(√(2+(√(2+(√(2+(√(2cos8x))))))))=?


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Question Number 212231 by Davidtim last updated on 07/Oct/24

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 c o s 8 x}}}} = ?$
Commented by Davidtim last updated on 07/Oct/24

$p l e a s e h e l p m e$
Commented by mr W last updated on 08/Oct/24

$t h e n a t f i r s t y o u s h o u l d h e l p o t h e r$ $p e o p l e t o u n d e r s t a n d w h a t y o u w a n t$ $o t h e r p e o p l e d o .$
Commented by Ghisom last updated on 08/Oct/24

$I {don}^{'} t think anybody can help here$ $what are we supposed to do ?$
Commented by Davidtim last updated on 08/Oct/24

$w h y ?! w e a l l s h o u l d h e l p e a c h o t h e r$ $t h i s A P P r e l a t e s t o h e l p i n g a n d e d u c a t i n g .$
Commented by Davidtim last updated on 08/Oct/24

$?$
Commented by Davidtim last updated on 08/Oct/24

$I f I c a n t s o l v e t h e p r o b l e m t h e n n e e d$ $y o u r h e l p .$
Commented by mr W last updated on 08/Oct/24

${i t}^{'} s n o t t h a t t h e y {d o n}^{'} t h e l p y o u, b u t$ $t h e y {d o n}^{'} t k n o w w h a t y o u e x p e c t .$ $i f y o u w a n t t o s i m p l y t h e e x p r e s s i o n,$ $i t h i n k y o u c a n n o t s i m p l y i t m o r e .$
Commented by Davidtim last updated on 08/Oct/24

$O k s i r I h a v e u p l o a d e d t h e p r o b l e m$ $a n d I h a v e e x p e c t e d o f s o l o t i o n t h e p r o b l e m .$ $I m e n t i o n a g a i n h e r e p l e a s e s o l v e t h i s .$ $\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 c o s 8 x}}}} = ?$
Commented by mr W last updated on 08/Oct/24

$i h a v e t o l d y o u t h a t t h i s e x p r e s s i o n$ $c a n n o t b e s i m p l i e d f u r t h e r .$
Commented by Davidtim last updated on 08/Oct/24

$o k s i r t h a n k s l o t s o f .$
Commented by TonyCWX08 last updated on 08/Oct/24

$Y o u g u v e u s a n e x p r e s s i o n a n d e x p e c t u s t o s i m p l i f y i t ? ?$
	Answered by ajfour last updated on 08/Oct/24
	$(√(2+(√(2+(√(2+(√(2cos8x))))))))=q {(q^2 −2)^2 −2}^2 =2+(√(2cos 8x)) let cos 8x=2cos^2 8θ ⇒ (q^2 −2)^2 =2+2cos 4θ ⇒ q^2 =2+2cos 2θ q=2cos θ = 2cos [(1/8)cos^(−1) (√((cos 8x)/2))]$
	$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 c o s 8 x}}}} = q$ ${{(q^{2} - 2)}^{2} - 2}^{2} = 2 + \sqrt{2 \cos 8 x}$ $l e t \cos 8 x = 2 \cos^{2} 8 θ$ $\Rightarrow {(q^{2} - 2)}^{2} = 2 + 2 \cos 4 θ$ $\Rightarrow q^{2} = 2 + 2 \cos 2 θ$ $q = 2 \cos θ$ $= 2 \cos [\frac{1}{8} \cos^{- 1} \sqrt{\frac{\cos 8 x}{2}}]$
	Commented by TonyCWX08 last updated on 08/Oct/24

	$N a h i t s w r o n g .$ $Y o u c o m p a r e d$ $c o s (8 x) a n d 2 {c o s}^{2} (8 θ), w h i c h i s t w o d i f f e r e n t t h i n g d .$
	Commented by ajfour last updated on 08/Oct/24

	$l e t s w a i t f o r o t h e r e x p e r t c o m m e n t s$
	Commented by Ghisom last updated on 09/Oct/24

	${it}^{'} s true, congratulations!$
	Commented by Frix last updated on 09/Oct/24

	$Before claiming this is wrong or right you$ $should check it . Plot it or simply plug in$ $some values$ $\Rightarrow$ ${It}^{'} s true .$
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