Find the value of ${{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right)}$

I am trying to solve:

${\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right)$

My solution is as follow:

$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right) $

Since:

$\csc{^{ - 1}}\left( {\sqrt 2 } \right) = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4};{\cos ^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$

Then:

$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + \frac{\pi }{4} + \frac{\pi }{6}} \right)$

I am not able to proceed further.

Solution 1:

Hint:

$$y=\dfrac{2+\sqrt3}4=\dfrac{(\sqrt3+1)^2}8$$

$$\sqrt y=\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4+\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi4-\dfrac\pi6\right)$$

Solution 2:

If $\alpha=\cos^{-1}\left(\sqrt{\frac{2+\sqrt3}4}\right)$, then $4\cos^2\alpha=2+\sqrt3$.

This means $\sqrt3=4\cos^2\alpha-2=2(2\cos^2\alpha-1)=2\cos2\alpha$.

$2\alpha=\cos^{-1}\frac{\sqrt3}2=\frac{\pi}6$. So $\alpha=\frac\pi{12}$.

So...

$T=\sin^{-1}\cot(\frac\pi2)=\sin^{-1}0=0$.

Solution 3:

\begin{align} \theta &= \cos^{-1}\sqrt{\frac{2+\sqrt{3}}{4}}\\ \cos \theta &= \sqrt{\frac{2+\sqrt{3}}{4}}\\ \cos^2 \theta &= \frac{2+\sqrt{3}}{4}\\ \frac 12 + \frac 12 \cos 2\theta &= \frac{2+\sqrt{3}}{4}\\[0.5em] 2 + 2 \cos 2\theta &= 2+ \sqrt 3\\ \cos 2\theta &= \frac{\sqrt 3}{2}\\ 2\theta &= \frac{\pi}{6}\\[1em] \theta &= \frac{\pi}{12} \end{align}