Question

If 0 < θ < π/8
$show \: that \: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos4 \theta} }} = 2cos \theta/2$
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Answer 1

Step-by-step explanation:

Given:-

• 0 < θ < π/8

To Prove:-

• $\rm\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos4 \theta} } } = 2cos \theta/2$

Formula Used:-

• 1 + cosθ = 2 cos² θ/2

Proof:-

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

$\rm= \sqrt{2 + \sqrt{2 + \sqrt{2 (1+ cos4 \theta)} } }$

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

$\rm= \sqrt{2 + \sqrt{2 + \sqrt{4 cos^{2} 2\theta} } }$

$\rm= \sqrt{2 + \sqrt{2 + 2cos2 \theta} }$

$\rm= \sqrt{2 + \sqrt{2 (1 + cos2 \theta)} }$

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

$\rm= \sqrt{2 + \sqrt{4cos ^{2} \theta} }$

$\rm= \sqrt{2 + 2cos \theta }$

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

$\rm= \sqrt{2 (2cos^{2} ( \theta/2) }$

$\rm= \sqrt{4cos^{2} ( \theta/2) }$

$\rm= 2cos( \theta/2)$

LHS = RHS

Hence Proved