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Chapter 4: Problem 71

Graph the functions \(f\) and \(g .\) Use the graphs to make a conjecture about the relationship between the functions. $$f(x)=\sin ^{2} x, \quad g(x)=\frac{1}{2}(1-\cos 2 x)$$

Short Answer

Expert verified
The graphs of the functions \(f(x)\) and \(g(x)\) are identical, implying that for all values of \(x\), \(f(x) = g(x)\).

Step by step solution

01

Graph the function \(f(x)=\sin ^{2} x\)

To graph the function \(f(x) = \sin^2 x\), start by graphing the function \(y = \sin x\). Then create a second identical graph and square every y-coordinate value. Do this for several values of \(x\) from \(-\pi\) to \(\pi\) at regular intervals. Plot these points on a coordinate grid to visualize the function \(f(x)\).
02

Graph the function \(g(x)=\frac{1}{2}(1-\cos 2x)\)

To graph the function \(g(x) = \frac{1}{2}(1 - \cos 2x)\), start by graphing the function \(y = \cos 2x\). Make sure to keep in mind that the period of this function is \(\pi\) instead of \(2\pi\). After graphing \(y = \cos 2x\), subtract each y-coordinate from 1 and divide by 2 for a number of points, then plot these points on the same coordinate grid as \(f(x)\).
03

Conjecture Relationship between \(f(x)\) and \(g(x)\)

Notice the points and shape of the graphs of \(f(x)\) and \(g(x)\). It appears that the two graphs are identical, suggesting that for all values of \(x\), \(f(x) = g(x)\). This can in fact be proven using the double-angle identity for cosine, which states that \( \cos 2x = 1 - 2\sin^2x \). Thus, \( \frac{1}{2}(1 - \cos 2x) = \sin^2x \).

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Most popular questions from this chapter

Use a graphing utility to graph the function. $$f(x)=\frac{\pi}{2}+\cos ^{-1}\left(\frac{1}{\pi}\right)$$

Use a graphing utility to graph the functions \(f(x)=\sqrt{x}\) and \(g(x)=6\) arctan \(x .\) For \(x>0,\) it appears that \(g>f .\) Explain why you know that there exists a positive real number \(a\) such that \(ga .\) Approximate the number \(a\).

Write the function in terms of the sine function by using the identity $$A \cos \omega t+B \sin \omega t=\sqrt{A^{2}+B^{2}} \sin \left(\omega t+\arctan \frac{A}{B}\right).$$ Use a graphing utility to graph both forms of the function. What does the graph imply? $$f(t)=3 \cos 2 t+3 \sin 2 t$$

Is a degree or a radian the greater unit of measure? Explain.

Sketch a graph of the function and compare the graph of \(g\) with the graph of \(f(x)=\arcsin x\). $$g(x)=\arcsin \frac{x}{2}$$
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