If θ<22(12)0, then 2+2+2+2cos4θ=

2+2+2+2cos4θ=2+2+2+2cos22θ=2+2+2cos2θ=2+2(1+cos2θ)=2+2.2cos2θ=2(1+cosθ)=2.2.cos2θ2=2.cosθ2

If θ<22(12)0, then 2+2+2+2cos4θ=

2+2+2+2cos4θ=2+2+2+2cos22θ=2+2+2cos2θ=2+2(1+cos2θ)=2+2.2cos2θ=2(1+cosθ)=2.2.cos2θ2=2.cosθ2

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If $θ < 22 {(\frac{1}{2})}^{0}$ , then $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 4 θ}}} =$

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$\frac{1}{4}$

$2 \cos θ$

$2 \cos \frac{θ}{2}$

$2 \cos \frac{θ}{4}$

answer is C.

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Detailed Solution

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 4 θ}}}$

$= \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos^{2} 2 θ}}}$

$= \sqrt{2 + \sqrt{2 + 2 \cos 2 θ}}$

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

$= \sqrt{2 + \sqrt{2.2 \cos^{2} θ}}$

$= \sqrt{2 (1 + \cos θ)}$

$= \sqrt{2.2. \cos^{2} \frac{θ}{2}}$

$= 2. \cos \frac{θ}{2}$

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If $θ < 22 {(\frac{1}{2})}^{0}$ , then $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 4 θ}}} =$