Question

if \[\sqrt{4-\sqrt{8+\sqrt{32{+\sqrt{768}}}}}=a \sqrt{2}\cos \frac{ 11 \pi }{ b }\] find a and b!

Answer 1

@abb0t ?
@preteenpony ?
@ganeshie8 ?
@hugsnotughs ?

Answer 2

@ganeshie8 ?@Abhisar ?
@bradely ?
@Compassionate ?
@D3xt3R ?

Answer 3

@Owlcoffee ?
@quickstudent ?
@radar ?
@SolomonZelman ?

Answer 4

Here's something to notice:
\[\begin{align*}\sqrt{4-\sqrt{8+\sqrt{32{+\sqrt{768}}}}}&=a \sqrt{2}\cos \frac{ 11 \pi }{ b }\\
\sqrt{\frac{1}{2}}\sqrt{4-\sqrt{8+\sqrt{32{+\sqrt{768}}}}}&=a\cos \frac{ 11 \pi }{ b }\\
\sqrt{2-\frac{1}{2}\sqrt{8+\sqrt{32{+\sqrt{768}}}}}&=a\cos \frac{ 11 \pi }{ b }\\
\sqrt{2-\sqrt{\frac{8}{4}+\frac{1}{4}\sqrt{32{+\sqrt{768}}}}}&=a\cos \frac{ 11 \pi }{ b }\\
\sqrt{2-\sqrt{2+\sqrt{\frac{32}{16}+\frac{1}{16}\sqrt{768}}}}&=a\cos \frac{ 11 \pi }{ b }\\
\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{\frac{768}{256}}}}}&=a\cos \frac{ 11 \pi }{ b }\\
\sqrt{2-\sqrt{2+\sqrt{2+\sqrt3}}}&=a\cos \frac{ 11 \pi }{ b }
\end{align*}\]
Not sure if that helps though...

Answer 5

Sorry, I don't know how to do this.

Answer 6

@mathstudent55 ?

Answer 7

@ganeshie8 ?

Answer 8

What if you use the fact that \(-1\le\cos\dfrac{11\pi}{b}\le1\)? You might be able to find some bounds for \(a\).

Answer 9

yes i guess so what should be our approach now?

Answer 10

Oh I have no idea, just throwing out some ideas ;)

Answer 11

i am trying to do the impossible task of simplifying left hand side
\(\large \sqrt{2+\sqrt{3}} = \dfrac{\sqrt{2} + \sqrt{6}}{2}\)

next level gives me 3 terms and i feel stuck

Answer 12

\[\large \sqrt{2-\frac{1}{4}\sqrt{4+\sqrt{2} + \sqrt{6} }} =a\cos \frac{ 11 \pi }{ b } \]

Answer 13

@mukushla

Answer 14

Is it even possible to find values for two unknowns with just the one equation? Or doe the periodic part have something to do with it?

Answer 15

anyone?

Answer 16

@ganeshie8 ?

Answer 17

@abb0t ?
@Compassionate ?
@dan815 ?
@esshotwired ?

Answer 18

@dan815 ?

Answer 19

can you use the same method that @ganeshie8 used repeatedly for the rest of the root
I dunno what that will give you