O lvl A Maths: Trigonometry

From http://www.sgforums.com/forums/2297/topics/339918

If $270^{o}<x<360^{o}$ ,

find $\small \left[\sqrt{2+\left(\sqrt{2+2\cos x}\right)} \right]$

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Answer:

$\left[\sqrt{2+\left(\sqrt{2+2\cos x}\right)} \right]$
$=\left[\sqrt{2+\left(\sqrt{2+4\cos^{2} \frac{1}{2}x -2}\right)} \right]$
$=\left[\sqrt{2+\left(\sqrt{4\cos^{2} \frac{1}{2}x}\right)} \right]$

Since $270^{o}<x<360^{o}$ ,
$\sqrt{4\cos^{2} \frac{1}{2}x}=-2\cos \frac{1}{2}x$ so that it is positive.

$\left[\sqrt{2+\left(\sqrt{4\cos^{2} \frac{1}{2}x}\right)} \right]$
$=\left[\sqrt{2-2\cos \frac{1}{2}x} \right]$
$=\left[\sqrt{2-\left( 2-4\sin^{2} \frac{1}{4}x}\right) \right]$
$=\left[\sqrt{4\sin^{2} \frac{1}{4}x} \right]$

Since $67.5^{o}<x<90^{o}$ ,
$\left[\sqrt{4\sin^{2} \frac{1}{4}x} \right]$
$=2\sin \frac{1}{4}x$

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O lvl A Maths: Trigonometry