Chapter 5.5, Problem 39E

To determine

To write:The expression ${\sin }^{2}x{\cos }^{4}x$ in first power of cosine and verify both the expressions by graphing utility.

Answer to Problem 39E

The expression ${\sin }^{2}x{\cos }^{4}x$ can be written as $\frac{1}{16}\left(1+\cos 2x-\cos 4x-\cos 2x\cos 4x\right)$ as a single power of cosine.

Explanation of Solution

Given information:

The expression is given as ${\sin }^{2}x{\cos }^{4}x$ .

Formula used:

Use the power reducing formula

  ${\cos }^{2}x=\frac{1+\cos 2x}{2}$ and ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ .

Calculation:

As per power reducing formula.

  ${\cos }^{2}x=\frac{1+\cos 2x}{2}$ and ${\sin }^{2}x=\frac{1-\cos 2x}{2}$

Substitute the value in the above expression.

  $\begin{array}{c}{\sin }^{2}x{\cos }^{4}x=\left(\frac{1-\cos 2x}{2}\right)\left(\frac{1+\cos 2x}{2}\right)\left(\frac{1+\cos 2x}{2}\right)\\ =\frac{1}{8}\left(1-{\cos }^{2}2x\right)\left(1+\cos 2x\right)\\ =\frac{1}{8}\left(1-\frac{1+\cos 4x}{2}\right)\left(1+\cos 2x\right)\end{array}$

Simplify further,

  $\begin{array}{c}{\sin }^{2}x{\cos }^{4}x=\frac{1}{8}\left(\frac{1-\cos 4x}{2}\right)\left(1+\cos 2x\right)\\ =\frac{1}{16}\left(1+\cos 2x-\cos 4x-\cos 4x\cos 2x\right)\end{array}$

Therefore the expression ${\sin }^{2}x{\cos }^{4}x$ can be written as $\frac{1}{16}\left(1+\cos 2x-\cos 4x-\cos 2x\cos 4x\right)$ as a single power of cosine.

Verify both the expressions using graphical utility:

Draw the graph for the expression ${\sin }^{2}x{\cos }^{4}x$ as shown in the Figure (1).

  

Figure (1)

Draw the graph for the expression $\frac{1}{16}\left(1+\cos 2x-\cos 4x-\cos 2x\cos 4x\right)$ as shown in the Figure (2).

  

Figure (2)

As both the expressions shows same values in Figure (1) and Figure (2) at each point.

Therefore both the expressions ${\sin }^{2}x{\cos }^{4}x$ and $\frac{1}{16}\left(1+\cos 2x-\cos 4x-\cos 2x\cos 4x\right)$ are same.