Question

If 0 < θ < $\frac{pi}{8}$, show that $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 cos 4\theta}}} = 2 cos(\frac{\theta}{2})$

Answer 1

LHS = $\sqrt{2+\sqrt{2+\sqrt{2+2cos4\theta}}}$

first of all, solve 2 + 2cos4θ or 2(1 + cos4θ)

we know, 1 + cos2x = 2cos²x

so, 1 + cos4θ = 2cos²2θ

now, 2(1 + cos4θ) = 4cos²2θ

taking square root both sides,

$\sqrt{2+2cos4\theta}$ = 2cos2θ

or, 2 + $\sqrt{2+2cos4\theta}$ = 2 + 2cos2θ

or, 2 + $\sqrt{2+2cos4\theta}$ = 2(1 + cos2θ) = 2(2cos²θ) = 4cos²θ

taking square root both sides,

$\sqrt{2+\sqrt{2+2cos4\theta}}$ = 2cosθ

or, 2 + $\sqrt{2+\sqrt{2+2cos4\theta}}$ = 2 + 2cosθ = 2(1 + cosθ)

or, 2 + $\sqrt{2+\sqrt{2+2cos4\theta}}$ = 2(2cos²θ/2) = 4cos²θ/2

talking square root both sides,

$\sqrt{2+\sqrt{2+\sqrt{2+2cos4\theta}}}$ = 2cosθ/2 [hence proved]

Answer 2

HELLO DEAR,



we know, 1 + cos2x = 2cos²x 

so, 1 + cos4θ = 2cos²2θ 

now,
2(1 + cos4θ) = 4cos²2θ 

taking square root both sides, 

√{2 + 2cos4θ} = 2cos2θ

=> 2 + √(2 + 2cos4θ) = 2(1 + cos 2θ)

taking square root both side

=> √{2 + √(2 + 2cos4θ)} = 2cosθ

=> 2 + √{2 + √(2 + 2cos4θ)} = 2(1 + cosθ) = 4cos²θ/2

taking square root both side

=> √[2 + √{2 + √(2 + 2cos4θ)}] = 2cosθ/2



I HOPE IT'S HELP YOU DEAR,
THANKS