Gwendolyn Willett

2021-12-31

Find the value of ${\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)$
I am trying to solve:
${\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{12}}{4}\right)+{\mathrm{csc}}^{-1}\left(\sqrt{2}\right)\right)$
My solution is as follow:
$T={\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+{\mathrm{cos}}^{-1}\left(\frac{\sqrt{12}}{4}\right)+{\mathrm{csc}}^{-1}\left(\sqrt{2}\right)\right)$
Since:

Then:
$T={\mathrm{sin}}^{-1}\mathrm{cot}\left({\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)+\frac{\pi }{4}+\frac{\pi }{6}\right)$
I am not able to proceed further.

Vivian Soares

Hint:
$y=\frac{y+\sqrt{3}}{4}=\frac{{\left(\sqrt{3}+1\right)}^{2}}{8}$
$\sqrt{y}=\frac{\sqrt{3}+1}{2\sqrt{2}}=\mathrm{cos}\frac{\pi }{6}\mathrm{cos}\frac{\pi }{4}+\mathrm{sin}\frac{\pi }{6}\mathrm{cos}\frac{\pi }{4}=\mathrm{cos}\left(\frac{\pi }{4}-\frac{\pi }{6}\right)$

Ella Williams

$\theta ={\mathrm{cos}}^{-1}\sqrt{\frac{2+\sqrt{3}}{4}}$
$\mathrm{cos}\theta =\sqrt{\frac{2+\sqrt{3}}{4}}$
${\mathrm{cos}}^{2}\theta =\frac{2+\sqrt{3}}{4}$
$\frac{12}{+}\frac{12}{\mathrm{cos}2}\theta =\frac{2+\sqrt{3}}{4}$
$2+2\mathrm{cos}2\theta =2+\sqrt{3}$
$\mathrm{cos}2\theta =\frac{\sqrt{3}}{2}$
$2\theta =\frac{\pi }{6}$
$\theta =\frac{\pi }{12}$

nick1337

If $\alpha ={\mathrm{cos}}^{-1}\left(\sqrt{\frac{2+\sqrt{3}}{4}}\right)$, then $4{\mathrm{cos}}^{2}\alpha =2+\sqrt{3}$
This means $\sqrt{3}=4{\mathrm{cos}}^{2}\alpha -2=2\left(2{\mathrm{cos}}^{2}\alpha -1\right)=2\mathrm{cos}2\alpha$
$2\alpha ={\mathrm{cos}}^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$. So $\alpha =\frac{\pi }{12}$
So $T={\mathrm{sin}}^{-1}\mathrm{cot}\left(\frac{\pi }{2}\right)={\mathrm{sin}}^{-1}0=0$

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