Chapter 5.2, Problem 46E

To determine

To explain : Why there is a zero of $y=\cos x$ between every two zeroes of $y=\sin x$ .

Answer to Problem 46E

The possible explanation is given below.

Explanation of Solution

Given information :

The two function $y=\cos x$ and $y=\sin x$ .

Calculation :

The Mean Value Theorem states that if $f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there exists at least one $c$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ .

Here, let $f\left(x\right)=\sin x$ , for the zeroes the $f\left(x\right)$ set $f\left(x\right)=0$ and solve:

  $\begin{array}{l}f\left(x\right)=\sin x\\ f\left(x\right)=0\\ \sin x=0\\ x=n\pi \text{ where is }n\text{ interger}\end{array}$

Therefore, consecutive zeroes will occur at $n\pi$ and $\left(n+1\right)\pi$ .

Now, since the function $f\left(x\right)=\sin x$ is continuous on the interval $\left[n\pi ,\left(n+1\right)\pi \right]$ and and differentiable on the interval $\left(n\pi ,\left(n+1\right)\pi \right)$ . Thus, according to mean value theorem there exists $c$ such that

  $f\text{'}\left(c\right)=\frac{f\left[\left(n+1\right)\pi \right]-f\left(n\pi \right)}{\left(n+1\right)\pi -n\pi }$ ...... (1)

Differentiate $f\left(x\right)=\sin x$ , then replace $x$ with $c$ :

  $\begin{array}{l}f\text{'}\left(x\right)=\cos x\\ f\text{'}\left(c\right)=\cos c\end{array}$

Substitute this value in equation (1)

  $\begin{array}{l}\cos c=\frac{0-0}{1}\\ \cos c=0\end{array}$

For some $c$ in between $\left[n\pi ,\left(n+1\right)\pi \right]$ .

Therefore,

  $y=\cos x$ will be zero for some $c$ in between two zeroes of $y=\sin x$ .