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When is $\cos (x) \geq \frac{1}{2}$?

I know the function repeats, so I know I should end up with an interval that allows for integer multiples.

e.g. something like this (but obviously not this exactly) $[0 + n, \pi + n]$.

user174606
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    Do you know when $\cos(x) = \frac{1}{2}$? – Benjamin Sep 08 '14 at 18:25
  • Draw a circle of radius $r=1$ and vertical line through $x=0.5$. Then draw the usual triangle used to define $\cos$ and arrive at some conclusion (later you can use the symmetry properties of $\cos$). – Avitus Sep 08 '14 at 18:29
  • $\cos(x) = \frac{1}{2}$ when $x = \frac{\pi}{3}$. – user174606 Sep 08 '14 at 18:35
  • @user174606 and when $x = -\frac{\pi}{3}$, if I'm not mistaken. Is that enough to see what intervals you are looking for? – Benjamin Sep 08 '14 at 18:44
  • Suppose that you have an equilateral triangle where the height has been penciled in. This creates 2 right angled triangles (one to the left of the height, and one to the right). If a side of the equilateral triangle is 2, what is the length of the base of one of the $half$ triangles? Actually draw the picture. – John Joy Sep 08 '14 at 20:37

4 Answers4

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First, we solve $\cos x=\frac12$, which is for $x=\frac{\pi}{3}$ and $x=\frac{5\pi}{3}$.

Then, we take a look at the unit circle.

enter image description here

The cosine value is the $x$-coordinate, so the question we ask is: "for which $x$ is the $x$-coordinate greater than or equal to $\frac12$?".

In the first quadrant, we have $0\leq x\leq\frac{\pi}{3}$ or $[0,\frac{\pi}{3}]$. In the fourth quadrant, we have $\frac{5\pi}{3}\leq x\leq2\pi$ or $[\frac{5\pi}{3},2\pi]$.

Now, we add the periodicity. We now that $\cos(x)=\cos(x+2\pi)$, so we can always add $+2\pi$ to our answer.

Therefore, the final answer is:

$\cos(x)\geq\frac12$ for $x\in[2\pi n,\frac{\pi}{3}+2\pi n]\cup[\frac{5\pi}{3}+2\pi n,2\pi+2\pi n]$ or, preferably for some, $x\in[-\frac{\pi}{3}+2\pi n, \frac{\pi}{3}+2\pi n]$, ($n\in\mathbb{Z}$)

rae306
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    This makes it look like there are two separate repeating intervals, but actually they come in pairs of subintervals that meet at a multiple of $2\pi.$ So this may be the most thorough and accessible explanation but I prefer the answers that come out to just one interval plus its repetitions. – David K Sep 08 '14 at 19:43
  • @David K, I also added the one interval notation. +1 now? ;-) – rae306 Sep 08 '14 at 20:03
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    Since the one interval was my only complaint I guess I have no alternative now, do I? :-) – David K Sep 08 '14 at 20:05
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$[-\frac{\pi}{3}+2\pi n, \frac{\pi}{3} + 2\pi n]$

On $[-\pi, \pi], \cos(x) > 1/2 \quad \iff \quad x\in[-\pi/3, \pi/3]$

Indeed, $\cos(\pi/3) = 1/2$, and you know that cosine is an even function.

Moreover, cosine is $2\pi$-periodic, and therefore have you can translate $[-\pi/3, \pi/3]$ by $2\pi n, n\in \mathbb{N}$ to get all possible answers.

UserX
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R2B2
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Some hints:

  • For what values of $x$ is the $x$ component of a point on the unit circle at least $1/2$? (Drawing this may help.)
  • You know the function repeats. How much do you add to the angle to go once around the circle?
  • How do you express this amount so that you can go around twice, three times, $n$ times? How about going around in the other direction?
John
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Counting the angles in turns ($360°$), and looking at the plot of the $\cos$ function ($\cos(\pm60°)=\frac12$), the intervals are $$[-\frac16+n,\frac16+n].$$

enter image description here