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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
 #1
avatar+7603 
+1

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 °

 

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

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

 

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

laugh  !

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

Thank you so much for your help!

Guest Sep 25, 2017
 #3
avatar+7603 
+1

Here are the graphs.

asinus  Sep 25, 2017

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