# 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.

• Observe that $2a^2-1=\sqrt{3}/2$ if $a=\sqrt{(2+\sqrt{3})/4}$. Commented Nov 22, 2021 at 6:50

Hint:

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

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

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...

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

\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}