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# Pre-calc problem

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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.

Jun 30, 2022

#1
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Jun 30, 2022
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I'd like to see someone answer this too.

Jun 30, 2022
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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
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Thanks but  please show some working Vinculum.

I have no idea how you got this.

Melody  Jun 30, 2022
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+33054
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Here's one way of showing it:

Alan  Jul 1, 2022
#6
+117766
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WOW

Thanks Alan,

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

Thanks!

Melody  Jul 1, 2022