3}\right)\right)}^2$

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

Substitute $\frac{\pi }{3}$ for $\theta$ in the property $\cos \left(2\pi -\frac{\pi }{3}\right)=\cos \frac{\pi }{3}$

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

Now substitute $\frac{1}{2}$ for $\cos \frac{\pi }{3}$ in the above equation.

  $\cos \left(\frac{5\pi }{3}\right)=\frac{1}{2}$

Now substitute $\frac{1}{2}$ for $\cos \left(\frac{5\pi }{3}\right)$ in the above identity.

  $\begin{array}{c}{\left[g\left(-\frac{5\pi }{3}\right)\right]}^2={\left(\cos \frac{5\pi }{3}\right)}^2\\ ={\left(\frac{1}{2}\right)}^2\\ =\frac{1}{4}\end{array}$

Therefore the value ${\left[g\left(-\frac{5\pi }{3}\right)\right]}^2$ is $\frac{1}{4}$ .

(d)

To determine

To calculate:The exact value of function $\left(fg\right)\left(\theta \right)$ for $\theta =-\frac{5\pi }{3}$ , where $f\left(\theta \right)=\sin \theta$ and $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 118E

The value $\left(fg\right)\left(-\frac{5\pi }{3}\right)$ 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 another property $\sin \left(2\pi -\theta \right)=-\sin \theta$ and $\cos \left(2\pi -\theta \right)=\cos \theta$ .

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.

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

Substitute $-\frac{5\pi }{3}$ for $\theta$ in the above equation it gives.

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

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

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

Therefore,

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

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

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

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

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

And,

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

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

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

Now substitute the above values in the formula.

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

Therefore the value $\left(fg\right)\left(-\frac{5\pi }{3}\right)$ 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 $\theta =-\frac{5\pi }{3}$ , where $f\left(\theta \right)=\sin \theta$ .

Answer to Problem 118E

The value $f\left(2\theta \right)$ for $\theta =-\frac{5\pi }{3}$ 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 identity $\sin \left(-\theta \right)=-\sin \theta$ and $\sin \left(2\pi +\theta \right)=\sin \theta$ .

Calculation:

As $\theta =-\frac{5\pi }{3}$ , so $2\theta =-\frac{10\pi }{3}$ .

Substitute the value $-\frac{10\pi }{3}$ for $2\theta$ in $f\left(2\theta \right)=\sin \left(2\theta \right)$

  $f\left(-\frac{10\pi }{3}\right)=\sin \left(-\frac{10\pi }{3}\right)$

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

  $\sin \left(-\frac{10\pi }{3}\right)=-\sin \frac{10\pi }{3}$

Therefore,

  $f\left(-\frac{10\pi }{3}\right)=-\sin \left(\frac{10\pi }{3}\right)$

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

Substitute $\frac{4\pi }{3}$ for $\theta$ in the property

  $\sin \left(2\pi +\frac{4\pi }{3}\right)=\sin \frac{4\pi }{3}$

  $\sin \left(\frac{10\pi }{3}\right)=\sin \frac{4\pi }{3}$

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

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

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

  $\sin \left(\frac{4\pi }{3}\right)=-\sin \frac{\pi }{3}$

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

  $\sin \left(\frac{10\pi }{3}\right)=-\frac{\sqrt{3}}{2}$

Therefore,

  $f\left(-\frac{10\pi }{3}\right)=-\left(-\frac{\sqrt{3}}{2}\right)$

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

(f)

To determine

To calculate:The exact value of function $g\left(-\theta \right)$ for $\theta =-\frac{5\pi }{3}$ , where $g\left(\theta \right)=\cos \theta$ .

Answer to Problem 118E

The value $g\left(\frac{5\pi }{3}\right)$ is $\frac{1}{4}$ .

Explanation of Solution

Given:

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

Formula Used:

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

Use another property $\cos \left(2\pi -\theta \right)=\cos \theta$ .

Calculation:

As $\theta =-\frac{5\pi }{3}$ , then $-\theta =\frac{5\pi }{3}$ .

Therefore,

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

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

Substitute $\frac{\pi }{3}$ for $\theta$ in the property $\cos \left(2\pi -\frac{\pi }{3}\right)=\cos \frac{\pi }{3}$

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

Now substitute $\frac{1}{2}$ for $\cos \frac{\pi }{3}$ in the above equation.

  $\cos \left(\frac{5\pi }{3}\right)=\frac{1}{2}$

Now substitute $\frac{1}{2}$ for $\cos \left(\frac{5\pi }{3}\right)$ in the above identity.

  $\begin{array}{c}{\left[g\left(\frac{5\pi }{3}\right)\right]}^2={\left(\cos \frac{5\pi }{3}\right)}^2\\ ={\left(\frac{1}{2}\right)}^2\\ =\frac{1}{4}\end{array}$

Therefore the value of $\cos \left(\frac{5\pi }{3}\right)$ is $\frac{1}{4}$ .