Question

If $\sqrt{4+\sqrt{8-\sqrt{32+\sqrt{768}}}}=a\sqrt{2}\cos \left(\frac{11\pi }{b}\right)$, where $a$ and $b$ are natural numbers then $\left(b-a\right)$ is divisible by:

Answer 1

Step 1: Simplify the given expression to required form

Given that $\sqrt{4+\sqrt{8-\sqrt{32+\sqrt{768}}}}=a\sqrt{2}\cos \left(\frac{11\pi }{b}\right)$

Let $\sqrt{768}=32\cos \left(\theta \right)$

$$\begin{gathered} \Rightarrow \sqrt{768}=32\cos \left(\theta \right) \\ \Rightarrow 16\sqrt{3}=32\cos \left(\theta \right) \\ \Rightarrow \cos \left(\theta \right)=\frac{\sqrt{3}}{2} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{6} \end{gathered}$$

$\therefore$ $\sqrt{768}=32\cos \left(\frac{\pi }{6}\right)$ …$\left[1\right]$

Consider $\sqrt{4+\sqrt{8-\sqrt{32+\sqrt{768}}}}$ ….$\left[2\right]$

Now, substitute $\left[1\right]$ in $\left[2\right]$, we get,

$\sqrt{4+\sqrt{8-\sqrt{32+\sqrt{768}}}}$$=\sqrt{4+\sqrt{8-\sqrt{32+32\cos \left(\frac{\mathrm{\pi }}{6}\right)}}}$

$=\sqrt{4+\sqrt{8-\sqrt{32\left(1+\cos \left(\frac{\mathrm{\pi }}{6}\right)\right)}}}$

$=\sqrt{4+\sqrt{8-\sqrt{32\left(2{\cos }^2\left(\frac{\mathrm{\pi }}{12}\right)\right)}}}\left[\because 1+\cos \left(x\right)=2{\cos }^2\left(\frac{x}{2}\right)\right]$

$$\begin{gathered} =\sqrt{4+\sqrt{8-8\cos \left(\frac{\mathrm{\pi }}{12}\right)}} \\ =\sqrt{4+\sqrt{8\left(1-\cos \left(\frac{\mathrm{\pi }}{12}\right)\right)}} \\ =\sqrt{4+\sqrt{8\left(2{\sin }^2\left(\frac{\mathrm{\pi }}{24}\right)\right)}}\left[\because 1-\cos \left(x\right)=2{\sin }^2\left(\frac{x}{2}\right)\right] \\ =\sqrt{4+4\sin \left(\frac{\mathrm{\pi }}{24}\right)} \\ =\sqrt{4+4\cos \left(\frac{\pi }{2}-\frac{\pi }{24}\right)}\left[\because \sin \left(x\right)=\cos \left(\frac{\pi }{2}-x\right)\right] \\ =\sqrt{4+4\cos \left(\frac{11\mathrm{\pi }}{24}\right)} \\ =\sqrt{4\left(1+\cos \left(\frac{11\mathrm{\pi }}{24}\right)\right)} \\ =\sqrt{4\left(2{\cos }^2\left(\frac{11\mathrm{\pi }}{48}\right)\right)} \\ =2\sqrt{2}\cos \left(\frac{11\mathrm{\pi }}{48}\right) \end{gathered}$$

Step 2: Solve for the required value

$\sqrt{4+\sqrt{8-\sqrt{32+\sqrt{768}}}}=a\sqrt{2}\cos \left(\frac{11\pi }{b}\right)$

$\Rightarrow$ $2\sqrt{2}\cos \left(\frac{11\mathrm{\pi }}{48}\right)=a\sqrt{2}\cos \left(\frac{11\pi }{b}\right)$

On comparing, we get, $a=2$ and $b=48$

$\therefore b-a=48-2=46$

We know that factors of $46$ are $1,2,23,46$

Hence the value of $\left(b-a\right)$ is divisible by $1,2,23,46$