Prove that? $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}}$ $sqrt(2+sqrt(2+sqrt(2+2cos8theta$ = $2 cos θ$ $2costheta$

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Prove that? $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}}$ $sqrt(2+sqrt(2+sqrt(2+2cos8theta$ = $2 cos θ$ $2costheta$

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Answer 1 · 2017-09-19T09:06:37

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}} = 2 cos θ$ $sqrt(2+sqrt(2+sqrt(2+2cos8theta)))=2costheta$

Explanation:

As $cos 2 A = 2 {cos}^{2} A - 1$ $cos2A=2cos^2A-1$ , we have

$cos 8 θ = 2 {cos}^{2} (4 θ) - 1$ $cos8theta=2cos^2(4theta)-1$ , $cos 4 θ = 2 {cos}^{2} (2 θ) - 1$ $cos4theta=2cos^2(2theta)-1$ and $cos 2 θ = 2 {cos}^{2} θ - 1$ $cos2theta=2cos^2theta-1$

i.e. $1 + cos 8 θ = 2 {cos}^{2} (4 θ)$ $1+cos8theta=2cos^2(4theta)$ , $1 + cos 4 θ = 2 {cos}^{2} (2 θ)$ $1+cos4theta=2cos^2(2theta)$ and $1 + cos 2 θ = 2 {cos}^{2} θ$ $1+cos2theta=2cos^2theta$

Hence $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}}$ $sqrt(2+sqrt(2+sqrt(2+2cos8theta)))$

= $\sqrt{2 + \sqrt{2 + \sqrt{2 (1 + cos 8 θ)}}}$ $sqrt(2+sqrt(2+sqrt(2(1+cos8theta))))$

= $\sqrt{2 + \sqrt{2 + \sqrt{2 \times 2 {cos}^{2} (4 θ)}}}$ $sqrt(2+sqrt(2+sqrt(2xx2cos^2(4theta))))$

= $\sqrt{2 + \sqrt{2 + 2 cos (4 θ)}}$ $sqrt(2+sqrt(2+2cos(4theta)))$

= $\sqrt{2 + \sqrt{2 (1 + cos (4 θ))}}$ $sqrt(2+sqrt(2(1+cos(4theta)))$

= $\sqrt{2 + \sqrt{2 \times 2 {cos}^{2} (2 θ)}}$ $sqrt(2+sqrt(2xx2cos^2(2theta)))$

= $\sqrt{2 + 2 cos 2 θ}$ $sqrt(2+2cos2theta)$

= $\sqrt{2 (1 + cos 2 θ)}$ $sqrt(2(1+cos2theta))$

= $\sqrt{2 \times 2 {cos}^{2} θ}$ $sqrt(2xx2cos^2theta)$

= $2 cos θ$ $2costheta$

Answer 2 · 2017-09-19T10:15:16

The identity is false in general, but true for $θ \in [- \frac{π}{8}, \frac{π}{8}]$ $theta in [-pi/8, pi/8]$

Explanation:

The given identity does not hold in general.

For example, with $θ = \frac{π}{6}$ $theta = pi/6$ we find:

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}}$ $sqrt(2+sqrt(2+sqrt(2+2cos8 theta)))$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos (\frac{4 π}{3})}}}$ $=sqrt(2+sqrt(2+sqrt(2+2cos((4pi)/3))))$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 (- \frac{1}{2})}}}$ $=sqrt(2+sqrt(2+sqrt(2+2(-1/2))))$

$= \sqrt{2 + \sqrt{2 + \sqrt{1}}}$ $=sqrt(2+sqrt(2+sqrt(1)))$

$= \sqrt{2 + \sqrt{3}} \approx 1.932$ $=sqrt(2+sqrt(3)) ~~ 1.932$

But:

$2 cos (θ) = 2 cos (\frac{π}{6}) = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3} \approx 1.732$ $2 cos(theta) = 2 cos(pi/6) = 2 (sqrt(3)/2) = sqrt(3) ~~ 1.732$

What is true?

A double angle formula for $cos$ $cos$ can be written:

$cos 2 θ = 2 {cos}^{2} θ - 1$ $cos 2theta = 2 cos^2 theta - 1$

So:

$cos 8 θ = 2 {cos}^{2} 4 θ - 1$ $cos 8 theta = 2 cos^2 4theta - 1$

and:

$2 + 2 cos 8 θ = 2 + 2 (2 {cos}^{2} 4 θ - 1) = 4 {cos}^{2} 4 θ$ $2+2cos 8 theta = 2+2(2 cos^2 4theta - 1) = 4 cos^2 4 theta$

So:

$\sqrt{2 + 2 cos 8 θ} = \sqrt{4 {cos}^{2} 4 θ} = | 2 cos 4 θ |$ $sqrt(2+2cos 8 theta) = sqrt(4 cos^2 4 theta) = abs(2 cos 4theta)$

So if $cos 4 θ \geq 0$ $cos 4 theta >= 0$ then:

$\sqrt{2 + 2 cos 8 θ} = 2 cos 4 θ$ $sqrt(2+2cos 8 theta) = 2 cos 4 theta$

Then:

$2 + 2 cos 4 θ = 2 + 2 (2 {cos}^{2} 2 θ - 1) = 4 {cos}^{2} 2 θ$ $2+2cos 4 theta = 2+2(2cos^2 2 theta - 1) = 4 cos^2 2 theta$

So:

$\sqrt{2 + 2 cos 4 θ} = \sqrt{4 {cos}^{2} 2 θ} = | 2 cos 2 θ |$ $sqrt(2+2cos 4 theta) = sqrt(4 cos^2 2 theta) = abs(2cos 2 theta)$

So if $cos 2 θ \geq 0$ $cos 2 theta >= 0$ then:

$\sqrt{2 + 2 cos 4 θ} = 2 cos 2 θ$ $sqrt(2+2cos 4 theta) = 2cos 2 theta$

Then:

$2 + 2 cos 2 θ = 2 + 2 (2 {cos}^{2} θ - 1) = 4 {cos}^{2} θ$ $2+2cos 2 theta = 2+2(2cos^2 theta - 1) = 4cos^2 theta$

So:

$\sqrt{2 + 2 cos 2 θ} = \sqrt{4 {cos}^{2} θ} = | 2 cos θ |$ $sqrt(2+2cos 2 theta) = sqrt(4 cos^2 theta) = abs(2 cos theta)$

So if $cos θ \geq 0$ $cos theta >= 0$ then:

$\sqrt{2 + 2 cos 2 θ} = 2 cos θ$ $sqrt(2+2cos 2 theta) = 2 cos theta$

So we have shown that provided all of:

$cos 4 θ \geq 0$ $cos 4 theta >= 0$
$cos 2 θ \geq 0$ $cos 2 theta >= 0$
$cos θ \geq 0$ $cos theta >= 0$

then:

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}} = 2 cos θ$ $sqrt(2+sqrt(2+sqrt(2+2cos 8 theta))) = 2 cos theta$

These conditions will be satisfied for $θ \in [- \frac{π}{8}, \frac{π}{8}]$ $theta in [-pi/8, pi/8]$

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Prove that? $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 8 θ}}}$ $sqrt(2+sqrt(2+sqrt(2+2cos8theta$ = $2 cos θ$ $2costheta$

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