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# cos(x)=1/sqrt2 in the intervall pi/12 and 7pi

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cos(x)=1/sqrt2 in the intervall pi/12 and 7pi

Guest Sep 25, 2017
edited by Guest  Sep 25, 2017
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#1
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cos(x)=1/sqrt2 in the intervall pi/12 and 7pi

Hello Guest !

$$cos(x)=\frac{1}{\sqrt{2}}$$

In a square with sides a and the diagonal $$d=a\times \sqrt{2}$$
is the cos of the angle between a and d
$$cos(x)=\frac{a}{d}=\frac{a}{a\times \sqrt{2}}=\frac{1}{\sqrt{2}}$$
This angle is known as 45 °.
So is

$$cos(45°)=\frac{1}{\sqrt{2}}$$
x = 45 °

$$cos(x)=\frac{1}{\sqrt{2}}$$

$$x=arccos(\frac{1}{\sqrt{2}})\\ \color{blue}x\ |\ period\ 2 \pi|\in\{\frac{1}{4}\pi\ ;\ 1\frac{3}{4}\pi\}$$

$$\large\mathbb{L}^{7\pi}_{\frac{\pi}{12}}=\{\frac{1}{4}\pi\ ;\ 2\frac{1}{4}\pi\ ;\ 4\frac{1}{4}\pi\ ;\ 6\frac{1}{4}\pi\ ;\ 1\frac{3}{4}\pi\ ;\ 3\frac{3}{4}\pi\ ;\ 5\frac{3}{4}\pi\}$$

!

asinus  Sep 25, 2017
edited by asinus  Sep 25, 2017
#2
+1

Thank you so much for your help!

Guest Sep 25, 2017
#3
+7159
+1

Here are the graphs.

asinus  Sep 25, 2017

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