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Calculate \(\arccos \sqrt{\cfrac{1+\sqrt{\cfrac{1-\sqrt{\cfrac{1-\sqrt{\cfrac{1+\cfrac{\sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}}.\)As usual, the output of an inverse trig function should be in radians.

Thanks in advance! 

 Jun 30, 2022
 #1
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The answer is 17*pi/32.

 Jun 30, 2022
 #2
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I'd like to see someone answer this too.  frown

 Jun 30, 2022
 #3
avatar+578 
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The above expression can be simplified to:

 \(\sqrt{\cfrac{1+\sqrt{\cfrac{1-\sqrt{\cfrac{1-\sqrt{\cfrac{1+\cfrac{\sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}} = \sqrt{ \frac{1 + \cos\left(\frac{13\pi}{48} \right)}{2}} \)

 

The above value is the cosine of \(13\pi/96\)

 

So the answer to the question is: \(13\pi/96\)

 

-Vinculum

 Jun 30, 2022
 #4
avatar+117746 
+1

 

Thanks but  please show some working Vinculum. 

I have no idea how you got this.

Melody  Jun 30, 2022
 #5
avatar+33050 
+3

Here's one way of showing it:

 

Alan  Jul 1, 2022
 #6
avatar+117746 
+1

WOW 

Thanks Alan,

That was a lot to display.  But I get it.  

 

Thanks!

Melody  Jul 1, 2022

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