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This trig question asks me to find an expression that describes the location of each of the following values for $y = \cos x$ where $n\in\mathbb{Z}$ and $x$ is in radians.

The question then asks for
a) $x$-intercepts b) maximum values c) minimum values

The answers to the questions

  • a) tn $= \pi/2 + n\pi$, $n\in\mathbb{Z}$
  • b) tn $= 2n\pi$, $n\in\mathbb{Z}$
  • c) tn $= -\pi + 2n\pi$, $n\in\mathbb{Z}$

The problem with this question is I don't really understand how you can use an expression to describe those values, so I want to know how they got those answers.

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up vote 2 down vote accepted

An $x$-intercept occurs with $\cos (x)=0$ (that is, when the function intersects the $x$-axis). To find all the solutions to this equation, think of the unit circle (see the picture below). enter image description here

From this picture, we see that $\cos (x)=0$ exactly when $x=\pi /2$ or $x=-\pi /2$ (for $x\in (-\pi /2,\pi /2]$). However, $\cos$ is $2\pi$-periodic, so $\pi /2+2\pi m$ and $-\pi /2+2\pi n$ for $m,n\in \mathbb{Z}$ will also be solutions, and in fact give us all the solutions. It is easy to see that every solution of one of these two forms can be written in the form $\pi /2+\pi n$ for $n\in \mathbb{Z}$.

We can similarly use this picture to determine where the function $\cos$ has maxima and minima. We see that $\cos$ has a single maximum (in the interval $(-\pi /2,\pi /2]$) at the value $x=0$, at which $\cos$ takes the value $1$. Using $2\pi$ periodicity again, the set of values at which $\cos$ achieves a maximum is of the form $2\pi n$ for $n\in \mathbb{Z}$. Similarly, in this interval, $\cos$ has a single minimum at $-\pi$, and hence a general minimum will be of the form $-\pi +2\pi n$ for $n\in \mathbb{Z}$.

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I had to give this answer a "+1" because it uses an illustration that I created myself and uploaded to Wikipedia. – Michael Hardy Nov 19 '11 at 20:09

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