Chapter 5, Problem 16PS

a.

To determine

Use the power-reducing formulas to write the function in terms of cosine to the first power.

Answer to Problem 16PS

  $\boxed{f\left(x\right)=\frac{3+\cos 4x}{4}}$

Explanation of Solution

Given information: Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Calculation:

Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Use the following trigonometric identities

  $\begin{array}{l}1+\cos 2\theta =2{\cos }^2\theta \\ 1-\cos 2\theta =2{\sin }^2\theta \end{array}$

Above expression is reduced to

  $\begin{array}{l}={\left(\frac{1-\cos 2x}{2}\right)}^2+{\left(\frac{1+\cos 2x}{2}\right)}^2\\ =\frac{1+{\cos }^{2}2x-2\cos 2x+1+{\cos }^{2}2x+2\cos 2x}{4}\\ =\frac{2+2{\cos }^{2}2x}{4}\\ \end{array}$

Or further simplify,

  $\begin{array}{l}f\left(x\right)=\frac{1+\frac{1+\cos 4x}{2}}{2}\\ f\left(x\right)=\frac{3+\cos 4x}{4}\end{array}$

Use TI-83 calculator with the following key steps, to draw the graph of expression

  $f\left(x\right)=\frac{3}{4}+\cos \frac{4x}{4}$

Enter $y_1=\frac{3}{4}+\cos \frac{4x}{4}$ and press the enter button.

  

Press $WINDOW$ button to set the range of window as shown below.

  

Press $GRAPH$ button, the diplay is shown below:

  

Hence, the function in terms of cosine is $\boxed{f\left(x\right)=\frac{3+\cos 4x}{4}}$

b.

To determine

Determine another way of rewriting the original function. Use a graphing utility to rule out incorrectly rewritten functions.

Answer to Problem 16PS

  $\boxed{f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}}$

Explanation of Solution

Given information:

Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Calculation:

Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

  $\begin{array}{l}=\left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\right)-2{\sin }^{2}x{\cos }^{2}x\\ ={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\end{array}$

Use the following trigonometric identity

  ${\sin }^{2}x+{\cos }^{2}x=1$

Above expression is reduced to

  $f\left(x\right)=1^2-\frac{4{\sin }^{2}x{\cos }^{2}x}{2}$

Or further simplify,

  $\boxed{f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}}$

Use TI-83 calculator with the following key steps, to draw the graph of expression

  $f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}$

Enter $y_1=1-\frac{{\sin }^{2}2x}{2}$ and press the enter button.

  

Press $WINDOW$ button to set the range of window as shown below.

  

Press $GRAPH$ button, the diplay is shown below:

  

Hence, the graph in part(a) is same as part(b).

c.

To determine

Add a trigonometric term to the original function so that it becomes a perfect square trinomial. Rewrite the function as a perfect square trinomial minus the term that you added. Use the graphing utility to rule out incorrectly rewritten functions.

Answer to Problem 16PS

  $\boxed{f\left(x\right)={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}$

Explanation of Solution

Given information: Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Calculation: Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

  $\begin{array}{l}=\left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\right)-2{\sin }^{2}x{\cos }^{2}x\\ ={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x\end{array}$

Use TI-83 calculator with the following key steps, to draw the graph of expression

  $f\left(x\right)$

  $={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x$

Enter $y_1={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x$ and press the enter button.

  

Press $WINDOW$ button to set the range of window as shown below.

  

Press $GRAPH$ button, the diplay is shown below:

  

Hence, the function as a perfect square trinomial minus the term is

  $\boxed{f\left(x\right)={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}$ .

d.

To determine

Rewrite the result of part (c) in terms of the $\sin$ of a double angle. Use the graphing utility to rule out incorrectly rewritten functions.

Answer to Problem 16PS

  $\boxed{f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}}$

Explanation of Solution

Given information: Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Calculation:

Consider the function.

  $f\left(x\right)={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x$

Use the following trigonometric identity

  ${\sin }^{2}x+{\cos }^{2}x=1$

Above expression is reduced to

  $f\left(x\right)=1^2-\frac{4{\sin }^{2}x{\cos }^{2}x}{2}$

Or further simplify,

  $\boxed{f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}}$

Hence, the function in terms of the $\sin$ of a double angle is $\boxed{f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}}$ .

e.

To determine

When you rewrite a trigonometric expression, the result may not be the same as a friend’s. Does this mean that one of you is wrong? Explain.

Answer to Problem 16PS

one can write a trigonometric in different ways.

Explanation of Solution

Given information: Consider the function.

  $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Calculation:

A trigonometric expression can be written in many different ways; the point to note is that all these expressions are equivalent.

For example, above expression was simplified in two different ways to yield two different yet equivalent results.

Hence, one can write a trigonometric in different ways.